This 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.
Subject: Foundations of Higher Mathematics · 119 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. Apply the subspace test to decide whether a subset is a subspace.
2. Explain why intersections of subspaces are always subspaces but unions usually are not.
3. Form the sum of two subspaces and recognize it as the smallest subspace containing both.
4. Characterize a direct sum three equivalent ways and apply the dimension formula.
Warm-up
Discussion prompt
Before we open Subspaces, Sums & Direct Sums: without looking back, what was the main idea of Vector Spaces: Axioms & Examples, 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 the eight vector-space axioms over a field, the consequences those axioms immediately force, and the zoo of genuine examples: tuples, matrices, polynomials, functions, the solution sets of homogeneous systems, and C regarded over different fields. It targets the misconceptions that a vector is an arrow in R^n, that any set of vectors is a space, that closure under scalar multiplication comes for free, and that the choice of scalar field does not affect dimension.
Concept
Fix a vector space V over a field F. A subspace is a subset of V that is itself a vector space, using the very same addition and scalar multiplication.
\( W \subseteq V \)
The key word is inside: W borrows the operations of V, it does not invent new ones.
subspace — A subset W of a vector space V that contains the zero vector and is closed under addition and scalar multiplication; it is then a vector space in its own right.
Counterexample
Discussion prompt
Fix a vector space V over a field F. A subspace is a subset of V that is itself a vector space, using the very same addition and scalar multiplication.
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 key word is inside: W borrows the operations of V, it does not invent new ones.
Concept
Two subspaces come with every vector space at no cost.
\( \{\mathbf{0}\} \quad \text{and} \quad V \)
The first is the zero subspace, holding only the origin. The second is the whole space. Both trivially pass every test.
trivial subspaces — The zero subspace, containing only the zero vector, and the entire space V itself. Any subspace strictly between them is called proper and nonzero.
Definition probe
Sort into buckets
Every line below is part of the definition of subspace or of trivial subspaces — one or the other, never both. Put each where it belongs.
Picture it
Figure (svg): Coordinate axes and a diagonal line all passing through the origin, each a subspace of the plane.
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:
In the plane the subspaces are exactly three shapes: just the origin, any straight line through the origin, and the whole plane.
Intuition
In the plane the subspaces are exactly three shapes: just the origin, any straight line through the origin, and the whole plane.
Figure (svg): Coordinate axes and a diagonal line all passing through the origin, each a subspace of the plane.
Every subspace must pass through the origin. A flat that misses the origin is an affine set, not a subspace.
Analogy
Discussion prompt
Explain Picture: subspaces are flats through the origin 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:
In the plane the subspaces are exactly three shapes: just the origin, any straight line through the origin, and the whole plane.
Concept
You rarely re-check all eight vector-space axioms. Instead you use the subspace test.
A nonempty subset W of V is a subspace exactly when both closure conditions hold. First, closure under addition:
\( \mathbf{u}, \mathbf{w} \in W \ \Rightarrow\ \mathbf{u} + \mathbf{w} \in W \)
Second, closure under scalar multiplication:
\( c \in F,\ \mathbf{w} \in W \ \Rightarrow\ c\,\mathbf{w} \in W \)
Nonemptiness is checked in practice by confirming the zero vector lies in W.
\( \mathbf{0} \in W \)
Explain it
Discussion prompt
Explain The subspace test: three conditions 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:
You rarely re-check all eight vector-space axioms. Instead you use the subspace test.
Intuition
The remaining axioms — associativity, commutativity, distributivity — are inherited automatically, because they already hold everywhere in V and W simply reuses them.
You only have to guarantee you never leave W. Closure under addition and scalar multiplication does exactly that.
Taking the scalar to be zero forces the zero vector in, and taking it to be negative one forces additive inverses in. So closure under scalars quietly hands you the rest.
Pattern
1. Check the zero vector is in W
Why: If the origin is missing, stop immediately: W is not a subspace. This also confirms W is nonempty.
2. Take two arbitrary elements of W and add them
Why: Show the sum still satisfies the defining condition of W. This is closure under addition.
3. Take an arbitrary scalar times an arbitrary element
Why: Show the result still lies in W. This is closure under scalar multiplication.
4. Conclude
Why: If all three hold, W is a subspace. If any fails, exhibit one concrete counterexample and you are done.
Ranking
Put in order
Put the moves of A line through the origin is a subspace 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. Set the parameter to zero to get the origin, so W is nonempty and contains the origin.
Worked example
Take the line through the origin with slope two in the plane.
\( W = \{ (t, 2t) : t \in \mathbb{R} \} \)
Confirm the zero vector is in W
Why: Set the parameter to zero to get the origin, so W is nonempty and contains the origin.
\( t = 0 \ \Rightarrow\ (0,0) \in W \)
Add two elements of W
Why: A general element has first coordinate free and second coordinate twice it; the sum keeps that shape.
\( (a,2a) + (b,2b) = (a+b,\ 2(a+b)) \in W \)
Scale an element of W
Why: Multiplying by any scalar preserves the ratio between the two coordinates.
\( c\,(a,2a) = (ca,\ 2(ca)) \in W \)
Verify against the subspace test
Why: Zero vector present, closed under addition, closed under scalars. All three conditions hold, so W is a subspace of the plane.
\( W \le \mathbb{R}^2 \)
Picture it
Animation
Shows: Each line of the worked example "A line through the origin is a subspace", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Zero vector present, closed under addition, closed under scalars. All three conditions hold, so W is a subspace of the plane.
