This deck presents a basis as the fusion of independence and spanning, then covers the unique-coordinate theorem, extending an independent set and shrinking a spanning set, the invariance-of-dimension theorem proved via the exchange lemma, and the Grassmann dimension formula. It targets the classic traps: calling a spanning but dependent set a basis, assuming that any n vectors in an n-dimensional space form a basis, confusing the number of vectors with the dimension, and forgetting that infinite-dimensional spaces exist.
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. State what a basis is and why it needs both independence and spanning.
2. Read off the unique coordinates of a vector relative to an ordered basis.
3. Extend an independent set to a basis and shrink a spanning set to a basis.
4. Prove any two bases of a space share the same size, and compute dimension.
5. Apply the dimension formula for the sum of two subspaces.
Warm-up
Discussion prompt
Before we open Basis & Dimension: without looking back, what was the main idea of Span & Linear Independence, 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 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.
Concept
This deck fuses the two ideas from the last two lectures into a single object.
A set spans the space when every vector is some finite linear combination of it.
\( \operatorname{span}(S) = V \)
A set is linearly independent when the only combination equal to the zero vector is the all-zeros one.
\( c_1\mathbf{v}_1 + \cdots + c_k\mathbf{v}_k = \mathbf{0} \ \Rightarrow\ c_1 = \cdots = c_k = 0 \)
Counterexample
Discussion prompt
A set spans the space when every vector is some finite linear combination of it.
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:
A set is linearly independent when the only combination equal to the zero vector is the all-zeros one.
Concept
A basis is a set that spans and is independent at the same time.
basis — A subset B of a vector space V that is linearly independent and spans V. Equivalently, a maximal independent set, or a minimal spanning set.
Spanning guarantees you can reach every vector; independence guarantees no vector in the set is wasted.
Analogy
Discussion prompt
Explain A basis does both jobs at once 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:
A basis is a set that spans and is independent at the same time.
Intuition
Think of a basis as just enough vectors — no more, no less.
Too few and you cannot reach every vector: the set fails to span. Too many and some vector is redundant: the set fails to be independent.
A basis is the sweet spot where you can reach everything and nothing repeats what the others already say.
Explain it
Discussion prompt
Explain The Goldilocks principle 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:
Think of a basis as just enough vectors — no more, no less.
Concept
The single most important consequence of being a basis is uniqueness of representation.
If B is a basis, every vector is a linear combination of B in exactly one way.
\( \mathbf{v} = c_1\mathbf{b}_1 + \cdots + c_n\mathbf{b}_n \quad \text{with the } c_i \text{ unique} \)
coordinates — The unique scalars expressing a vector in a fixed ordered basis. They are the vector's address in that basis; spanning supplies existence, independence supplies uniqueness.
Definition probe
Sort into buckets
Every line below is part of the definition of basis or of coordinates — one or the other, never both. Put each where it belongs.
Ranking
Put in order
Put the moves of A basis forces unique coordinates 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. Assume the same vector is written two ways in the basis; the goal is to force the coefficients to agree.
Worked example
Claim: if B is a basis, the coordinates of any vector are unique. Prove it.
\( B = \{\mathbf{b}_1, \ldots, \mathbf{b}_n\} \)
Suppose a vector has two expansions
Why: Assume the same vector is written two ways in the basis; the goal is to force the coefficients to agree.
\( \mathbf{v} = \sum_i c_i \mathbf{b}_i = \sum_i d_i \mathbf{b}_i \)
Subtract the two expansions
Why: Moving everything to one side, the difference of the expansions is the zero vector.
\( \sum_i (c_i - d_i)\mathbf{b}_i = \mathbf{0} \)
Apply independence
Why: The basis is independent, so the only combination equal to zero is the all-zeros one; every coefficient must vanish.
\( c_i - d_i = 0 \quad \text{for every } i \)
Verify the conclusion
Why: Each coefficient of the first expansion equals the matching one of the second, so the two were identical. Existence came from spanning, uniqueness from independence.
\( c_i = d_i \ \Rightarrow\ \text{the representation is unique} \)
Picture it
Animation
Shows: Each line of the worked example "A basis forces unique coordinates", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Each coefficient of the first expansion equals the matching one of the second, so the two were identical. Existence came from spanning, uniqueness from independence.
Intuition
A basis turns abstract vectors into ordinary tuples of numbers.
Fixing a basis of size n records any vector as its list of coordinates — a faithful, reversible encoding.
\( \mathbf{v} \ \longleftrightarrow\ (c_1, \ldots, c_n) \)
In programming terms this is serialization: the basis is the schema, the coordinate tuple is the stored record, and you can always reconstruct the object exactly.
Concept
The most familiar basis is the standard basis of the space of n-tuples over a field.
\[ \mathbf{e}_i = (0,\ldots,0,\underset{i}{1},0,\ldots,0) \]
Each vector has a single one in position i and zeros elsewhere; there are exactly n of them.
Step zero
Discussion prompt
The standard vectors are a basis of 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: Check spanning
Answer:
Worked example
Show the three standard vectors form a basis of three-space.
