This deck covers linear combinations and the span of a set as the smallest subspace containing it, then draws the sharp line between spanning and independence. It builds the Steinitz exchange lemma as the engine behind a well-defined notion of dimension, ties independence to unique representation and to row reduction, and dismantles the classic traps: confusing spanning with independence, testing three or more vectors a pair at a time, and slipping the zero vector into an independent set.
Subject: Foundations of Higher Mathematics · 116 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
By the end of this deck you will be able to:
1. Write vectors as linear combinations and describe the span of a set.
2. Recognize span as the smallest subspace containing a set, and prove it.
3. Decide whether a set is linearly independent, three different ways.
4. Use the Steinitz exchange lemma to bound independent sets by spanning sets.
5. Avoid the standard traps: confusing spanning with independence, and testing triples by pairs.
Warm-up
Discussion prompt
Before we open Span & Linear Independence: without looking back, what was the main idea of Subspaces, Sums & Direct Sums, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck gives the subspace test, explains why an intersection of subspaces is a subspace but a union usually is not, and presents the sum as the smallest subspace containing both. It then covers the three equivalent faces of a direct sum and the Grassmann dimension formula. It targets the classic traps: taking unions to be subspaces, mistaking spanning for directness, and forgetting the intersection correction term.
Concept
Fix a vector space over a field. Given finitely many vectors, a linear combination is what you get by scaling each one and adding the results.
\[ c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \cdots + c_n \mathbf{v}_n \]
The scalars are the weights or coefficients. Only finitely many vectors and finitely many additions are ever involved.
linear combination — A vector of the form c1 v1 + ... + cn vn, where the vi are vectors and the ci are scalars from the field. The weights ci may be any scalars, including zero.
Counterexample
Discussion prompt
Fix a vector space over a field. Given finitely many vectors, a linear combination is what you get by scaling each one and adding the results.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
The scalars are the weights or coefficients. Only finitely many vectors and finitely many additions are ever involved.
Concept
One combination is always available: set every weight to zero. It produces the zero vector no matter what the vectors are.
\[ 0\,\mathbf{v}_1 + 0\,\mathbf{v}_2 + \cdots + 0\,\mathbf{v}_n = \mathbf{0} \]
This is the trivial combination. A nontrivial combination is one where at least one weight is not zero. The whole theory of independence turns on which combinations can equal the zero vector.
trivial combination — The linear combination in which every coefficient is zero; it always yields the zero vector. Any combination with at least one nonzero coefficient is called nontrivial.
Definition probe
Sort into buckets
Every line below is part of the definition of linear combination or of trivial combination — one or the other, never both. Put each where it belongs.
Intuition
Think of each vector as an ingredient and each weight as how much of it you pour in. A linear combination is the finished mixture.
Negative weights are allowed, so you can also pour ingredients back out. Zero weight means you skip that ingredient entirely.
The question that drives this whole topic is: which target vectors can you reach by some recipe, and can a target be reached in more than one way?
Analogy
Discussion prompt
Explain A combination is a recipe by analogy to something with no Foundations of Higher Mathematics in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Think of each vector as an ingredient and each weight as how much of it you pour in. A linear combination is the finished mixture.
Ranking
Put in order
Put the moves of Worked example: writing a vector as a combination into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. We seek scalars x and y with x a + y b equal to the target.
Worked example
In the plane, express the target vector as a linear combination of two given vectors.
\[ \text{target } (5,4), \quad \mathbf{a}=(1,0), \quad \mathbf{b}=(1,2) \]
Set up the unknown weights
Why: We seek scalars x and y with x a + y b equal to the target. Write the two coordinate equations.
\[ x(1,0) + y(1,2) = (5,4) \]
Read off the coordinate equations
Why: Matching first and second coordinates turns the vector equation into two scalar equations.
\[ x + y = 5, \qquad 2y = 4 \]
Solve the system
Why: The second equation gives y directly; substitute into the first to get x.
\[ y = 2, \qquad x = 5 - 2 = 3 \]
Verify the combination reproduces the target
Why: Substitute the weights back and add. It must equal (5,4) exactly.
\[ 3(1,0) + 2(1,2) = (3,0) + (2,4) = (5,4)\ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "writing a vector as a combination", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Substitute the weights back and add. It must equal (5,4) exactly.
Concept
Collect together every vector you can build from a set as a linear combination. That collection is the span of the set.
\[ \operatorname{span}(S) = \{\, c_1 \mathbf{v}_1 + \cdots + c_n \mathbf{v}_n : n \ge 0,\ \mathbf{v}_i \in S,\ c_i \in F \,\} \]
span — The span of a set S is the set of all finite linear combinations of vectors drawn from S. By convention the span of the empty set is the zero subspace.
Explain it
Discussion prompt
Explain The span of a set is all its linear combinations to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Collect together every vector you can build from a set as a linear combination. That collection is the span of the set.
Concept
The set being spanned may be infinite, but each single element of the span uses only finitely many of its vectors.
There is no notion of an infinite sum here: we have not defined limits or convergence, only the algebraic operations of the vector space.
So the span of an infinite set is still built one finite recipe at a time. This is the algebraic span, distinct from any topological closure you might meet later.
Intuition
Start standing at the origin with your given vectors as allowed moves, each usable in any scaled amount, forward or backward, any finite number of times.
The span is the complete set of destinations you can land on. Nothing outside it is reachable; everything inside it is.
This is why span always contains the origin: the empty recipe, using no moves, leaves you at the start.
Picture it
Figure (svg): A single arrow from the origin and the full line through the origin that its scalar multiples trace out.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
With a single nonzero vector, the only recipes are scalar multiples of it. Sweeping the scalar over all values traces a straight line through the origin.
Intuition
With a single nonzero vector, the only recipes are scalar multiples of it. Sweeping the scalar over all values traces a straight line through the origin.
Figure (svg): A single arrow from the origin and the full line through the origin that its scalar multiples trace out.