Step zero
Discussion prompt
A plane through the origin is a subspace — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Check the zero vector
Answer:
Worked example
Take the set of points in space whose coordinates sum to zero.
\( W = \{ (x,y,z) \in \mathbb{R}^3 : x + y + z = 0 \} \)
Check the zero vector
Why: The coordinates of the origin sum to zero, so the origin satisfies the defining equation.
\( 0 + 0 + 0 = 0 \)
Add two solutions
Why: If each triple has coordinates summing to zero, the coordinatewise sums add to zero as well.
\( (x_1+x_2)+(y_1+y_2)+(z_1+z_2) = 0 + 0 = 0 \)
Scale a solution
Why: Multiplying the defining equation by a scalar keeps the right-hand side zero.
\( c x + c y + c z = c(x+y+z) = c\cdot 0 = 0 \)
Verify the conclusion
Why: All three test conditions hold. W is a plane through the origin, hence a subspace of three-space.
\( W \le \mathbb{R}^3 \)
Picture it
Animation
Shows: Each line of the worked example "A plane through the origin is a subspace", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The coordinates of the origin sum to zero, so the origin satisfies the defining equation.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Claim: the line with slope two and intercept one is a subspace, since adding points on it looks fine.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: A subspace must contain the origin.
The honest subspace is the parallel line through the origin — drop the intercept.
Why: A subspace must contain the origin. On this line the second coordinate is twice the first plus one, which cannot be zero when the first coordinate is zero.
Trap
Claim: the line with slope two and intercept one is a subspace, since adding points on it looks fine.
\( L = \{ (t,\ 2t+1) : t \in \mathbb{R} \} \)
Test the zero vector first
Why: A subspace must contain the origin. On this line the second coordinate is twice the first plus one, which cannot be zero when the first coordinate is zero.
\( 2\cdot 0 + 1 = 1 \neq 0 \ \Rightarrow\ (0,0) \notin L \)
Addition escapes too
Why: Adding two points of the line doubles the intercept, landing on the parallel line with intercept two, not on the original.
\( (0,1)+(1,3) = (1,4),\quad 2\cdot 1 + 1 = 3 \neq 4 \)
The honest subspace is the parallel line through the origin — drop the intercept.
\( W = \{ (t,\ 2t) : t \in \mathbb{R} \} \)
This one passes the test
Why: The origin lies on W and both closures hold. Only flats through the origin can be subspaces; the line L was merely an affine shift of W.
Notation
Annotate
From Trap: a flat that misses the origin — read this one piece at a time. What is each part doing?
On: \( L = \{ (t,\ 2t+1) : t \in \mathbb{R} \} \)
Elimination
Eliminate the wrong options
Which of these subsets of the plane is a subspace?
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: This is a line through the origin: it contains the origin and is closed under addition and scalar multiplication, so it passes the subspace test.
Check
Pick the subset of the plane that is a subspace.
Check your understanding
Which of these subsets of the plane is a subspace?
Answer: A
Why: This is a line through the origin: it contains the origin and is closed under addition and scalar multiplication, so it passes the subspace test.
Concept
This deserves its own billing because it is the fastest disqualifier there is.
Closure under scalar multiplication, applied with the scalar zero, forces the zero vector into any nonempty subspace.
\( 0 \cdot \mathbf{w} = \mathbf{0} \in W \)
So the very first thing to check for any candidate subspace is whether the origin belongs. If it does not, you can stop.
Estimation
Predict first
Consider the solution set of a homogeneous linear system, written as a matrix times a vector equal to the zero vector.
Commit before you compute: what does Homogeneous solutions form a 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 the contrast
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 homogeneous set passed all three tests and is a subspace; the inhomogeneous set fails at the origin and is only an affine translate of it.
Worked example
Consider the solution set of a homogeneous linear system, written as a matrix times a vector equal to the zero vector.
\( W = \{ \mathbf{x} \in \mathbb{R}^n : A\mathbf{x} = \mathbf{0} \} \)
The zero vector solves it
Why: The matrix times the zero vector is the zero vector, so the origin is in W.
\( A\mathbf{0} = \mathbf{0} \)
Sums of solutions solve it
Why: Matrix multiplication is linear, so the sum of two solutions is again a solution.
\( A(\mathbf{x}+\mathbf{y}) = A\mathbf{x} + A\mathbf{y} = \mathbf{0} \)
Scalar multiples solve it
Why: Pulling the scalar through the matrix keeps the right-hand side zero.
\( A(c\mathbf{x}) = c\,A\mathbf{x} = \mathbf{0} \)
Contrast the inhomogeneous case
Why: If the right-hand side is nonzero, the origin fails the equation, so that solution set is not a subspace.
\( A\mathbf{x} = \mathbf{b},\ \mathbf{b} \neq \mathbf{0} \ \Rightarrow\ \mathbf{0} \notin \text{solutions} \)
Verify the contrast
Why: The homogeneous set passed all three tests and is a subspace; the inhomogeneous set fails at the origin and is only an affine translate of it.
Picture it
Animation
Shows: Each line of the worked example "Homogeneous solutions form a subspace", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The homogeneous set passed all three tests and is a subspace; the inhomogeneous set fails at the origin and is only an affine translate of it.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Claim: the integer grid in the plane is a subspace, since adding two integer points gives an integer point.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Multiply an integer point by one half and you leave the grid: half of the point one-zero is not an integer point.