\( \mathbf{e}_1=(1,0,0),\ \mathbf{e}_2=(0,1,0),\ \mathbf{e}_3=(0,0,1) \)
Check spanning
Why: Any triple is reached by scaling the standard vectors by its own entries and adding.
\( (x,y,z) = x\mathbf{e}_1 + y\mathbf{e}_2 + z\mathbf{e}_3 \)
Check independence
Why: Setting a combination to zero and reading coordinatewise forces each coefficient to vanish.
\( a\mathbf{e}_1+b\mathbf{e}_2+c\mathbf{e}_3=(a,b,c)=\mathbf{0} \ \Rightarrow\ a=b=c=0 \)
Verify both conditions
Why: Spanning and independence both hold, so the standard vectors are a basis and the dimension of three-space is three.
\( \dim \mathbb{R}^3 = 3 \)
Picture it
Animation
Shows: Each line of the worked example "The standard vectors are a basis of three-space", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Any triple is reached by scaling the standard vectors by its own entries and adding.
Concept
Once we care about coordinates, the order of the basis vectors matters.
ordered basis — A basis together with a fixed order on its vectors, so that a vector's coordinates can be listed as an ordered tuple.
Listing the coordinates relative to an ordered basis B gives the coordinate vector.
\( [\mathbf{v}]_B = (c_1, \ldots, c_n) \)
Intuition
The same vector wears different coordinates in different bases.
The arrow in space never moves; only the ruler you measure it against changes. Coordinates are relative, the vector itself is absolute.
This is why a vector and its coordinate tuple are not the same thing: the tuple only means something once a basis is named.
Estimation
Predict first
Find the coordinates of a vector in a nonstandard basis of the plane.
Commit before you compute: what does Coordinates in a nonstandard basis 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 by substitution
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. Plugging the coefficients back reproduces the original vector, so the coordinate vector is correct.
Worked example
Find the coordinates of a vector in a nonstandard basis of the plane.
\( B=\{(1,1),\ (1,-1)\},\qquad \mathbf{v}=(5,3) \)
Set up the coordinate equation
Why: Write the vector as a combination of the basis vectors with unknown coefficients.
\( a(1,1)+b(1,-1)=(5,3) \)
Split into components
Why: Matching first and second coordinates turns the vector equation into a linear system.
\( a+b=5,\qquad a-b=3 \)
Solve the system
Why: Adding the equations isolates a; subtracting isolates b.
\( a=4,\qquad b=1 \)
Verify by substitution
Why: Plugging the coefficients back reproduces the original vector, so the coordinate vector is correct.
\( 4(1,1)+1(1,-1)=(5,3),\qquad [\mathbf{v}]_B=(4,1) \)
Picture it
Animation
Shows: Each line of the worked example "Coordinates in a nonstandard basis", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Plugging the coefficients back reproduces the original vector, so the coordinate vector is correct.
Missing information
Discussion prompt
Find the coordinates of a quadratic in the monomial basis of degree-two polynomials.
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 monomial basis makes coordinates the coefficients, listed by increasing degree.
Worked example
Find the coordinates of a quadratic in the monomial basis of degree-two polynomials.
\( B=\{1,\ x,\ x^2\},\qquad p(x)=5-2x+3x^2 \)
Read the coefficients in order
Why: The monomial basis makes coordinates the coefficients, listed by increasing degree.
\( p = 5\cdot 1 + (-2)\cdot x + 3\cdot x^2 \)
Verify the reconstruction
Why: Rebuilding the polynomial from the coordinate tuple returns the original, so the coordinates are correct.
\( [p]_B=(5,-2,3) \)
Picture it
Animation
Shows: Each line of the worked example "Coordinates of a polynomial", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Rebuilding the polynomial from the coordinate tuple returns the original, so the coordinates are correct.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Claim: coordinates are just a bag of numbers, so the order of the basis is irrelevant.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Treating the basis as unordered, one writes the coordinates of the vector the same way regardless of which basis is named.
Order is part of an ordered basis; swapping basis vectors permutes the coordinates.
Why: Treating the basis as unordered, one writes the coordinates of the vector the same way regardless of which basis is named.
Trap
Claim: coordinates are just a bag of numbers, so the order of the basis is irrelevant.
\( B=\{(1,0),(0,1)\},\qquad B'=\{(0,1),(1,0)\} \)
Reuse the same tuple for both bases
Why: Treating the basis as unordered, one writes the coordinates of the vector the same way regardless of which basis is named.
\( [(3,7)]_{B}=(3,7)\ \text{copied as}\ (3,7)\ \text{for } B' \)
Order is part of an ordered basis; swapping basis vectors permutes the coordinates.
\( [(3,7)]_{B}=(3,7),\qquad [(3,7)]_{B'}=(7,3) \)
Coordinates are an ordered tuple
Why: Because the second basis lists the vertical axis first, the coordinate that was second moves into first place. The tuple is ordered, not a set.
Notation
Annotate
From Trap: order of basis vectors matters — read this one piece at a time. What is each part doing?