With two vectors that do not lie on one line, the recipes fill an entire plane through the origin.
Step zero
Discussion prompt
Worked example: describing spans in three-space — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Span of u alone
Answer:
Worked example
Describe the span of one vector, then of two vectors, in real three-space.
\[ \mathbf{u}=(1,0,0), \qquad \mathbf{w}=(0,1,0) \]
Span of u alone
Why: Every combination is a single scalar times u, so we get the set of all scalar multiples of u.
\[ \operatorname{span}\{\mathbf{u}\} = \{(t,0,0): t \in \mathbb{R}\} \]
Recognize it as a line
Why: This is exactly the first coordinate axis: a line through the origin.
Span of u and w together
Why: Combinations s u + t w sweep the first two coordinates freely while the third stays zero.
\[ \operatorname{span}\{\mathbf{u},\mathbf{w}\} = \{(s,t,0): s,t \in \mathbb{R}\} \]
Verify a sample point lies in the plane and off it
Why: Check that (2,3,0) is a valid combination and (0,0,1) is not, confirming the description.
\[ (2,3,0)=2\mathbf{u}+3\mathbf{w}\ \checkmark, \qquad (0,0,1)\notin\operatorname{span}\{\mathbf{u},\mathbf{w}\}\ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "describing spans in three-space", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Check that (2,3,0) is a valid combination and (0,0,1) is not, confirming the description.
Concept
When the span of a set equals the entire space, we say the set spans the space, or is a spanning set.
\[ \operatorname{span}(S) = V \]
spanning set — A set S with span equal to the whole space V. Equivalently, every vector of V can be written as some finite linear combination of vectors in S.
Concept
For any set, its span passes the subspace test automatically.
It contains the zero vector, and the sum of two combinations is again a combination, and a scalar times a combination is again a combination.
\[ \mathbf{0}\in\operatorname{span}(S), \qquad \operatorname{span}(S)+\operatorname{span}(S)\subseteq\operatorname{span}(S) \]
Concept
Span is not just any subspace containing the set. It is the smallest one: it sits inside every subspace that contains the set.
\[ S \subseteq W \text{ and } W \text{ a subspace} \ \Longrightarrow\ \operatorname{span}(S) \subseteq W \]
This is the clean, basis-free way to think about span: it is the closure of the set under the vector-space operations.
Estimation
Predict first
Prove that the span of a set is the smallest subspace that contains it. This has two parts.
Commit before you compute: what does Worked example: span is the smallest containing subspace come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify against a concrete instance
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Take S the single vector (1,0) and W any subspace containing it.
Worked example
Prove that the span of a set is the smallest subspace that contains it. This has two parts.
Part one: the span contains the set and is a subspace
Why: Each vector v in S equals the one-term combination 1 times v, so S sits inside its span, and we already know the span is a subspace.
\[ \mathbf{v} = 1\cdot\mathbf{v} \in \operatorname{span}(S) \]
Part two: any subspace containing the set contains every combination
Why: Let W be a subspace with S inside it. W is closed under scaling and addition, so it must contain every finite linear combination of vectors from S.
\[ \mathbf{v}_i \in W \ \Rightarrow\ c_1\mathbf{v}_1 + \cdots + c_n\mathbf{v}_n \in W \]
Combine the two inclusions
Why: Every element of the span lies in W, so the span is contained in W. Since W was an arbitrary containing subspace, span is the smallest.
\[ \operatorname{span}(S) \subseteq W \]
Verify against a concrete instance
Why: Take S the single vector (1,0) and W any subspace containing it. W must contain every (t,0), which is exactly the span, confirming minimality.
\[ S=\{(1,0)\},\quad \operatorname{span}(S)=\{(t,0)\}\subseteq W\ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "span is the smallest containing subspace", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Take S the single vector (1,0) and W any subspace containing it. W must contain every (t,0), which is exactly the span, confirming minimality.
Pattern
1. Write a general linear combination
Why: Introduce a scalar weight for each vector and form the weighted sum symbolically.
2. Let the weights range over all scalars
Why: The span is the set of outputs as the weights vary freely over the field.
3. Simplify to a clean description
Why: Reduce the parametric family to a recognizable object: a line, a plane, a coordinate condition, or the whole space.
4. To test membership, solve for the weights
Why: A target lies in the span exactly when the resulting linear system for the weights has at least one solution.
Elimination
Eliminate the wrong options
Which vector lies in the span of the single vector (1,2)?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: A vector is in the span exactly when its second coordinate is twice its first, matching the pattern (t, 2t). For (2,4) we have 4 equal to 2 times 2, so it equals 2 times (1,2). The others fail this test.
Check
The span of a single vector in the plane is the line of all its scalar multiples. Use that to decide membership.
\[ \operatorname{span}\{(1,2)\} = \{(t,2t): t \in \mathbb{R}\} \]
Check your understanding
Which vector lies in the span of the single vector (1,2)?
Answer: A
Why: A vector is in the span exactly when its second coordinate is twice its first, matching the pattern (t, 2t). For (2,4) we have 4 equal to 2 times 2, so it equals 2 times (1,2). The others fail this test.
Concept
A set is linearly dependent when some nontrivial combination of its vectors equals the zero vector: at least one weight is not zero, yet the sum collapses to the origin.
\[ c_1\mathbf{v}_1 + \cdots + c_n\mathbf{v}_n = \mathbf{0} \quad\text{with some } c_i \ne 0 \]
linearly dependent — A set of vectors is linearly dependent if the zero vector can be written as a nontrivial linear combination of them, that is, with at least one nonzero coefficient.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of linear combination, trivial combination, span, spanning set, linearly dependent as Span & Linear Independence uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Concept
A set is linearly independent when the only way a combination of its vectors can equal the zero vector is the trivial way, with every weight zero.
\[ c_1\mathbf{v}_1 + \cdots + c_n\mathbf{v}_n = \mathbf{0} \ \Longrightarrow\ c_1 = \cdots = c_n = 0 \]
linearly independent — A set is linearly independent if the zero vector has only the trivial representation: every linear combination equal to zero forces all coefficients to be zero.