A subspace must absorb every real scalar, not merely integer combinations.
Why: Multiply an integer point by one half and you leave the grid: half of the point one-zero is not an integer point.
Trap
Claim: the integer grid in the plane is a subspace, since adding two integer points gives an integer point.
\( W = \{ (m,n) : m,n \in \mathbb{Z} \} \)
Scalar closure fails
Why: Multiply an integer point by one half and you leave the grid: half of the point one-zero is not an integer point.
\( \tfrac{1}{2}(1,0) = (\tfrac{1}{2},0) \notin W \)
A subspace must absorb every real scalar, not merely integer combinations.
The honest subspace is the whole line
Why: Allowing all real multiples of the point one-zero fills the horizontal axis, which is closed under addition and under every real scalar.
\( \mathbb{R}\,(1,0) = \{ (t,0) : t \in \mathbb{R} \} \)
Concept
Take two subspaces of the same space. Their intersection is the set of vectors lying in both at once.
\( U \cap W = \{ \mathbf{v} : \mathbf{v} \in U \ \text{and}\ \mathbf{v} \in W \} \)
This is always a subspace — no exceptions, no conditions.
Intuition
If a vector obeys the rules of U and also the rules of W, then any operation lands you back in U, because U is closed, and back in W, because W is closed. So you stay in both.
The two sets of constraints stack; they never conflict. That is why the intersection survives untouched.
Missing information
Discussion prompt
Let U and W be subspaces of V. Show their intersection is a subspace.
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:
Each subspace contains the origin, so the origin lies in their intersection.
Worked example
Let U and W be subspaces of V. Show their intersection is a subspace.
\( U \le V, \quad W \le V \)
The zero vector is in both
Why: Each subspace contains the origin, so the origin lies in their intersection.
\( \mathbf{0} \in U \ \text{and}\ \mathbf{0} \in W \ \Rightarrow\ \mathbf{0} \in U \cap W \)
Closed under addition
Why: Take two vectors in the intersection. They lie in U so their sum does, and they lie in W so their sum does; hence the sum is in both.
\( \mathbf{u},\mathbf{v} \in U\cap W \ \Rightarrow\ \mathbf{u}+\mathbf{v} \in U \cap W \)
Closed under scalars
Why: A scalar multiple of a vector that lies in both U and W stays in U and stays in W.
\( c\mathbf{u} \in U \cap W \)
Verify all three conditions
Why: The origin is present and both closures hold, so the intersection passes the subspace test. The argument never used that there were only two subspaces.
Picture it
Animation
Shows: Each line of the worked example "Prove the intersection is a subspace", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The origin is present and both closures hold, so the intersection passes the subspace test. The argument never used that there were only two subspaces.
Concept
The proof used nothing about the number of subspaces. Intersecting any collection, even infinitely many, still yields a subspace.
\( \bigcap_{i \in I} U_i \ \le\ V \)
This is what makes the next idea precise: the smallest subspace containing a set is the intersection of all subspaces that contain it.
Prediction
Predict first
For subspaces U and W of three-space, which statement is always true?
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: Their intersection is a subspace of three-space
Why: The intersection always contains the origin and is closed under addition and scalar multiplication, so it is a subspace no matter how U and W are positioned.
Check
Let U and W be subspaces of three-space. Which statement is always true?
Check your understanding
For subspaces U and W of three-space, which statement is always true?
Answer: A
Why: The intersection always contains the origin and is closed under addition and scalar multiplication, so it is a subspace no matter how U and W are positioned.
Concept
The union of two subspaces is the set of vectors lying in at least one of them.
\( U \cup W = \{ \mathbf{v} : \mathbf{v} \in U \ \text{or}\ \mathbf{v} \in W \} \)
Unlike the intersection, this is usually not a subspace.
Picture it
Figure (svg): The horizontal and vertical axes forming a cross, with a diagonal arrow from the origin to the point one-one, which lies on neither axis.
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:
Picture the two coordinate axes. A vector on the first axis plus a vector on the second points diagonally — off both axes. The sum escaped the union.
Intuition
Figure (svg): The horizontal and vertical axes forming a cross, with a diagonal arrow from the origin to the point one-one, which lies on neither axis.
Picture the two coordinate axes. A vector on the first axis plus a vector on the second points diagonally — off both axes. The sum escaped the union.
Step zero
Discussion prompt
The union of the two axes is not a subspace — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Pick one vector from each axis
Answer:
Worked example
Let U be the horizontal axis and W the vertical axis in the plane.
\( U = \{(t,0) : t \in \mathbb{R}\},\quad W = \{(0,s) : s \in \mathbb{R}\} \)
Pick one vector from each axis
Why: The point one-zero lies in U and zero-one lies in W, so both belong to the union.
\( (1,0) \in U \subseteq U\cup W,\quad (0,1) \in W \subseteq U \cup W \)
Add them
Why: Their sum is the diagonal vector, which lies on neither axis.
\( (1,0)+(0,1) = (1,1) \)
Check membership of the sum
Why: The point one-one has both coordinates nonzero, so it sits on neither the horizontal nor the vertical axis.
\( (1,1) \notin U \cup W \)
Verify the failure
Why: We exhibited two elements of the union whose sum leaves it, so closure under addition fails and the union is not a subspace.