On: \( B=\{(1,0),(0,1)\},\qquad B'=\{(0,1),(1,0)\} \)
Anomaly
Predict first
A student writes this, and it looks reasonable:
Claim: these three vectors span the plane, so they form a basis of it.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Every plane vector is reachable, so the spanning box is ticked and one is tempted to declare a basis.
A basis must ALSO be independent, so drop the redundant vector.
Why: Every plane vector is reachable, so the spanning box is ticked and one is tempted to declare a basis.
Trap
Claim: these three vectors span the plane, so they form a basis of it.
\( S=\{(1,0),(0,1),(1,1)\} \)
They do span
Why: Every plane vector is reachable, so the spanning box is ticked and one is tempted to declare a basis.
\( \operatorname{span}(S)=\mathbb{R}^2 \)
But independence fails
Why: The third vector is the sum of the first two, giving a nontrivial combination equal to zero.
\( (1,0)+(0,1)-(1,1)=(0,0) \)
A basis must ALSO be independent, so drop the redundant vector.
\( B=\{(1,0),(0,1)\} \)
Now both conditions hold
Why: Two independent vectors that still span the plane. Three vectors can never be independent in a two-dimensional space, so spanning alone is never enough.
Break the constraint
Discussion prompt
The rule this trap just fixed:
Two independent vectors that still span the plane. Three vectors can never be independent in a two-dimensional space, so spanning alone is never enough.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Answer:
Every plane vector is reachable, so the spanning box is ticked and one is tempted to declare a basis.
Concept
A finite spanning set can always be trimmed down to a basis.
If some vector in the set is a combination of the others, delete it; the smaller set still spans. Repeat until no vector is redundant.
\[ \text{spanning set} \ \supseteq\ \text{basis} \]
Intuition
Peeling off a redundant vector never shrinks the span.
If a vector was already a combination of the others, everything it could reach was already reachable without it. So deleting it costs nothing.
You stop peeling exactly when no vector is redundant any more — and that is precisely the moment the set becomes independent, hence a basis.
Fill the middle
Fill in the blanks
From Shrink a spanning set to a basis — finish the line. Write what belongs on the right of the equals sign before you look.
(2,0) = 2(1,0)
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The second vector is twice the first, so it contributes nothing new to the span.
Worked example
Trim this spanning set of the plane down to a basis.
\( S=\{(1,0),\ (2,0),\ (0,1)\} \)
Spot a redundant vector
Why: The second vector is twice the first, so it contributes nothing new to the span.
\( (2,0)=2(1,0) \)
Delete it and keep spanning
Why: Removing a vector that was already a combination of the rest leaves the span unchanged.
\( \operatorname{span}\{(1,0),(0,1)\}=\mathbb{R}^2 \)
Check what remains is independent
Why: The two survivors are not multiples of one another, so no further deletion is possible.
\( a(1,0)+b(0,1)=(a,b)=\mathbf{0} \ \Rightarrow\ a=b=0 \)
Verify it is a basis
Why: The trimmed set both spans and is independent, so it is a basis, and its size two is the dimension of the plane.
\( \dim \mathbb{R}^2 = 2 \)
Picture it
Animation
Shows: Each line of the worked example "Shrink a spanning set to a basis", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The two survivors are not multiples of one another, so no further deletion is possible.
Concept
The dual move to trimming is growing an independent set.
If an independent set does not yet span, some vector lies outside its span; adjoin that vector and the enlarged set stays independent. Repeat until it spans.
\[ \text{independent set} \ \subseteq\ \text{basis} \]
Concept
Both moves are powered by one counting lemma from the previous lecture.
exchange lemma — If a set of independent vectors sits inside a space spanned by another set, then the independent set is no larger than the spanning set; each independent vector can be swapped in for a spanning one.
In one line: an independent set can never outnumber a spanning set.
\( |\text{independent}| \ \le\ |\text{spanning}| \)
Step zero
Discussion prompt
Extend one vector to a basis of 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: Adjoin a vector outside the current span
Answer:
Worked example
Grow this single independent vector into a basis of three-space.
\( \{\,(1,1,0)\,\} \)
Adjoin a vector outside the current span
Why: The vector (1,0,0) is not a scalar multiple of (1,1,0), so the pair stays independent.
\( \{(1,1,0),\ (1,0,0)\} \)
Adjoin a third vector off their plane
Why: The first two span a plane inside the coordinate hyperplane; (0,0,1) has a nonzero third entry, so it lies outside that plane.
\( \{(1,1,0),\ (1,0,0),\ (0,0,1)\} \)
Confirm independence by determinant
Why: The three vectors as rows give a nonzero determinant, so none is a combination of the others.
\( \det\begin{pmatrix} 1&1&0\\ 1&0&0\\ 0&0&1 \end{pmatrix} = -1 \neq 0 \)
Verify a basis of three-space
Why: Three independent vectors in a three-dimensional space automatically span it, so this is a basis extending the original vector.
\( \dim \mathbb{R}^3 = 3 \)
Picture it
Animation
Shows: Each line of the worked example "Extend one vector to a basis of three-space", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Three independent vectors in a three-dimensional space automatically span it, so this is a basis extending the original vector.
