Linear-Algebra-Done-Right · Unit 4 · Video 3 · Interactive Practice

The Dimension of a Sum of Two Subspaces

IKey Formulas

Statement	Name	Role
$\dim(V_1 + V_2) = \dim V_1 + \dim V_2 - \dim(V_1 \cap V_2)$	Dimension of a sum	$V_1, V_2$ subspaces of a finite-dimensional vector space
$v_1, \dots, v_m,\ u_1, \dots, u_j,\ w_1, \dots, w_k$ is a basis of $V_1 + V_2$	The combined list	$v$ 's: basis of $V_1 \cap V_2$ ; add $u$ 's for $V_1$ , $w$ 's for $V_2$
$\#(S_1 \cup S_2) = \#S_1 + \#S_2 - \#(S_1 \cap S_2)$	Counting formula for finite sets	$\#$ plays $\dim$ ; $\cup$ plays $+$
$\dim(V_1 + V_2) = \dim V_1 + \dim V_2 \iff V_1 \cap V_2 = \{0\}$	No overlap	For two subspaces: exactly when $V_1 + V_2$ is direct

Key Insight: The combined list has length $m + j + k = (m + j) + (m + k) - m$ . Adding $\dim V_1 = m + j$ and $\dim V_2 = m + k$ counts the $m$ vectors $v_1, \dots, v_m$ of the intersection twice; subtracting $\dim(V_1 \cap V_2) = m$ removes one copy.

IITwo Subspaces of R³, Counted

Two different planes in $\mathbf{R}^3$ have dimensions adding to $4$ , yet their sum is $\mathbf{R}^3$ : the shared line counts twice.

V₁ V₂

IIIOne Basis of the Intersection, Extended Twice

Extending one basis of $V_1 \cap V_2$ into $V_1$ and into $V_2$ yields a basis of $V_1 + V_2$ .

Step 1

Step 1 · Basis of the intersection

$V_1 = \{(x, y, z) : x + y + z = 0\}$ and $V_2 = \{(x, x, y)\}$ . A vector $(x, x, y)$ lies in $V_1$ exactly when $2x + y = 0$ :

V_1 \cap V_2 = \{(x, x, -2x)\} = \operatorname{span}(v_1),

with $v_1 = (1, 1, -2)$ , so $m = 1$ .

IVThree or More Subspaces

Pairwise disjoint finite sets always have counts that add; lines meeting pairwise only in $0$ need not have dimensions that add.

💡 Challenge: exactly two of the ten triples of these lines fail to form a direct sum, although every pair of distinct lines meets only in $0$ . Find both.

VQuiz Questions

Problem 1 · Apply the Formula

Given: $V_1$ and $V_2$ are subspaces of $\mathbf{F}^7$ with $\dim V_1 = 4$ , $\dim V_2 = 5$ and $\dim(V_1 \cap V_2) = 3$ . Find $\dim(V_1 + V_2)$ .

✅ Correct!

\dim(V_1 + V_2) = 4 + 5 - 3 = 6

: the three basis vectors of

V_1 \cap V_2

are counted once, not twice.

❌ The overlap is counted twice.

4 + 5

counts each of the

3

basis vectors of

V_1 \cap V_2

once in the basis of

V_1

and again in the basis of

V_2

; subtract

\dim(V_1 \cap V_2) = 3

once. (Also,

9 > 7 = \dim \mathbf{F}^7

❌ Not quite. Since

4 + 5 > 7

, the intersection cannot be

\{0\}

, but that does not make

V_1 + V_2

all of

\mathbf{F}^7

. The formula gives

4 + 5 - 3 = 6

❌ Wrong sign. The dimension of the intersection is subtracted, not added. A subspace of

\mathbf{F}^7

has dimension at most

7

, so

12

is impossible.

Show solution

By the dimension-of-a-sum formula,

\dim(V_1 + V_2) = \dim V_1 + \dim V_2 - \dim(V_1 \cap V_2) = 4 + 5 - 3 = 6.

In the proof's terms: a basis $v_1, v_2, v_3$ of $V_1 \cap V_2$ ( $m = 3$ ) extends by $j = 1$ vector to a basis of $V_1$ and by $k = 2$ vectors to a basis of $V_2$ . The combined list $v_1, v_2, v_3, u_1, w_1, w_2$ is a basis of $V_1 + V_2$ of length $m + j + k = 3 + 1 + 2 = 6$ .