Picture it
Animation
Shows: Each line of the worked example "The union of the two axes is not a subspace", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The point one-one has both coordinates nonzero, so it sits on neither the horizontal nor the vertical axis.
Fill the middle
Fill in the blanks
From Trap: is a union of subspaces a subspace? — finish the line. Write what belongs on the right of the equals sign before you look.
(1,0)+(0,1) = (1,1) \notin U \cup W
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The union of the two axes contains one-zero and zero-one but not their sum one-one, so closure under addition fails.
Trap
Claim: the union of two subspaces is a subspace, since each piece is.
Counterexample from the axes
Why: The union of the two axes contains one-zero and zero-one but not their sum one-one, so closure under addition fails.
\( (1,0)+(0,1) = (1,1) \notin U \cup W \)
The union is a subspace exactly when one of the two subspaces contains the other.
The nested case works
Why: If U sits inside W then the union is just W, already a subspace. Otherwise you can always add one vector from each and escape.
\( U \subseteq W \ \Rightarrow\ U \cup W = W \)
Translation
\( U \subseteq W \ \Rightarrow\ U \cup W = W \)
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.
Check
For subspaces U and W of V, complete the statement.
Check your understanding
The union of two subspaces U and W is a subspace if and only if:
Answer: A
Why: A union of two subspaces is closed under addition only when one contains the other; then the union equals the larger subspace. Otherwise a cross sum escapes it.
Concept
To repair the union, do not just collect the two sets — collect every vector you can build by adding one from each.
\( U + W = \{ \mathbf{u} + \mathbf{w} : \mathbf{u} \in U,\ \mathbf{w} \in W \} \)
sum of subspaces — The set of all sums of a vector from U and a vector from W. It is the smallest subspace containing both U and W.
Intuition
The union only lets you stand on one piece at a time. The sum lets you take a step along U and then a step along W, reaching everything in between.
That extra freedom is exactly what closes up the gaps the union left open.
Concept
The sum is not merely some subspace containing U and W — it is the smallest one.
\( U + W = \bigcap \{ S \le V : U \subseteq S,\ W \subseteq S \} \)
Any subspace that contains both U and W must contain all of their pairwise sums, so it contains the sum. And the sum is itself such a subspace, so it is the least one.
Concept
The subspaces of V sit in a structure with two operations: intersection acts as a greatest lower bound and sum acts as a least upper bound.
\( U \wedge W = U \cap W, \qquad U \vee W = U + W \)
Intersection is the largest subspace inside both; sum is the smallest subspace containing both. That is exactly a lattice, ordered by inclusion.
Fill the middle
Fill in the blanks
From Prove the sum is a subspace — finish the line. Write what belongs on the right of the equals sign before you look.
(\mathbf(\mathbf{u}_1+\mathbf{u}_2)+(\mathbf{w}_1+\mathbf{w}_2)_1+\mathbf____1)+(\mathbf____2+\mathbf____2) = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Write the origin as the origin plus the origin, one term drawn from each subspace.
Worked example
Show the sum of two subspaces is itself a subspace.
\( U \le V, \quad W \le V \)
The zero vector
Why: Write the origin as the origin plus the origin, one term drawn from each subspace.
\( \mathbf{0} = \mathbf{0} + \mathbf{0} \in U + W \)
Closed under addition
Why: Add two elements of the sum and regroup so the U parts combine and the W parts combine, using that each subspace is closed.
\( (\mathbf{u}_1+\mathbf{w}_1)+(\mathbf{u}_2+\mathbf{w}_2) = (\mathbf{u}_1+\mathbf{u}_2)+(\mathbf{w}_1+\mathbf{w}_2) \)
Closed under scalars
Why: Distribute the scalar across the two parts; each part stays inside its own subspace.
\( c(\mathbf{u}+\mathbf{w}) = c\mathbf{u} + c\mathbf{w} \in U + W \)
Verify the test
Why: Origin present, both closures hold, so the sum is a subspace. It contains U by taking the W part to be zero, and contains W by taking the U part to be zero.
Picture it
Animation
Shows: Each line of the worked example "Prove the sum is a subspace", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Origin present, both closures hold, so the sum is a subspace. It contains U by taking the W part to be zero, and contains W by taking the U part to be zero.
Pattern
1. Write a spanning set for U and one for W
Why: Say U is spanned by one list of vectors and W by another.
2. Pool the two spanning sets
Why: The union of the two lists spans the sum, because every element of the sum is a combination from U plus a combination from W.
3. Reduce to a basis
Why: Discard dependent vectors to read off a basis and the dimension of the sum.
\( U + W = \operatorname{span}\big(\text{spanning set of } U \ \cup\ \text{spanning set of } W\big) \)
Hypothesis
Predict first
The sum of two distinct lines fills the plane 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: Pool the spanning vectors
Why: The sum is spanned by the two direction vectors together.
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
Let U be the horizontal axis and W the line through the origin with slope one.
\( U = \operatorname{span}\{(1,0)\},\quad W = \operatorname{span}\{(1,1)\} \)
Pool the spanning vectors
Why: The sum is spanned by the two direction vectors together.
\( U + W = \operatorname{span}\{(1,0),(1,1)\} \)
Check independence
Why: The two vectors are not multiples of each other, so they are independent and span a two-dimensional space.
\( \det\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix} = 1 \neq 0 \)
Identify the sum
Why: Two independent vectors in the plane span the whole plane.