Concept
The exchange lemma has an immediate, powerful consequence.
In any single space, the size of every independent set is at most the size of every spanning set.
\( |\text{any independent set}| \ \le\ |\text{any spanning set}| \)
This one inequality is the whole reason dimension is a well-defined number, as the next slides show.
Concept
Here is the theorem that makes dimension meaningful.
Any two bases of the same finite-dimensional space have exactly the same number of vectors.
invariance of dimension — The theorem that all bases of a finite-dimensional vector space have equal size; that common size is well-defined and does not depend on which basis you choose.
Intuition
Why must two bases match in size? Each basis is both independent and spanning.
So one basis, viewed as independent, is bounded above by the other, viewed as spanning. Then swap the roles and the bound runs the other way.
Two bounds in opposite directions leave only one possibility: the two sizes are equal.
Estimation
Predict first
Prove it. Let B and C both be bases of the same space, with sizes m and n.
Commit before you compute: what does Any two bases have the same size 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 equality
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. Two inequalities in opposite directions pin the sizes together, so every basis of the space has the same number of vectors.
Worked example
Prove it. Let B and C both be bases of the same space, with sizes m and n.
\( |B|=m,\qquad |C|=n \)
Bound m by n
Why: B is independent and C spans, so by the exchange lemma the independent set B is no larger than the spanning set C.
\( m \le n \)
Swap the roles
Why: Now C is independent and B spans, so the same lemma gives the reverse bound.
\( n \le m \)
Verify equality
Why: Two inequalities in opposite directions pin the sizes together, so every basis of the space has the same number of vectors.
\( m \le n \ \text{and}\ n \le m \ \Rightarrow\ m=n \)
Picture it
Animation
Shows: Each line of the worked example "Any two bases have the same size", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Two inequalities in opposite directions pin the sizes together, so every basis of the space has the same number of vectors.
Concept
Because all bases share a size, we can name that size.
dimension — The number of vectors in any basis of a finite-dimensional vector space. Well-defined by invariance of dimension.
The zero space has the empty set as its only basis, so its dimension is zero.
\( \dim \{\mathbf{0}\} = 0 \)
Pattern
1. Write a spanning set for the space
Why: Start from any generating set you can describe, even a wasteful one.
2. Remove redundant vectors
Why: Delete any vector that is a combination of the others; the span is unchanged until the set is independent.
3. Confirm independence and spanning
Why: What survives is a basis: it still spans and now has no redundancy.
4. Count
Why: The number of surviving vectors is the dimension of the space.
Concept
A few standard spaces and their dimensions are worth memorizing.
| Space | A natural basis | Dimension |
|---|---|---|
| Space of n-tuples | the standard vectors | n |
| Polynomials of degree at most n | the monomials up to degree n | n + 1 |
| m by n matrices | the matrix units | m n |
Notice the degree-two polynomials have dimension three, not two — the constant term counts.
Comparison
Comparison matrix
From The dimension zoo: refill the A natural basis column from what you know. The rest of the table is as it appeared.
| Space | A natural basis | Dimension |
|---|---|---|
| Space of n-tuples | the standard vectors | n |
| Polynomials of degree at most n | the monomials up to degree n | n + 1 |
| m by n matrices | the matrix units | m n |
Ranking
Put in order
Put the moves of The monomials are a basis of the quadratics 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. Every quadratic is exactly a combination of the three monomials via its coefficients.
Worked example
Show the three monomials are a basis of the degree-two polynomials.
\( B=\{1,\ x,\ x^2\} \)
Check spanning
Why: Every quadratic is exactly a combination of the three monomials via its coefficients.
\( a+bx+cx^2 = a\cdot 1 + b\cdot x + c\cdot x^2 \)
Check independence
Why: A polynomial is the zero polynomial only when every coefficient is zero, so no nontrivial combination vanishes.
\( a+bx+cx^2 = 0 \ \text{for all } x \ \Rightarrow\ a=b=c=0 \)
Verify the dimension
Why: Both conditions hold, so the monomials form a basis and the space of quadratics has dimension three.
\( \dim \mathcal{P}_2 = 3 \)
Picture it
Animation
Shows: Each line of the worked example "The monomials are a basis of the quadratics", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Every quadratic is exactly a combination of the three monomials via its coefficients.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Claim: three vectors in a three-dimensional space must be a basis.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The set has three vectors and the space is three-dimensional, so it looks like a basis.
Matching the count is necessary but not sufficient — you must also confirm independence.
Why: The set has three vectors and the space is three-dimensional, so it looks like a basis.
Trap
Claim: three vectors in a three-dimensional space must be a basis.
\( S=\{(1,0,0),(0,1,0),(1,1,0)\} \)
Count and declare
Why: The set has three vectors and the space is three-dimensional, so it looks like a basis.
\( |S| = 3 = \dim \mathbb{R}^3 \)
But they are dependent
Why: The third vector is the sum of the first two; all three lie in the plane where the last coordinate is zero and can never reach a vector pointing up.