Intuition
Dependence says one of your ingredients is redundant: some combination of the others already reproduces it, so you could throw it away without shrinking the span.
Independence says every ingredient pulls in a genuinely new direction. Remove any one and the span strictly shrinks.
So independence is a statement about efficiency: no fat, no redundancy, nothing you could have derived from the rest.
Concept
If the zero vector is one of your vectors, the set is automatically dependent. Put weight one on the zero vector and weight zero on everything else.
\[ 1\cdot\mathbf{0} + 0\cdot\mathbf{v}_2 + \cdots + 0\cdot\mathbf{v}_n = \mathbf{0} \]
This is a nontrivial combination, because the weight on the zero vector is not zero, yet it sums to the origin. So the zero vector can never belong to an independent set.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Claiming this set is independent because the two nonzero vectors point in different directions.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The other two are independent, so the eye jumps to independent.
A nontrivial combination reaching zero exists, so the set is dependent.
Why: The other two are independent, so the eye jumps to independent. But independence is about the whole set, and the zero vector is a member.
Trap
Claiming this set is independent because the two nonzero vectors point in different directions.
\[ \{\mathbf{0},\ (1,0),\ (0,1)\} \]
The mistake: ignoring the zero vector
Why: The other two are independent, so the eye jumps to independent. But independence is about the whole set, and the zero vector is a member.
\[ 1\cdot\mathbf{0} + 0\cdot(1,0) + 0\cdot(0,1) = \mathbf{0} \]
A nontrivial combination reaching zero exists, so the set is dependent.
\[ \text{weight } 1 \text{ on } \mathbf{0} \text{ is nonzero} \Rightarrow \text{dependent} \]
The fix: check every vector, including the zero vector
Why: Any set that contains the zero vector is dependent, full stop. Drop the zero vector and the remaining two are independent.
\[ \{(1,0),\ (0,1)\} \text{ is independent} \]
Notation
Annotate
From Trap: sneaking the zero vector into an independent set — read this one piece at a time. What is each part doing?
On: \( \{\mathbf{0},\ (1,0),\ (0,1)\} \)
Missing information
Discussion prompt
Decide whether this set of three vectors in three-space is independent.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
Compare the vectors coordinatewise. The second is exactly double the first in every coordinate.
Worked example
Decide whether this set of three vectors in three-space is independent.
\[ \mathbf{v}_1=(1,2,1),\ \mathbf{v}_2=(2,4,2),\ \mathbf{v}_3=(0,1,3) \]
Scan for an obvious relation
Why: Compare the vectors coordinatewise. The second is exactly double the first in every coordinate.
\[ \mathbf{v}_2 = 2\,\mathbf{v}_1 \]
Rewrite as a combination equal to zero
Why: Move both to one side to expose a nontrivial recipe for the zero vector; the third vector gets weight zero.
\[ 2\,\mathbf{v}_1 - 1\,\mathbf{v}_2 + 0\,\mathbf{v}_3 = \mathbf{0} \]
Conclude dependence
Why: The weights 2, minus 1, 0 are not all zero, so a nontrivial combination reaches the origin. The set is dependent.
Verify the relation numerically
Why: Compute the combination coordinate by coordinate; it must be the zero vector exactly.
\[ 2(1,2,1) - (2,4,2) = (2,4,2)-(2,4,2) = (0,0,0)\ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "spotting a hidden multiple", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Compute the combination coordinate by coordinate; it must be the zero vector exactly.
Concept
There is an equivalent, often more useful, description of dependence: at least one vector in the set can be built from the remaining ones.
Start from a nontrivial relation and isolate a vector whose weight is nonzero; divide by that weight.
\[ c_k \ne 0 \ \Rightarrow\ \mathbf{v}_k = -\tfrac{1}{c_k}\!\sum_{i \ne k} c_i \mathbf{v}_i \]
This is exactly the redundancy picture: that vector adds nothing new, since it already lives in the span of the others.
Fill the middle
Fill in the blanks
From Worked example: solving for the dependence relation — finish the line. Write what belongs on the right of the equals sign before you look.
\mathbf2\,\mathbf{v}_1 + \mathbf{v}_2_3 = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Look for scalars a, b, c, not all zero, with a v1 + b v2 + c v3 equal to the zero vector; write the coordinate equations.
Worked example
Three vectors in the plane must be dependent. Find an explicit relation, then solve for one vector.
\[ \mathbf{v}_1=(1,1),\ \mathbf{v}_2=(2,3),\ \mathbf{v}_3=(4,5) \]
Set up the zero combination
Why: Look for scalars a, b, c, not all zero, with a v1 + b v2 + c v3 equal to the zero vector; write the coordinate equations.
\[ a + 2b + 4c = 0, \qquad a + 3b + 5c = 0 \]
Eliminate a
Why: Subtract the first equation from the second to remove a and get a relation between b and c.
\[ b + c = 0 \ \Rightarrow\ b = -c \]
Back-substitute and pick a parameter
Why: Insert b equals minus c into the first equation to solve for a in terms of c, then choose c equal to one.
\[ a = -2c,\quad c=1 \ \Rightarrow\ (a,b,c)=(-2,-1,1) \]
Rearrange into the third vector
Why: The relation minus 2 v1 minus v2 plus v3 equals zero solves for v3 as a combination of the first two.
\[ \mathbf{v}_3 = 2\,\mathbf{v}_1 + \mathbf{v}_2 \]
Verify the solved relation
Why: Compute 2 v1 plus v2 and confirm it equals v3 exactly.
\[ 2(1,1)+(2,3) = (2,2)+(2,3) = (4,5) = \mathbf{v}_3\ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "solving for the dependence relation", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Compute 2 v1 plus v2 and confirm it equals v3 exactly.