\( U + W = \mathbb{R}^2 \)
Verify by hitting a target
Why: A general point is reached explicitly, confirming the sum is all of the plane.
\( (x,y) = (x-y)(1,0) + y(1,1) \)
Picture it
Animation
Shows: Each line of the worked example "The sum of two distinct lines fills the plane", 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 each other, so they are independent and span a two-dimensional space.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Claim: the sum is just the union of U and W collected together.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The union of the two axes is only the cross shape, but the sum is the entire plane, which is far larger.
The sum contains the union but also adds every cross term.
Why: The union of the two axes is only the cross shape, but the sum is the entire plane, which is far larger.
Trap
Claim: the sum is just the union of U and W collected together.
Count the axes example
Why: The union of the two axes is only the cross shape, but the sum is the entire plane, which is far larger.
\( U \cup W \subsetneq U + W = \mathbb{R}^2 \)
The sum contains the union but also adds every cross term.
Every diagonal appears in the sum
Why: The point one-one, absent from the union, equals one-zero plus zero-one, so it lives in the sum.
\( (1,1) = (1,0)+(0,1) \in U + W \)
Break the constraint
Discussion prompt
The rule this trap just fixed:
The point one-one, absent from the union, equals one-zero plus zero-one, so it lives in the sum.
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:
The union of the two axes is only the cross shape, but the sum is the entire plane, which is far larger.
Commit first
Predict first
In three-space, with U the span of one-zero-zero and W the span of zero-one-zero, what is the sum of U and W?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: The plane of all points with third coordinate zero
Why: The sum is spanned by one-zero-zero and zero-one-zero, two independent vectors, so it is the two-dimensional plane of points whose third coordinate is zero.
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
In three-space, let U be the span of one-zero-zero and W the span of zero-one-zero.
Check your understanding
In three-space, with U the span of one-zero-zero and W the span of zero-one-zero, what is the sum of U and W?
Answer: A
Why: The sum is spanned by one-zero-zero and zero-one-zero, two independent vectors, so it is the two-dimensional plane of points whose third coordinate is zero.
Concept
A sum is called direct when the two subspaces meet only at the origin.
\( U \oplus W \ \text{means} \ U + W \ \text{with} \ U \cap W = \{\mathbf{0}\} \)
direct sum — A sum U + W in which the intersection of U and W contains only the zero vector. It is written with the circled-plus symbol.
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 subspace, trivial subspaces, sum of subspaces, direct sum as Subspaces, Sums & Direct Sums uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Intuition
If the two pieces shared a nonzero vector, that vector could be described using either piece, creating ambiguity. A trivial intersection removes the ambiguity.
Direct means each vector of the sum has one and only one way to be split into a U part and a W part.
Concept
Here is the central theorem, stated as an equivalence.
\( U \cap W = \{\mathbf{0}\} \iff \text{every } \mathbf{v} \in U+W \text{ splits uniquely as } \mathbf{u}+\mathbf{w} \)
Trivial intersection and unique decomposition are two names for the same condition.
Intuition
Once each vector splits uniquely, the U part and the W part behave like a pair of coordinates. This is the abstract version of writing a plane vector as its horizontal part plus its vertical part.
Ranking
Put in order
Put the moves of The plane is the direct sum of its axes 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. Any point splits as its horizontal part plus its vertical part, so the sum is the whole plane.
Worked example
Show the plane is the direct sum of the horizontal and vertical axes.
\( U = \{(t,0)\},\quad W = \{(0,s)\} \)
The sum is everything
Why: Any point splits as its horizontal part plus its vertical part, so the sum is the whole plane.
\( (x,y) = (x,0)+(0,y) \)
The intersection is trivial
Why: A point on both axes has second coordinate zero (on U) and first coordinate zero (on W), so it is the origin.
\( U \cap W = \{(0,0)\} \)
The decomposition is unique
Why: Matching coordinates in two candidate splittings forces them to agree.
\( (x,0)+(0,y) = (a,0)+(0,b) \ \Rightarrow\ x=a,\ y=b \)
Verify both defining conditions
Why: The sum is all of the plane and the intersection is only the origin, so the sum is direct; uniqueness followed automatically.
\( \mathbb{R}^2 = U \oplus W \)
Picture it
Animation
Shows: Each line of the worked example "The plane is the direct sum of its axes", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The sum is all of the plane and the intersection is only the origin, so the sum is direct; uniqueness followed automatically.
Estimation
Predict first
Prove one direction of the theorem: a trivial intersection makes every splitting unique.
Commit before you compute: what does Trivial intersection forces unique decomposition 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 uniqueness
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. Both differences are the zero vector, so the two U parts agree and the two W parts agree; the decompositions were identical.
Worked example
Prove one direction of the theorem: a trivial intersection makes every splitting unique.
\( U \cap W = \{\mathbf{0}\} \)
Assume two decompositions
Why: Suppose a vector equals one U-plus-W splitting and also a second one.
\( \mathbf{u}_1+\mathbf{w}_1 = \mathbf{u}_2+\mathbf{w}_2 \)
Rearrange onto each side
Why: Move the U terms to the left and the W terms to the right; the left side lies in U and the right side lies in W, yet they are equal.
\( \mathbf{u}_1-\mathbf{u}_2 = \mathbf{w}_2-\mathbf{w}_1 \)
Land in the intersection
Why: That common vector lies in both U and W, hence in the intersection, which is only the origin.