\( (1,0,0)+(0,1,0)=(1,1,0) \)
Matching the count is necessary but not sufficient — you must also confirm independence.
\( B=\{(1,0,0),(0,1,0),(0,0,1)\} \)
Replace the redundant vector
Why: Swapping the dependent vector for one with a nonzero third entry gives three independent vectors that now reach every direction.
\( \operatorname{span}(B) = \mathbb{R}^3 \)
Translation
\( S=\{(1,0,0),(0,1,0),(1,1,0)\} \)
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 a set already has the right number of vectors, one check does the whole job.
In an n-dimensional space, any n linearly independent vectors automatically span, so they are a basis.
\[ |S| = \dim V \ \text{and } S \text{ independent} \ \Rightarrow\ S \text{ is a basis} \]
Concept
The mirror statement holds for spanning.
In an n-dimensional space, any n vectors that span are automatically independent, so they too are a basis.
\[ |S| = \dim V \ \text{and } S \text{ spans} \ \Rightarrow\ S \text{ is a basis} \]
Elimination
Eliminate the wrong options
Which of these sets is a basis of three-space?
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: B
Why: Set B is the standard basis: three vectors that are independent and span three-space. Its size equals the dimension, three, with no redundant vector, so it satisfies both basis conditions.
Check
Read each set carefully — count, independence, and spanning all matter.
Check your understanding
Which of these sets is a basis of three-space?
Answer: B
Why: Set B is the standard basis: three vectors that are independent and span three-space. Its size equals the dimension, three, with no redundant vector, so it satisfies both basis conditions.
Concept
A subspace is itself a vector space, so it has a dimension, and that dimension is bounded.
Any independent set inside a subspace is also independent in the whole space, so it cannot be larger than the ambient dimension.
\( W \subseteq V \ \Rightarrow\ \dim W \le \dim V \)
Moreover a basis of the subspace can always be extended to a basis of the whole space.
Concept
Reaching the ambient dimension is surprisingly rigid.
If a subspace has the same dimension as the finite-dimensional space it lives in, it must be the whole space.
\( W \subseteq V,\ \dim W = \dim V \ \Rightarrow\ W = V \)
A basis of the subspace is then an independent set of full size, hence already a basis of everything.
Missing information
Discussion prompt
Find a basis and the dimension of this plane in three-space.
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:
Treat two coordinates as free and let the equation determine the third.
Worked example
Find a basis and the dimension of this plane in three-space.
\( W = \{(x,y,z) : x + y + z = 0\} \)
Solve the defining equation for one variable
Why: Treat two coordinates as free and let the equation determine the third.
\( x = -y - z \)
Write the general solution
Why: Substitute and group the free variables to expose one vector per free variable.
\( (-y-z,\ y,\ z) = y(-1,1,0) + z(-1,0,1) \)
Read off the spanning vectors
Why: The coefficients of the free variables form a spanning set of the plane.
\( W = \operatorname{span}\{(-1,1,0),\ (-1,0,1)\} \)
Verify independence and dimension
Why: The two vectors are not multiples of one another, so they are independent and form a basis; the plane has dimension two.
\( \dim W = 2 \)
Picture it
Animation
Shows: Each line of the worked example "The dimension of a plane through the origin", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The two vectors are not multiples of one another, so they are independent and form a basis; the plane has dimension two.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Claim: the span of these three vectors is three-dimensional because there are three of them.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Three vectors are written down, so one guesses the span has dimension three.
Dimension counts independent directions, not listed vectors; reduce to a basis first.
Why: Three vectors are written down, so one guesses the span has dimension three.
Trap
Claim: the span of these three vectors is three-dimensional because there are three of them.
\( S=\{(1,2),\ (2,4),\ (3,6)\} \)
Count the listed vectors
Why: Three vectors are written down, so one guesses the span has dimension three.
\( |S| = 3 \)
But they are all parallel
Why: Each vector is a scalar multiple of the first, so the span is a single line.
\( (2,4)=2(1,2),\qquad (3,6)=3(1,2) \)
Dimension counts independent directions, not listed vectors; reduce to a basis first.
\( \operatorname{span}(S) = \operatorname{span}\{(1,2)\} \)
Reduce to a basis
Why: Only one independent direction survives, so the span is a line of dimension one, not three.
\( \dim \operatorname{span}(S) = 1 \)
Notation
Annotate
From Trap: counting vectors instead of dimension — read this one piece at a time. What is each part doing?
On: \( \dim \operatorname{span}(S) = 1 \)
Step zero
Discussion prompt
A basis for the solution space of a system — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Combine the equations
Answer:
Worked example
Find a basis and the dimension of the solutions of this homogeneous system in four-space.
\( \begin{aligned} x_1+x_2+x_3+x_4 &= 0 \\ x_1-x_2+x_3-x_4 &= 0 \end{aligned} \)
Combine the equations
Why: Adding and subtracting the two equations isolates the pivot variables in terms of the rest.
\( x_1 = -x_3,\qquad x_2 = -x_4 \)
Identify the free variables
Why: The non-pivot variables are free; the pivot variables are forced by them.