Concept
The smallest case is one vector. A one-element set is independent precisely when that vector is not the zero vector.
\[ c\,\mathbf{v} = \mathbf{0} \text{ with } \mathbf{v} \ne \mathbf{0} \ \Longrightarrow\ c = 0 \]
If the vector is nonzero, the only scalar that kills it is zero. If the vector is the zero vector, any scalar works, so the set is dependent.
Concept
For a pair, dependence has a simple geometric meaning: the two vectors lie on a single line through the origin.
\[ \{\mathbf{u},\mathbf{w}\} \text{ dependent} \iff \mathbf{u} = \lambda \mathbf{w} \text{ or } \mathbf{w} = \lambda \mathbf{u} \]
This pairwise multiple test is correct for two vectors. The crucial warning, coming up, is that it does not extend to three or more.
Step zero
Discussion prompt
Worked example: independence of a pair — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Assume a combination equals zero
Answer:
Worked example
Decide whether this pair in three-space is independent.
\[ \mathbf{u}=(1,2,3), \qquad \mathbf{w}=(2,-1,0) \]
Assume a combination equals zero
Why: Suppose a u plus b w equals the zero vector and read off the three coordinate equations.
\[ a+2b=0,\quad 2a-b=0,\quad 3a=0 \]
Solve the forced system
Why: The third equation gives a equals zero; substituting into the first gives b equals zero.
\[ 3a=0 \Rightarrow a=0 \Rightarrow b=0 \]
Conclude independence
Why: Only the trivial weights reach zero, so the pair is independent. Equivalently, neither is a scalar multiple of the other.
Verify by the multiple test
Why: Check no single scalar turns u into w; the first coordinate would force the scalar to be 2, but then the second coordinate fails.
\[ 2\cdot(1,2,3)=(2,4,6)\ne(2,-1,0)\ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "independence of a pair", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Check no single scalar turns u into w; the first coordinate would force the scalar to be 2, but then the second coordinate fails.
Prediction
Predict first
Which of these sets of vectors in the plane is linearly independent?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: The set with (1,0) and (0,1)
Why: The vectors (1,0) and (0,1) are not scalar multiples of each other, so the only combination equal to zero is the trivial one. Each of the other sets has a built-in nontrivial relation, making it dependent.
Check
Test each set against the definition: does only the trivial combination reach the zero vector? Watch for multiples and for the zero vector.
Check your understanding
Which of these sets of vectors in the plane is linearly independent?
Answer: B
Why: The vectors (1,0) and (0,1) are not scalar multiples of each other, so the only combination equal to zero is the trivial one. Each of the other sets has a built-in nontrivial relation, making it dependent.
Concept
Here is the payoff of independence. If a set is independent, then every vector in its span can be written as a combination in exactly one way.
Suppose a vector had two representations. Subtract them: the difference is a combination of the set equal to the zero vector.
\[ \sum c_i \mathbf{v}_i = \sum d_i \mathbf{v}_i \ \Rightarrow\ \sum (c_i - d_i)\mathbf{v}_i = \mathbf{0} \]
By independence every coefficient of that difference is zero, so the two representations had identical weights all along. Uniqueness follows.
\[ c_i - d_i = 0 \ \Rightarrow\ c_i = d_i \]
Intuition
Uniqueness is what makes coordinates meaningful. If the weights were not unique, a vector could have several different address labels and no single well-defined coordinate tuple.
So an independent spanning set turns every vector into one and only one list of numbers. That is exactly the coordinate encoding a computer stores.
Dependence breaks this: with a redundant vector, the same point has infinitely many recipes, and the coordinate system becomes ambiguous.
Hypothesis
Predict first
Worked example: the monomials are independent is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Assume a combination is the zero polynomial
Why: Suppose the weighted sum equals the zero function, meaning it evaluates to zero for every input x.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Show that the three monomials are linearly independent in the space of polynomials of degree at most two.
\[ \{1,\ x,\ x^2\} \]
Assume a combination is the zero polynomial
Why: Suppose the weighted sum equals the zero function, meaning it evaluates to zero for every input x.
\[ a\cdot 1 + b\,x + c\,x^2 = 0 \ \text{ for all } x \]
Evaluate at a first point
Why: Plug in x equal to zero. Every term with x vanishes, isolating the constant weight.
\[ x=0:\quad a = 0 \]
Use two more points
Why: With a known to be zero, evaluate at x equal to one and x equal to minus one to get two equations in b and c.
\[ b+c=0,\qquad -b+c=0 \ \Rightarrow\ b=0,\ c=0 \]
Conclude independence
Why: All three weights are forced to zero, so the only representation of the zero polynomial is trivial: the monomials are independent.
Verify via the coefficient viewpoint
Why: A polynomial is the zero polynomial exactly when every coefficient is zero, which directly gives a, b, c all zero and confirms the result.
\[ a\cdot 1 + b\,x + c\,x^2 \equiv 0 \iff a=b=c=0\ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "the monomials are independent", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A polynomial is the zero polynomial exactly when every coefficient is zero, which directly gives a, b, c all zero and confirms the result.
Ranking
Put in order
Put the moves of Worked example: the standard unit vectors are independent into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Weight each unit vector and add; the result reads off directly as a coordinate tuple.
Worked example
Show the three standard unit vectors of three-space are independent.
\[ \mathbf{e}_1=(1,0,0),\ \mathbf{e}_2=(0,1,0),\ \mathbf{e}_3=(0,0,1) \]
Form a general combination
Why: Weight each unit vector and add; the result reads off directly as a coordinate tuple.
\[ a\,\mathbf{e}_1 + b\,\mathbf{e}_2 + c\,\mathbf{e}_3 = (a,b,c) \]
Set the combination to zero
Why: Requiring the tuple to be the zero vector forces each coordinate to vanish.
\[ (a,b,c) = (0,0,0) \]
Read off the weights
Why: Each coordinate equation is simply one weight equals zero, so all weights are zero.