\( \mathbf{u}_1-\mathbf{u}_2 = \mathbf{w}_2-\mathbf{w}_1 \in U \cap W = \{\mathbf{0}\} \)
Verify uniqueness
Why: Both differences are the zero vector, so the two U parts agree and the two W parts agree; the decompositions were identical.
\( \mathbf{u}_1 = \mathbf{u}_2,\quad \mathbf{w}_1 = \mathbf{w}_2 \)
Picture it
Animation
Shows: Each line of the worked example "Trivial intersection forces unique decomposition", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Both differences are the zero vector, so the two U parts agree and the two W parts agree; the decompositions were identical.
Concept
There are two flavors. Internal: U and W already sit inside one space V and their sum happens to be direct. External: you build a brand-new space from two separate spaces by pairing their vectors.
\( U \oplus W \ (\text{external}) = \{ (\mathbf{u},\mathbf{w}) : \mathbf{u}\in U,\ \mathbf{w}\in W \} \)
The two notions line up: the external construction matches the internal one, up to relabeling, whenever the pieces intersect trivially.
Step zero
Discussion prompt
The external direct sum is a vector space — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Scalars act slotwise
Answer:
Worked example
Build the external direct sum by pairing vectors and operating slot by slot.
\( (\mathbf{u}_1,\mathbf{w}_1)+(\mathbf{u}_2,\mathbf{w}_2) = (\mathbf{u}_1+\mathbf{u}_2,\ \mathbf{w}_1+\mathbf{w}_2) \)
Scalars act slotwise
Why: Multiply both slots by the scalar; each slot stays inside its own space.
\( c(\mathbf{u},\mathbf{w}) = (c\mathbf{u},\ c\mathbf{w}) \)
Locate copies of U and W
Why: The pairs with second slot zero form a copy of U; those with first slot zero form a copy of W.
\( \tilde U = \{(\mathbf{u},\mathbf{0})\},\quad \tilde W = \{(\mathbf{0},\mathbf{w})\} \)
Verify it realizes an internal direct sum
Why: Every pair splits uniquely as a U-copy plus a W-copy, and the two copies meet only at the zero pair, so the external build is an internal direct sum of those copies.
\( \tilde U \cap \tilde W = \{(\mathbf{0},\mathbf{0})\} \)
Picture it
Animation
Shows: Each line of the worked example "The external direct sum is a vector space", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Every pair splits uniquely as a U-copy plus a W-copy, and the two copies meet only at the zero pair, so the external build is an internal direct sum of those copies.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Claim: since U and W together span the plane, their sum is direct.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: U sits entirely inside W, so their intersection is all of U, not just the origin.
Spanning controls the sum; only a trivial intersection makes it direct.
Why: U sits entirely inside W, so their intersection is all of U, not just the origin.
Trap
Claim: since U and W together span the plane, their sum is direct.
\( U = \operatorname{span}\{(1,0)\},\ W = \operatorname{span}\{(1,0),(0,1)\} \)
Check the intersection
Why: U sits entirely inside W, so their intersection is all of U, not just the origin.
\( U \cap W = U \neq \{\mathbf{0}\} \)
Decomposition is not unique
Why: The vector one-zero can be taken from U or drawn from W, giving two different splittings of the same vector.
Spanning controls the sum; only a trivial intersection makes it direct.
Use genuinely complementary pieces
Why: Replace W by the vertical axis. Now the pieces span the plane and meet only at the origin, so the sum is direct.
\( \operatorname{span}\{(1,0)\} \cap \operatorname{span}\{(0,1)\} = \{\mathbf{0}\} \)
Notation
Annotate
From Trap: spanning is not enough for directness — read this one piece at a time. What is each part doing?
On: \( \operatorname{span}\{(1,0)\} \cap \operatorname{span}\{(0,1)\} = \{\mathbf{0}\} \)
Concept
A subspace W is a complement of U when the whole space is the direct sum of U and W.
\( V = U \oplus W \)
In finite dimensions every subspace has a complement: extend a basis of U to a basis of V, and let the freshly added vectors span W. Complements are generally not unique.
Intuition
Start with a basis of U. Because it is independent, you can keep adding vectors until it becomes a basis of the entire space. The vectors you added span a complement of U.
Different choices of added vectors give different complements — there is no canonical one without extra structure such as an inner product.
Missing information
Discussion prompt
Let V be all real functions on the line, U the even functions, and W the odd functions.
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:
Write any function as its symmetric part plus its antisymmetric part.
Worked example
Let V be all real functions on the line, U the even functions, and W the odd functions.
\( U = \{f : f(-x)=f(x)\},\quad W = \{g : g(-x)=-g(x)\} \)
Every function splits
Why: Write any function as its symmetric part plus its antisymmetric part.
\( f(x) = \underbrace{\tfrac{f(x)+f(-x)}{2}}_{\text{even}} + \underbrace{\tfrac{f(x)-f(-x)}{2}}_{\text{odd}} \)
Each part lands where claimed
Why: Replacing the input by its negative fixes the first part and flips the sign of the second, so they are even and odd respectively.
The intersection is trivial
Why: A function that is both even and odd equals both its reflection and the negative of its reflection, forcing it to be zero everywhere.