\( x_3, x_4 \ \text{are free} \)
Write the general solution by free variable
Why: Every solution is a combination of one vector for each free variable.
\( (-x_3,\ -x_4,\ x_3,\ x_4) = x_3(-1,0,1,0) + x_4(0,-1,0,1) \)
Verify the basis and dimension
Why: Each basis vector satisfies both equations, and the two are independent, so the solution space has dimension two.
\( \dim(\text{solution space}) = 2 \)
Picture it
Animation
Shows: Each line of the worked example "A basis for the solution space of a system", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Each basis vector satisfies both equations, and the two are independent, so the solution space has dimension two.
Intuition
The number of free variables equals the dimension of the solution space.
Each free variable is a knob you can turn independently; turning one traces out a single basis direction. Pivot variables are forced, so they add no freedom.
This is why a homogeneous system with more unknowns than equations always has a nonzero solution: at least one knob remains free to turn.
Check
Count the free variables, not the equations.
Check your understanding
A single equation x1 + x2 + x3 + x4 = 0 defines a subspace of four-space. What is its dimension?
Answer: B
Why: One independent equation on four unknowns leaves three free variables, and the solution space has one basis vector per free variable, so its dimension is three (unknowns minus rank, four minus one).
Hypothesis
Predict first
Basis and dimension of a span 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: Discard the obvious multiple
Why: The second vector is exactly twice the first, so it is redundant and can be dropped without changing the span.
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
Find the dimension of the span of these three vectors in three-space.
\( S=\{(1,2,3),\ (2,4,6),\ (1,0,1)\} \)
Discard the obvious multiple
Why: The second vector is exactly twice the first, so it is redundant and can be dropped without changing the span.
\( (2,4,6) = 2(1,2,3) \)
Test the remaining two
Why: The first and third vectors are not scalar multiples of each other, so they are independent.
\( (1,2,3),\ (1,0,1) \ \text{are independent} \)
Verify the basis and dimension
Why: The two independent survivors span the same set as the original three, so they are a basis of the span and its dimension is two.
\( \dim \operatorname{span}(S) = 2 \)
Picture it
Animation
Shows: Each line of the worked example "Basis and dimension of a span", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The two independent survivors span the same set as the original three, so they are a basis of the span and its dimension is two.
Prediction
Predict first
In the ordered basis { (1,1), (1,-1) } of the plane, what is the coordinate vector of (2,6)?
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: (4, -2)
Why: Solving a(1,1)+b(1,-1)=(2,6) gives the equations a+b=2 and a-b=6, so a=4 and b=-2; the coordinate vector is (4, -2), and substituting back returns (2,6).
Check
Set up and solve the coordinate equations.
Check your understanding
In the ordered basis { (1,1), (1,-1) } of the plane, what is the coordinate vector of (2,6)?
Answer: A
Why: Solving a(1,1)+b(1,-1)=(2,6) gives the equations a+b=2 and a-b=6, so a=4 and b=-2; the coordinate vector is (4, -2), and substituting back returns (2,6).
Concept
Fixing an ordered basis of size n identifies the space with the space of n-tuples.
The coordinate map sends each vector to its coordinate tuple; it is reversible and respects addition and scaling.
\( [\,\cdot\,]_B : V \ \xrightarrow{\ \sim\ }\ F^n \)
So any two spaces of the same dimension over the same field are structurally identical — the same space wearing different clothes.
Intuition
Dimension is the single number that classifies a finite-dimensional space up to relabeling.
Same field and same dimension means an exact structural match: a bijection preserving every vector-space operation. Nothing else about the space can tell the two apart.
This echoes earlier units — bijections classified sets by size, isomorphisms classified groups; here dimension classifies vector spaces.
Concept
Recall two ways to combine subspaces from the previous lecture.
Their intersection collects vectors lying in both; their sum collects every sum of a vector from each.
\( U \cap W, \qquad U + W = \{\mathbf{u}+\mathbf{w} : \mathbf{u}\in U,\ \mathbf{w}\in W\} \)
Both are subspaces, and the sum is the smallest subspace containing both of them.
Concept
There is an exact accounting relating the dimensions of a sum and an intersection.
dimension formula — For finite-dimensional subspaces U and W, the dimension of their sum equals the sum of their dimensions minus the dimension of their intersection.
\[ \dim(U+W) = \dim U + \dim W - \dim(U \cap W) \]
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 basis, ordered basis, invariance of dimension, dimension, dimension formula as Basis & Dimension uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Intuition
Why the subtraction? Because a basis of the intersection gets counted twice.
Vectors living in both subspaces are counted once when you measure the first and again when you measure the second. Adding the two dimensions double-counts exactly the overlap.
Removing that overlap once fixes the double count — the same inclusion-exclusion idea as counting elements in two overlapping sets.
Estimation
Predict first
Apply the formula to two coordinate planes.
Commit before you compute: what does Two coordinate planes in three-space 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 the ambient space
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. The sum has dimension three inside three-space, so the two planes together span everything.