Verify uniqueness of coordinates
Why: Since the combination equals its own weight tuple, every vector has exactly one representation, confirming both independence and that these vectors give standard coordinates.
\[ a\,\mathbf{e}_1+b\,\mathbf{e}_2+c\,\mathbf{e}_3=\mathbf{0}\iff a=b=c=0\ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "the standard unit vectors are independent", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Since the combination equals its own weight tuple, every vector has exactly one representation, confirming both independence and that these vectors give standard coordinates.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Concluding these three are independent because no two of them are scalar multiples of each other.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Every pair here is genuinely non-parallel, so the pairwise multiple test passes.
A three-term relation exists even though no two vectors are multiples.
Why: Every pair here is genuinely non-parallel, so the pairwise multiple test passes. But independence of a set is not the same as pairwise independence.
Trap
Concluding these three are independent because no two of them are scalar multiples of each other.
\[ \mathbf{v}_1=(1,0),\ \mathbf{v}_2=(0,1),\ \mathbf{v}_3=(1,1) \]
The mistake: the pairwise test only works for two vectors
Why: Every pair here is genuinely non-parallel, so the pairwise multiple test passes. But independence of a set is not the same as pairwise independence.
\[ \text{no } \mathbf{v}_i \text{ is a multiple of a single other} \]
A three-term relation exists even though no two vectors are multiples.
\[ \mathbf{v}_1 + \mathbf{v}_2 - \mathbf{v}_3 = (1,0)+(0,1)-(1,1) = (0,0) \]
The fix: test the whole set at once
Why: Independence requires no nontrivial combination of all the vectors reaching zero. Here weights 1, 1, minus 1 do it, so the triple is dependent despite passing every pairwise check.
\[ 1\,\mathbf{v}_1 + 1\,\mathbf{v}_2 + (-1)\,\mathbf{v}_3 = \mathbf{0} \]
Translation
\( \mathbf{v}_1 + \mathbf{v}_2 - \mathbf{v}_3 = (1,0)+(0,1)-(1,1) = (0,0) \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Concept
If one vector in a set is a combination of the others, deleting it leaves the span untouched.
\[ \mathbf{v}_k \in \operatorname{span}(S \setminus \{\mathbf{v}_k\}) \ \Longrightarrow\ \operatorname{span}(S) = \operatorname{span}(S \setminus \{\mathbf{v}_k\}) \]
This is the engine of trimming: you can keep discarding redundant vectors, never shrinking the span, until nothing redundant remains, which is exactly an independent spanning set.
Estimation
Predict first
Show that removing the redundant vector keeps the span equal to the whole plane.
Commit before you compute: what does Worked example: deleting a dependent vector come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify with the redundant vector as a target
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Confirm the deleted vector is still reachable from the trimmed set, so nothing was lost.
Worked example
Show that removing the redundant vector keeps the span equal to the whole plane.
\[ S=\{(1,0),\ (0,1),\ (1,1)\} \]
Identify the redundancy
Why: The third vector is the sum of the first two, so it already lies in the span of the other two.
\[ (1,1) = (1,0)+(0,1) \]
Show the smaller set still spans the plane
Why: The first two vectors already reach every point of the plane, since any target is a combination of them.
\[ (x,y) = x(1,0)+y(0,1) \]
Argue the two spans are equal
Why: Removing a vector can only shrink or preserve a span; since the smaller set already spans everything, the two spans coincide.
\[ \operatorname{span}(S)=\operatorname{span}\{(1,0),(0,1)\}=\mathbb{R}^2 \]
Verify with the redundant vector as a target
Why: Confirm the deleted vector is still reachable from the trimmed set, so nothing was lost.
\[ (1,1)=1(1,0)+1(0,1)\in\operatorname{span}\{(1,0),(0,1)\}\ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "deleting a dependent vector", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Confirm the deleted vector is still reachable from the trimmed set, so nothing was lost.
Intuition
Think of a spanning set as a description of the space. Dependence means the description is compressible: you can shorten it without losing any reach.
Independence means the description is already fully compressed: every vector carries information the others cannot supply.
This is the exact tension a basis resolves, and it is why independence plus spanning is the sweet spot you are heading toward next deck.
Concept
Stack your vectors as the rows of a matrix. Row reduce it. The number of nonzero rows that remain, the pivots, is the number of independent directions.
If any row reduces to all zeros, the original vectors were dependent: that zero row is a nontrivial relation among them.
So a purely mechanical, always-terminating procedure decides independence. This is the computational face of the whole topic.
Intuition
Each elimination step subtracts multiples of earlier rows from later ones, which is exactly building linear combinations.
A row that collapses to zero was therefore expressible from the rows above it: a caught redundancy. A row that survives with a fresh leading entry is genuinely new.
The count of surviving pivots is the size of the largest independent subset, a number you will soon name the rank.
Step zero
Discussion prompt
Worked example: independence by row reduction — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Eliminate below the first pivot
Answer:
Worked example
Use row reduction to test the earlier triple, now as rows of a matrix.
\[ \begin{bmatrix} 1 & 2 & 1 \\ 2 & 4 & 2 \\ 0 & 1 & 3 \end{bmatrix} \]
Eliminate below the first pivot
Why: Subtract two times row one from row two to clear the first column below the pivot.
\[ R_2 \to R_2 - 2R_1 = (0,0,0) \]
Spot the zero row
Why: Row two has collapsed entirely to zero, which signals a dependency captured during elimination.
\[ \begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 1 & 3 \end{bmatrix} \]
Count the pivots
Why: Only two nonzero rows survive, giving two pivots for three vectors, so the set is dependent with rank two.
Verify the zero row is the relation
Why: The step that zeroed row two was row two minus twice row one, which rearranges to the dependence relation among the original vectors.
\[ R_2 - 2R_1 = \mathbf{0} \iff \mathbf{v}_2 = 2\,\mathbf{v}_1\ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "independence by row reduction", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The step that zeroed row two was row two minus twice row one, which rearranges to the dependence relation among the original vectors.
Pattern
1. Write the zero combination
Why: Set a general weighted sum of the vectors equal to the zero vector; the weights are the unknowns.