\( f(-x)=f(x)=-f(x) \ \Rightarrow\ f \equiv 0 \)
Verify the direct sum
Why: The sum reaches every function and the intersection is only the zero function, so the space of functions is the direct sum of even and odd parts.
\( V = U \oplus W \)
Picture it
Animation
Shows: Each line of the worked example "Even plus odd is a direct sum", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The sum reaches every function and the intersection is only the zero function, so the space of functions is the direct sum of even and odd parts.
Prediction
Predict first
Is three-space the direct sum of U (the plane with third coordinate zero) and W (the span of zero-one-one)?
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: Yes: the intersection is only the origin and the dimensions add to three
Why: The only multiple of zero-one-one lying in the plane is the origin, so the intersection is trivial; with dimensions two and one summing to three, the direct sum fills all of three-space.
Check
In three-space let U be the plane of points with third coordinate zero, and W the span of zero-one-one.
Check your understanding
Is three-space the direct sum of U (the plane with third coordinate zero) and W (the span of zero-one-one)?
Answer: A
Why: The only multiple of zero-one-one lying in the plane is the origin, so the intersection is trivial; with dimensions two and one summing to three, the direct sum fills all of three-space.
Concept
The dimensions of a sum obey an inclusion-exclusion law, sometimes named after Grassmann.
\( \dim(U+W) = \dim U + \dim W - \dim(U \cap W) \)
The overlap is subtracted once so it is not counted twice. The full proof, by extending a basis of the intersection, appears in the basis-and-dimension deck.
Intuition
It mirrors the counting rule for finite sets: the size of a union is the sum of the sizes minus the size of the overlap. Here 'size' is dimension and 'union' becomes the sum.
Estimation
Predict first
Take two distinct planes through the origin in three-space.
Commit before you compute: what does Two planes through the origin in 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 with a concrete pair
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 plane with third coordinate zero and the plane with second coordinate zero meet exactly in the first axis, a line, confirming the intersection has dimension one.
Worked example
Take two distinct planes through the origin in three-space.
\( \dim U = \dim W = 2,\quad U \neq W \)
The sum fills space
Why: Two distinct planes together contain three independent directions, so their sum is all of three-space.
\( \dim(U+W) = 3 \)
Apply the formula
Why: Rearrange the dimension formula to solve for the dimension of the intersection.
\( \dim(U\cap W) = \dim U + \dim W - \dim(U+W) = 2+2-3 = 1 \)
Interpret the answer
Why: A one-dimensional intersection is a line: two distinct planes through the origin meet exactly in a line.
Verify with a concrete pair
Why: The plane with third coordinate zero and the plane with second coordinate zero meet exactly in the first axis, a line, confirming the intersection has dimension one.
\( \{z=0\} \cap \{y=0\} = \{(t,0,0)\} \)
Picture it
Animation
Shows: Each line of the worked example "Two planes through the origin in space", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The plane with third coordinate zero and the plane with second coordinate zero meet exactly in the first axis, a line, confirming the intersection has dimension one.
Concept
When the sum is direct the overlap term vanishes, so the dimensions simply add.
\( V = U \oplus W \ \Rightarrow\ \dim V = \dim U + \dim W \)
This is the dimension bookkeeping behind splitting a space into independent pieces.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Claim: for two planes in three-space the sum has dimension two plus two, hence four.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: No subspace of three-space can have dimension four; the naive addition ignored the shared line.
Subtract the dimension of the intersection.
Why: No subspace of three-space can have dimension four; the naive addition ignored the shared line.
Trap
Claim: for two planes in three-space the sum has dimension two plus two, hence four.
\( \dim(U+W) \stackrel{?}{=} 2 + 2 = 4 \)
Impossible in three-space
Why: No subspace of three-space can have dimension four; the naive addition ignored the shared line.
\( \dim(U+W) \le 3 \)
Subtract the dimension of the intersection.
Use the full formula
Why: With a one-dimensional overlap the sum has dimension two plus two minus one, which is three — consistent with three-space.
\( \dim(U+W) = 2 + 2 - 1 = 3 \)
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.
Step zero
Discussion prompt
Two two-dimensional subspaces of space must meet — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Bound the sum
Answer:
Worked example
Prove any two two-dimensional subspaces of three-space share a nonzero vector.
\( \dim U = \dim W = 2 \ \text{in}\ \mathbb{R}^3 \)
Bound the sum
Why: The sum is a subspace of three-space, so its dimension is at most three.
\( \dim(U+W) \le 3 \)
Solve for the intersection
Why: Rearranged, the intersection dimension is at least two plus two minus three.
\( \dim(U \cap W) = 2 + 2 - \dim(U+W) \ge 4 - 3 = 1 \)
Conclude
Why: A dimension of at least one means the intersection contains a nonzero vector.
\( \dim(U \cap W) \ge 1 \)
Verify the bound is tight
Why: Two distinct planes meet in a line of dimension one and two equal planes meet in a plane of dimension two, so the bound of at least one is achieved and cannot be improved.
Picture it
Animation
Shows: Each line of the worked example "Two two-dimensional subspaces of space must meet", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Two distinct planes meet in a line of dimension one and two equal planes meet in a plane of dimension two, so the bound of at least one is achieved and cannot be improved.
Check
In four-space, suppose U has dimension three, W has dimension two, and their intersection has dimension one.
Check your understanding
In four-space, U has dimension three, W has dimension two, and their intersection has dimension one. What is the dimension of the sum?