Worked example
Apply the formula to two coordinate planes.
\( U=\{(x,y,0)\},\qquad W=\{(x,0,z)\} \)
Record the two dimensions
Why: Each coordinate plane is spanned by two standard vectors, so each has dimension two.
\( \dim U = 2,\qquad \dim W = 2 \)
Find the intersection
Why: A vector in both planes has second and third coordinates zero, leaving exactly the first axis.
\( U \cap W = \{(x,0,0)\},\qquad \dim(U\cap W) = 1 \)
Apply the dimension formula
Why: Substitute the three dimensions into the Grassmann formula.
\( \dim(U+W) = 2 + 2 - 1 = 3 \)
Verify against the ambient space
Why: The sum has dimension three inside three-space, so the two planes together span everything.
\( U + W = \mathbb{R}^3 \)
Picture it
Animation
Shows: Each line of the worked example "Two coordinate planes in three-space", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The sum has dimension three inside three-space, so the two planes together span everything.
Step zero
Discussion prompt
A sum inside four-space — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Note the shared vector
Answer:
Worked example
Apply the formula to two overlapping subspaces of four-space.
\( U=\operatorname{span}\{\mathbf{e}_1,\mathbf{e}_2\},\qquad W=\operatorname{span}\{\mathbf{e}_2,\mathbf{e}_3\} \)
Note the shared vector
Why: Both subspaces contain the second standard vector, so their intersection is at least a line.
\( \mathbf{e}_2 \in U \cap W \)
Confirm the intersection is exactly that line
Why: No other independent vector lies in both subspaces, so the intersection is one-dimensional.
\( \dim(U \cap W) = 1 \)
Verify with the dimension formula
Why: The formula gives three, matching the span of the three distinct standard vectors involved.
\( \dim(U+W) = 2 + 2 - 1 = 3 \)
Picture it
Animation
Shows: Each line of the worked example "A sum inside four-space", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The formula gives three, matching the span of the three distinct standard vectors involved.
Check
Think about how two distinct planes through the origin meet.
Check your understanding
U and W are two different planes through the origin in three-space, each of dimension two. What is the dimension of their sum U + W?
Answer: B
Why: Two distinct planes through the origin in three-space meet in a line, so the intersection has dimension one, and the formula gives dim(U+W) = 2 + 2 - 1 = 3, filling all of three-space.
Concept
Not every vector space has a finite basis.
infinite-dimensional — A vector space with no finite spanning set; equivalently, one that contains independent sets of arbitrarily large size.
The space of all polynomials is the standard example: no finite collection of polynomials can produce every degree.
Ranking
Put in order
Put the moves of The polynomials are infinite-dimensional 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. Assume finitely many polynomials span all of the space, and let the largest degree among them be a fixed number.
Worked example
Prove the space of all polynomials has no finite basis.
\( \mathcal{P} = \{\text{all polynomials}\} \)
Suppose a finite spanning set existed
Why: Assume finitely many polynomials span all of the space, and let the largest degree among them be a fixed number.
\( \deg p_i \le d \quad \text{for every } i \)
Every combination stays bounded in degree
Why: A linear combination cannot exceed the largest degree present, so nothing beyond that degree is reachable.
\( \deg\Big(\textstyle\sum_i c_i p_i\Big) \le d \)
Exhibit a missing polynomial
Why: The next monomial lies in the space but outside the span, contradicting the assumption of spanning.
\( x^{\,d+1} \notin \operatorname{span}\{p_1,\ldots,p_n\} \)
Verify the conclusion
Why: No finite set can span, while the infinite family of monomials is independent, so the space is infinite-dimensional.
\( \dim \mathcal{P} = \infty \)
Picture it
Animation
Shows: Each line of the worked example "The polynomials are infinite-dimensional", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: No finite set can span, while the infinite family of monomials is independent, so the space is infinite-dimensional.
Concept
Even infinite-dimensional spaces have bases, of a subtler kind.
Hamel basis — A possibly infinite linearly independent spanning set, where spanning still means every vector is a FINITE linear combination of basis elements.
That every vector space has such a basis is a theorem equivalent to the axiom of choice, proved using Zorn's lemma.
Explain it
Discussion prompt
Explain Every space still has a basis 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:
Even infinite-dimensional spaces have bases, of a subtler kind.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Claim: every vector space has a finite basis, so dimension is always an ordinary number.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: One selects the monomials up to some fixed degree, hoping they span every polynomial.
Some spaces are genuinely infinite-dimensional, and their basis is infinite.
Why: One selects the monomials up to some fixed degree, hoping they span every polynomial.
Trap
Claim: every vector space has a finite basis, so dimension is always an ordinary number.
\( \text{assume every } V \text{ is finite-dimensional} \)
Pick a finite candidate for the polynomials
Why: One selects the monomials up to some fixed degree, hoping they span every polynomial.
\( \{1,\ x,\ x^2,\ \ldots,\ x^d\} \)
It fails at the next degree
Why: Whatever degree bound you fix, the following monomial escapes the span.