2. Turn it into a homogeneous linear system
Why: Match coordinates, or stack the vectors as rows or columns of a matrix; the relation becomes a system in the weights.
3. Row reduce and count pivots
Why: Independence holds exactly when the only solution is all weights zero, that is, when there is a pivot for every vector.
4. Read the verdict
Why: A free variable, or equivalently a zero row, exhibits a nontrivial relation and proves dependence; otherwise the set is independent.
Check
You reduce a matrix whose three rows are three vectors in three-space and obtain a form with two pivots and one full row of zeros.
Check your understanding
What does two pivots and one zero row tell you about the three original vectors?
Answer: B
Why: A zero row means one row was a combination of the others, so the elimination caught a nontrivial relation. With only two pivots for three vectors, the set is dependent and spans at most a two-dimensional subspace.
Concept
Now the structural heart of the topic. The exchange lemma compares any independent set against any spanning set.
Suppose a set of vectors spans the space, and a second set is independent. Then the independent set is no larger than the spanning set, and its vectors can be swapped in one at a time for vectors of the spanning set, keeping a spanning set throughout.
\[ \{\mathbf{u}_1,\dots,\mathbf{u}_k\}\text{ indep.}, \ \{\mathbf{w}_1,\dots,\mathbf{w}_m\}\text{ spans} \ \Longrightarrow\ k \le m \]
exchange lemma — If a space is spanned by m vectors, then any linearly independent set has at most m vectors, and each independent vector can replace a spanning vector while preserving the spanning property.
Intuition
Take an independent vector. Because the old set spans, that vector is some combination of the old set, with at least one nonzero weight.
Pick a spanning vector that appears with a nonzero weight and swap it out for your independent vector. Solving the relation the other way shows the discarded vector is recovered, so nothing is lost: the new set still spans.
Repeat. Each independent vector consumes one slot of the spanning set, which is exactly why you can never have more independent vectors than spanning vectors.
Fill the middle
Fill in the blanks
From Worked example: one exchange step — finish the line. Write what belongs on the right of the equals sign before you look.
\mathbf1\,\mathbf{e}_1 + 1\,\mathbf{e}_2 = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Because the old set spans, u is a combination of the standard vectors; find the weights.
Worked example
Start from a spanning set of the plane and exchange in an independent vector.
\[ \text{spanning } \{\mathbf{e}_1,\mathbf{e}_2\}, \qquad \text{independent } \mathbf{u}=(1,1) \]
Express the new vector in the old set
Why: Because the old set spans, u is a combination of the standard vectors; find the weights.
\[ \mathbf{u} = 1\,\mathbf{e}_1 + 1\,\mathbf{e}_2 \]
Choose a vector with a nonzero weight to remove
Why: The weight on the first standard vector is nonzero, so we may solve the relation for that vector and discard it.
\[ \mathbf{e}_1 = \mathbf{u} - \mathbf{e}_2 \]
Form the swapped set
Why: Replace the removed vector by u. The new set still reaches everything the old one did, since the removed vector is recovered from the new set.
\[ \{\mathbf{u},\ \mathbf{e}_2\} = \{(1,1),(0,1)\} \]
Verify the swapped set still spans the plane
Why: Recover both standard vectors from the new set; if both are reachable, the new set spans everything the old set did.
\[ \mathbf{e}_2=(0,1),\quad \mathbf{e}_1=\mathbf{u}-\mathbf{e}_2=(1,0)\ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "one exchange step", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Recover both standard vectors from the new set; if both are reachable, the new set spans everything the old set did.
Concept
The exchange lemma delivers one clean inequality that will justify dimension: no independent set can outnumber a spanning set.
\[ |\text{independent set}| \ \le\ |\text{spanning set}| \]
Turn it around and it says more: in a space spanned by a fixed number of vectors, any collection with strictly more vectors than that number is forced to be dependent.
Missing information
Discussion prompt
Use the counting corollary to prove that any three vectors in the plane are dependent.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
The two standard vectors span the whole plane, so any spanning count is at most two.
Worked example
Use the counting corollary to prove that any three vectors in the plane are dependent.
\[ \mathbf{a},\mathbf{b},\mathbf{c} \in \mathbb{R}^2 \]
Name a spanning set of size two
Why: The two standard vectors span the whole plane, so any spanning count is at most two.
\[ \operatorname{span}\{\mathbf{e}_1,\mathbf{e}_2\} = \mathbb{R}^2 \]
Apply the bound
Why: An independent set can have at most as many vectors as a spanning set, so at most two vectors can be independent here.
\[ |\text{independent}| \le 2 \]
Conclude for three vectors
Why: Three exceeds the bound of two, so three vectors in the plane cannot be independent; they must be dependent.
Verify on a concrete triple
Why: Exhibit an explicit nontrivial relation for a sample of three vectors, confirming dependence directly.
\[ 2\mathbf{e}_1 + 3\mathbf{e}_2 - (2,3) = \mathbf{0}\ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "three vectors in the plane must be dependent", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Exhibit an explicit nontrivial relation for a sample of three vectors, confirming dependence directly.
Intuition
Imagine two different independent spanning sets for the same space. Each spans, and each is independent.
Run the exchange bound once each way: the first is no bigger than the second, and the second is no bigger than the first. So they have exactly the same size.
That common size is the dimension, and the exchange lemma is precisely what guarantees it does not depend on which spanning-and-independent set you chose. Next deck makes this official.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Reasoning that because these three vectors reach every point of the plane, they must be independent.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Spanning is about reaching everything; independence is about doing so without redundancy.
The set spans the plane, yet it is dependent because of a nontrivial relation.
Why: Spanning is about reaching everything; independence is about doing so without redundancy. A set can reach everything and still carry a redundant vector.
Trap
Reasoning that because these three vectors reach every point of the plane, they must be independent.
\[ \{(1,0),\ (0,1),\ (1,1)\} \]
The mistake: fusing two different properties
Why: Spanning is about reaching everything; independence is about doing so without redundancy. A set can reach everything and still carry a redundant vector.