Answer: A
Why: By the dimension formula the sum has dimension three plus two minus one, which is four, so the sum is all of four-space.
Ranking
Put in order
Put the moves of Deciding a direct-sum decomposition 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. A vector in both has third coordinate zero (in U) and first two coordinates zero (in W), so only the origin qualifies.
Worked example
Let U be the plane with third coordinate zero and W the vertical axis in three-space.
\( U = \{(x,y,0)\},\quad W = \{(0,0,z)\} \)
The intersection
Why: A vector in both has third coordinate zero (in U) and first two coordinates zero (in W), so only the origin qualifies.
\( U \cap W = \{\mathbf{0}\} \)
The dimensions add correctly
Why: Two plus one equals three, matching the dimension of the whole space.
\( \dim U + \dim W = 2 + 1 = 3 = \dim \mathbb{R}^3 \)
The sum is everything
Why: A trivial intersection together with the matching total dimension forces the sum to be the whole space.
\( U + W = \mathbb{R}^3 \)
Verify a decomposition
Why: The point two-five-seven splits uniquely as two-five-zero plus zero-zero-seven, confirming the direct sum.
\( (2,5,7) = (2,5,0)+(0,0,7) \)
Picture it
Animation
Shows: Each line of the worked example "Deciding a direct-sum decomposition", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The point two-five-seven splits uniquely as two-five-zero plus zero-zero-seven, confirming the direct sum.
Concept
Sums extend to any finite list of subspaces: collect one vector from each piece and add them all.
\( U_1 + U_2 + \cdots + U_k = \{ \mathbf{u}_1 + \cdots + \mathbf{u}_k : \mathbf{u}_i \in U_i \} \)
The sum is direct when every vector splits uniquely, equivalently when each piece meets the sum of all the others only at the origin.
Intuition
For three or more subspaces, checking that each pair meets trivially is strictly weaker than genuine independence.
The correct test is that each subspace avoids the sum of all the others except at the origin. This is the same subtlety as pairwise-independent versus jointly-independent vectors.
Explain it
Discussion prompt
Explain Pairwise trivial is not enough for many pieces 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:
For three or more subspaces, checking that each pair meets trivially is strictly weaker than genuine independence.
Pattern
Subspace?
Why: Confirm the origin is present and both closures hold, or exhibit a failing sum or scalar multiple.
Sum?
Why: Pool spanning sets; the span of the pooled list is the sum. Its dimension comes from the dimension formula.
Direct?
Why: Confirm the intersection is only the origin, equivalently that every vector decomposes uniquely, equivalently that the dimensions add.
Complement?
Why: To split the whole space as a direct sum, extend a basis of U to a basis of the space; the added vectors span a complement.
Elimination
Eliminate the wrong options
For finite-dimensional U and W, which condition is NOT equivalent to the sum U + W being direct?
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: Directness is captured by trivial intersection, unique decomposition, or additive dimension; spanning V is a separate condition about the sum being all of V and says nothing about whether the sum is direct.
Check
Let U and W be finite-dimensional subspaces. Three of these characterize a direct sum; one does not.
Check your understanding
For finite-dimensional U and W, which condition is NOT equivalent to the sum U + W being direct?
Answer: A
Why: Directness is captured by trivial intersection, unique decomposition, or additive dimension; spanning V is a separate condition about the sum being all of V and says nothing about whether the sum is direct.
Concept
Decomposing a space as a direct sum is how you separate independent degrees of freedom: symmetric and antisymmetric parts of a signal, even and odd components, or the independent registers of a machine state.
Each summand can be processed on its own and reassembled without loss, precisely because the decomposition is unique.
Analogy
Discussion prompt
Explain CS tie: splitting a state space 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:
Each summand can be processed on its own and reassembled without loss, precisely because the decomposition is unique.
Intuition
A direct-sum decomposition is a lossless split: no information is duplicated across the pieces and none is lost, exactly because the intersection is trivial while the pieces together span.
Counterexample
Discussion prompt
A direct-sum decomposition is a lossless split: no information is duplicated across the pieces and none is lost, exactly because the intersection is trivial while the pieces together span.
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
Test 1: intersection
Why: Show the intersection of U and W is only the origin.
Test 2: uniqueness
Why: Show that a repeated decomposition forces the U parts and the W parts to coincide.
Test 3: dimension
Why: Show the dimension of U plus the dimension of W equals the dimension of the sum; in finite dimensions this is equivalent to the other two tests.
Real world
Discussion prompt
Outside this lesson: where does Subspaces, Sums & Direct Sums 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 Three equivalent tests for a direct sum 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 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.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Recipe: is a set W a subspace? · Recipe: describe a sum by spanning sets · Master decision recipe · Three equivalent tests for a direct sum · A subspace is a vector space living inside another. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
A subspace is a subset that contains the origin and is closed under addition and scalars — a flat through the origin. Intersections are always subspaces; unions almost never are.
The sum U + W is the smallest subspace containing both, built from one-from-each additions. It is direct when the pieces meet only at the origin, equivalently when every vector splits uniquely, equivalently when the dimensions add.
| You want to know | Use |
|---|---|
| Is W a subspace? | Origin present, closed under + and scalars |
| Is a union a subspace? | Only if one piece contains the other |
| What is U + W? | Span of the pooled spanning sets |
| Is the sum direct? | Intersection is only the origin |
| Dimension of U + W? | dim U + dim W minus dim of the intersection |
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