\( x^{\,d+1} \notin \operatorname{span}\{1,\ldots,x^d\} \)
Some spaces are genuinely infinite-dimensional, and their basis is infinite.
\( \{1,\ x,\ x^2,\ x^3,\ \ldots\} \)
Use the full infinite family
Why: The complete set of monomials is an independent Hamel basis of the polynomials, but it is not finite.
\( \dim \mathcal{P} = \infty \)
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.
Intuition
A crucial subtlety: even with infinitely many basis vectors, each vector uses only finitely many of them.
You never add infinitely many terms — that would require a notion of limit, which belongs to analysis, not algebra. A Hamel basis stays purely algebraic.
This is exactly what separates a Hamel basis from the bases met later in analysis, where convergent infinite sums are allowed.
Analogy
Discussion prompt
Explain Finite combinations, even with infinite bases 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:
A crucial subtlety: even with infinitely many basis vectors, each vector uses only finitely many of them.
Pattern
1. Count against the dimension
Why: If the count differs from the dimension it cannot be a basis: too few miss spanning, too many force dependence.
2. If the count matches, check one condition
Why: For exactly the right number of vectors, independence alone (or spanning alone) already forces a basis.
3. Otherwise check both directly
Why: When the dimension is unknown, verify independence and spanning separately.
4. Sanity-check with coordinates
Why: A genuine basis represents every vector in exactly one way; a failure of uniqueness signals dependence.
Real world
Discussion prompt
Outside this lesson: where does Basis & Dimension 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 Recipe: is a given set a basis? 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 presents a basis as the fusion of independence and spanning, then covers the unique-coordinate theorem, extending an independent set and shrinking a spanning set, the invariance-of-dimension theorem proved via the exchange lemma, and the Grassmann dimension formula. It targets the classic traps: calling a spanning but dependent set a basis, assuming that any n vectors in an n-dimensional space form a basis, confusing the number of vectors with the dimension, and forgetting that infinite-dimensional spaces exist.
Commit first
Predict first
You have a linearly independent set of two vectors in three-space. Which operation turns it into a basis?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: Adjoin one vector lying outside the span of the two
Why: Two independent vectors span only a plane, so you must extend by adjoining a third vector chosen outside their span; that keeps the set independent and brings the count up to the dimension three.
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
Decide which operation reaches a basis from where you are.
Check your understanding
You have a linearly independent set of two vectors in three-space. Which operation turns it into a basis?
Answer: B
Why: Two independent vectors span only a plane, so you must extend by adjoining a third vector chosen outside their span; that keeps the set independent and brings the count up to the dimension three.
Prediction
Predict first
Which statement about the space of all polynomials is correct?
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: It is infinite-dimensional, with the monomials a basis in which each polynomial is a finite combination
Why: Although each individual polynomial has finite degree, no finite set spans all degrees, so the space is infinite-dimensional; the monomials form a Hamel basis and each polynomial is a finite combination of them.
Check
Reconcile finite degree with infinite dimension.
Check your understanding
Which statement about the space of all polynomials is correct?
Answer: B
Why: Although each individual polynomial has finite degree, no finite set spans all degrees, so the space is infinite-dimensional; the monomials form a Hamel basis and each polynomial is a finite combination of them.
Elimination
Eliminate the wrong options
W is a subspace of a five-dimensional space V, and dim W = 5. What can you conclude?
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 subspace with the same dimension as the finite-dimensional whole must equal it: a basis of W is an independent set of five vectors in a five-dimensional space, hence a basis of V, so W spans all of V.
Check
Use the equal-dimension rule for subspaces.
Check your understanding
W is a subspace of a five-dimensional space V, and dim W = 5. What can you conclude?
Answer: A
Why: A subspace with the same dimension as the finite-dimensional whole must equal it: a basis of W is an independent set of five vectors in a five-dimensional space, hence a basis of V, so W spans all of V.
Concept
Step back and place today's ideas inside the whole bridge course.
Three threads keep recurring: structure-preserving maps, quotients by an equivalence, and a well-defined invariant. Dimension is the invariant thread — the vector-space cousin of a set's cardinality and a group's order.
The coordinate map is the structure-preserving thread, turning every n-dimensional space into the one standard tuple space.
Counterexample
Discussion prompt
Step back and place today's ideas inside the whole bridge course.
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 coordinate map is the structure-preserving thread, turning every n-dimensional space into the one standard tuple space.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Recipe: find a basis and the dimension · Recipe: is a given set a basis? · Recall: span and linear independence · A basis does both jobs at once · The Goldilocks principle. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
You can now build bases, recognize them, and compute dimension with confidence.
A basis is independent and spanning, so it gives every vector unique coordinates. All bases of a space share one size, and that size is the dimension.
| Idea | What to remember |
|---|---|
| Basis | independent AND spanning; gives unique coordinates |
| Dimension | the common size of every basis; a well-defined invariant |
| Matching count | n independent or n spanning vectors in n dimensions is automatically a basis |
| Sum formula | dim(U+W) = dim U + dim W - dim(U intersect W) |
| Infinite-dim | the polynomials need an infinite Hamel basis |
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