The set spans the plane, yet it is dependent because of a nontrivial relation.
\[ (1,0)+(0,1)-(1,1)=(0,0) \]
The fix: keep spanning and independence separate
Why: Spanning can hold with too many vectors; independence can hold with too few. Only when both hold at once do you have a basis.
Notation
Annotate
From Trap: assuming a spanning set must be independent — read this one piece at a time. What is each part doing?
On: \( (1,0)+(0,1)-(1,1)=(0,0) \)
Anomaly
Predict first
A student writes this, and it looks reasonable:
Believing that adding a vector to a set must enlarge its span.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The eye sees a second vector and assumes new directions, but the added vector may already lie inside the current span.
The added vector is a multiple of the first, so it is already in the span; the span does not grow.
Why: The eye sees a second vector and assumes new directions, but the added vector may already lie inside the current span.
Trap
Believing that adding a vector to a set must enlarge its span.
\[ \{(1,2)\} \ \rightsquigarrow\ \{(1,2),\ (2,4)\} \]
The mistake: ignoring whether the new vector is already reachable
Why: The eye sees a second vector and assumes new directions, but the added vector may already lie inside the current span.
The added vector is a multiple of the first, so it is already in the span; the span does not grow.
\[ (2,4)=2(1,2)\in\operatorname{span}\{(1,2)\} \]
The fix: a vector enlarges the span only if it is outside it
Why: Adding a dependent vector leaves the span exactly the same line. Only a vector not already in the span adds new reach.
\[ \operatorname{span}\{(1,2),(2,4)\}=\operatorname{span}\{(1,2)\} \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Concept
Independence is inherited downward. If a set is independent, so is any subset of it.
A nontrivial relation among the subset would extend, by padding with zero weights, to a nontrivial relation among the whole set, contradicting its independence.
\[ B \subseteq A,\ A \text{ independent} \ \Longrightarrow\ B \text{ independent} \]
The mirror statement for spanning goes the other way: independence shrinks safely, while spanning grows safely.
Estimation
Predict first
Given that the three standard unit vectors are independent, show a two-element subset is independent.
Commit before you compute: what does Worked example: a subset of an independent set come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the subset directly
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Check that the only combination of the subset giving zero is trivial, confirming independence of the subset.
Worked example
Given that the three standard unit vectors are independent, show a two-element subset is independent.
\[ A=\{\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\},\qquad B=\{\mathbf{e}_1,\mathbf{e}_3\} \]
Assume a relation on the subset
Why: Suppose a combination of the subset equals zero; we want to force its weights to vanish.
\[ a\,\mathbf{e}_1 + c\,\mathbf{e}_3 = \mathbf{0} \]
Pad it to a relation on the full set
Why: Add the missing vector with weight zero; this is now a combination of the full independent set equal to zero.
\[ a\,\mathbf{e}_1 + 0\,\mathbf{e}_2 + c\,\mathbf{e}_3 = \mathbf{0} \]
Invoke independence of the full set
Why: Since the full set is independent, every weight in this padded relation is zero, in particular a and c.
\[ a = 0,\ c = 0 \]
Verify the subset directly
Why: Check that the only combination of the subset giving zero is trivial, confirming independence of the subset.
\[ a\,\mathbf{e}_1 + c\,\mathbf{e}_3 = (a,0,c) = \mathbf{0} \iff a=c=0\ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "a subset of an independent set", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Check that the only combination of the subset giving zero is trivial, confirming independence of the subset.
Concept
Spanning is inherited upward. If a set spans the space, then any larger set containing it also spans.
\[ S \subseteq T,\ \operatorname{span}(S)=V \ \Longrightarrow\ \operatorname{span}(T)=V \]
Adding vectors can only enlarge or preserve the span, and it was already the whole space, so it stays the whole space. Extra vectors never hurt spanning, only independence.
Explain it
Discussion prompt
Explain Every superset of a spanning set still spans to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Spanning is inherited upward. If a set spans the space, then any larger set containing it also spans.
Commit first
Predict first
In real three-space, which one of these statements is true?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: A spanning set can have four vectors.
Why: Three-space is spanned by three standard vectors, so any independent set has at most three vectors, while spanning sets may have three or more. Four vectors can still span, they will just be dependent.
The rating matters as much as the answer: confident-and-wrong is the combination that survives revision, because nothing about it feels like it needs revisiting.
Check
Keep the two bounds straight: an independent set is capped by a spanning set's size, and a spanning set is floored by an independent set's size.
Check your understanding
In real three-space, which one of these statements is true?
Answer: A
Why: Three-space is spanned by three standard vectors, so any independent set has at most three vectors, while spanning sets may have three or more. Four vectors can still span, they will just be dependent.
Concept
You have met the two halves. A set can span without being independent, and be independent without spanning.
The prize is a set that does both at once: it reaches every vector, and it does so without redundancy. That is a basis, the subject of the next deck.
The exchange lemma already told you why any two such sets share one size. That shared size is the dimension.
Analogy
Discussion prompt
Explain Independent and spanning together: the target by analogy to something with no Foundations of Higher Mathematics in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
You have met the two halves. A set can span without being independent, and be independent without spanning.
Step zero
Discussion prompt
Worked example: a set that is both independent and spanning — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Check independence
Answer:
Worked example
Show this pair is independent and spans the plane.
\[ \{(1,0),\ (1,1)\} \]
Check independence
Why: Neither vector is a scalar multiple of the other, so as a pair they are independent.
\[ (1,1) \ne \lambda(1,0) \text{ for any } \lambda \]
Check spanning by solving for a general target
Why: Write an arbitrary point as a combination and solve; a solution for every target proves spanning.
\[ (x,y) = a(1,0) + b(1,1) \]
Solve the weights
Why: Match coordinates: the second gives b, then the first gives a. A solution always exists.
\[ b = y,\qquad a = x - y \]
Verify the general combination
Why: Substitute the weights back and confirm the sum is the arbitrary target, proving the pair spans the plane.
\[ (x-y)(1,0) + y(1,1) = (x-y+y,\ y) = (x,y)\ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "a set that is both independent and spanning", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Neither vector is a scalar multiple of the other, so as a pair they are independent.
Prediction
Predict first
If a set S is linearly independent and a vector v lies in its span, how many ways can v be written as a combination of S?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: Exactly one.
Why: Two representations would differ to give a nontrivial combination equal to zero, which independence forbids. So a vector in the span of an independent set has one and only one set of weights.
Check
Recall the uniqueness theorem: independence controls how many recipes a reachable vector can have.
Check your understanding
If a set S is linearly independent and a vector v lies in its span, how many ways can v be written as a combination of S?
Answer: A
Why: Two representations would differ to give a nontrivial combination equal to zero, which independence forbids. So a vector in the span of an independent set has one and only one set of weights.
Concept
For a strong computer scientist, the whole topic reduces to one algorithm: Gaussian elimination on the matrix of your vectors.
The number of pivots is the rank, the size of the largest independent subset and the dimension of the span. Independence is rank equal to the number of vectors; spanning is rank equal to the dimension of the space.
This makes both questions decidable in cubic time, and connects span and independence to the same elimination that solves linear systems.
Counterexample
Discussion prompt
For a strong computer scientist, the whole topic reduces to one algorithm: Gaussian elimination on the matrix of your vectors.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Pattern
To describe a span
Why: Parametrize all weighted sums and simplify to a line, plane, coordinate condition, or the whole space.
To test membership in a span
Why: Solve the linear system asking whether the target is a combination; a solution means yes.
To test independence
Why: Set a general combination to zero, row reduce, and check that the only solution is all weights zero, one pivot per vector.
To exhibit a dependence
Why: If a free variable appears, read the nontrivial relation off the reduced system and solve one vector in terms of the others.
To compare sizes
Why: Use the exchange bound: any independent set has at most as many vectors as any spanning set.
Real world
Discussion prompt
Outside this lesson: where does Span & Linear Independence actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Full decision flow: span and independence is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck covers linear combinations and the span of a set as the smallest subspace containing it, then draws the sharp line between spanning and independence. It builds the Steinitz exchange lemma as the engine behind a well-defined notion of dimension, ties independence to unique representation and to row reduction, and dismantles the classic traps: confusing spanning with independence, testing three or more vectors a pair at a time, and slipping the zero vector into an independent set.
Ranking
Put in order
Put the moves of Worked example: capstone with four vectors into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Three-space is spanned by three vectors, so four vectors must be dependent before any computation.
Worked example
Analyze four vectors in three-space: find a dependence and extract a spanning subset that is independent.
\[ \mathbf{v}_1=(1,0,0),\ \mathbf{v}_2=(0,1,0),\ \mathbf{v}_3=(1,1,0),\ \mathbf{v}_4=(0,0,1) \]
Bound by counting
Why: Three-space is spanned by three vectors, so four vectors must be dependent before any computation.
\[ |\{\mathbf{v}_1,\dots,\mathbf{v}_4\}| = 4 > 3 \]
Exhibit the explicit relation
Why: The third vector is the sum of the first two, giving a nontrivial combination equal to zero.
\[ \mathbf{v}_1 + \mathbf{v}_2 - \mathbf{v}_3 = \mathbf{0} \]
Discard the redundant vector
Why: Since the third is a combination of the others, removing it preserves the span; the remaining three are the first, second, and fourth.
\[ \{\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_4\} \]
Confirm the trimmed set is independent and spans
Why: The first, second, and fourth vectors are the standard unit vectors, which are independent and span three-space.
\[ \{\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\} = \{\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_4\} \]
Verify the dependence and the recovery of the dropped vector
Why: Check the relation numerically and confirm the discarded vector is still reachable from the trimmed set, so nothing was lost.
\[ \mathbf{v}_1+\mathbf{v}_2 = (1,1,0) = \mathbf{v}_3 \in \operatorname{span}\{\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_4\}\ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "capstone with four vectors", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Check the relation numerically and confirm the discarded vector is still reachable from the trimmed set, so nothing was lost.
Elimination
Eliminate the wrong options
Which statement about these three vectors is correct?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: C
Why: The third vector equals the first plus the second, so the combination v1 plus v2 minus v3 is a nontrivial relation equal to zero. That makes the set dependent, and its span is only the plane where the third coordinate is zero.
Check
Apply everything at once to a specific triple in three-space.
\[ \mathbf{v}_1=(1,0,0),\ \mathbf{v}_2=(0,1,0),\ \mathbf{v}_3=(1,1,0) \]
Check your understanding
Which statement about these three vectors is correct?
Answer: C
Why: The third vector equals the first plus the second, so the combination v1 plus v2 minus v3 is a nontrivial relation equal to zero. That makes the set dependent, and its span is only the plane where the third coordinate is zero.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Recipe: describing the span of a set · Recipe: test a finite set for independence · Full decision flow: span and independence · A linear combination mixes vectors with scalar weights · Trivial versus nontrivial combinations. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
A linear combination scales and adds finitely many vectors. The span of a set is all its finite combinations, and it is the smallest subspace containing the set.
A set is independent when only the trivial combination reaches the zero vector. Equivalently: no vector is a combination of the others, and every vector in the span has a unique representation.
Spanning and independence are different and opposite pressures: spanning wants enough vectors, independence wants few enough. Test independence by row reduction, and never test a triple by checking pairs.
The Steinitz exchange lemma caps any independent set by any spanning set, which is exactly what will make dimension well-defined in the next deck on Basis and Dimension.
| Question | How to answer it |
|---|---|
| Is v in span(S)? | Solve for weights; a solution means yes |
| Is S independent? | Row reduce; a pivot for every vector |
| Is S a basis? | Independent and spanning at once |
| How big can an independent set be? | At most the size of any spanning set |
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