Problem 2 · Two Subspaces That Must Meet

Given: $V_1$ and $V_2$ are subspaces of $\mathbf{F}^5$ with $\dim V_1 = \dim V_2 = 3$ . Which statement must be true?

✅ Correct!

\dim(V_1 \cap V_2) = 6 - \dim(V_1 + V_2) \geq 6 - 5 = 1

, and a subspace of dimension at least

1

is not

\{0\}

❌ Not forced. If

V_1 = V_2

, then

V_1 + V_2 = V_1

has dimension

3

, not

5

❌ Impossible here. If

V_1 \cap V_2 = \{0\}

, the dimensions would add:

\dim(V_1 + V_2) = 3 + 3 = 6

. But

V_1 + V_2

is a subspace of

\mathbf{F}^5

, so its dimension is at most

5

❌ $1$ is only the smallest possible value. With

V_1 = \{(x_1, x_2, x_3, 0, 0)\}

and

V_2 = \{(0, x_2, x_3, x_4, 0)\}

the intersection

\{(0, x_2, x_3, 0, 0)\}

has dimension

2

; with

V_1 = V_2

it has dimension

3

Show solution

$V_1 + V_2$ is a subspace of $\mathbf{F}^5$ , so $\dim(V_1 + V_2) \leq 5$ . Rearranging the formula,

\dim(V_1 \cap V_2) = \dim V_1 + \dim V_2 - \dim(V_1 + V_2) \geq 3 + 3 - 5 = 1.

The only subspace of dimension $0$ is $\{0\}$ , so $V_1 \cap V_2$ contains a nonzero vector. All three values $1, 2, 3$ occur:

$V_1 = \{(x_1, x_2, x_3, 0, 0)\}$ , $V_2 = \{(0, 0, x_3, x_4, x_5)\}$ : intersection $\{(0, 0, x_3, 0, 0)\}$ , dimension $1$ , and $V_1 + V_2 = \mathbf{F}^5$ .
$V_1 = \{(x_1, x_2, x_3, 0, 0)\}$ , $V_2 = \{(0, x_2, x_3, x_4, 0)\}$ : dimension $2$ .
$V_1 = V_2$ : dimension $3$ .

So (a) and (c) can fail, and (b) never holds.

Problem 3 · Two Subspaces of F⁴

Given: in $\mathbf{F}^4$ , let $V_1 = \{(a, b, c, d) : a + b = 0\}$ and $V_2 = \{(0, b, c, c) : b, c \in \mathbf{F}\}$ . Find $\dim(V_1 \cap V_2)$ and $\dim(V_1 + V_2)$ .

Dimension of the intersection

Dimension of the sum

✅ Correct!

V_1 \cap V_2 = \{(0, 0, c, c)\}

has dimension

1

, so

\dim(V_1 + V_2) = 3 + 2 - 1 = 4

and

V_1 + V_2 = \mathbf{F}^4

❌ Not quite.

(0, 0, 1, 1)

lies in both: it has the form

(0, b, c, c)

, and

0 + 0 = 0

puts it in

V_1

. So

V_1 \cap V_2 \neq \{0\}

❌ Too big. A vector

(0, b, c, c)

V_2

lies in

V_1

only when

0 + b = 0

. That leaves the single free scalar

c

V_1 \cap V_2 = \{(0, 0, c, c)\}

❌ Too big.

V_1 \cap V_2

is a subspace of

V_2

, and

\dim V_2 = 2

(basis

(0, 1, 0, 0), (0, 0, 1, 1)

), so the intersection has dimension at most

2

❌ The overlap is counted twice.

3 + 2 = 5

forgets to subtract

\dim(V_1 \cap V_2) = 1

; and no subspace of

\mathbf{F}^4

has dimension

5

❌ Not quite.

\dim(V_1 + V_2) = 3 = \dim V_1

would force

V_1 + V_2 = V_1

, so

V_2 \subseteq V_1

. But

(0, 1, 0, 0) \in V_2

has

0 + 1 \neq 0

. The formula gives

3 + 2 - 1 = 4

❌ Too small.

V_1 + V_2

contains

V_1

, so its dimension is at least

\dim V_1 = 3

. The formula gives

3 + 2 - 1 = 4

Show solution

Dimensions of the pieces. A vector of $V_1$ has $b = -a$ : $(a, -a, c, d) = a(1, -1, 0, 0) + c(0, 0, 1, 0) + d(0, 0, 0, 1)$ , and these three vectors are linearly independent, so $\dim V_1 = 3$ . Similarly $(0, b, c, c) = b(0, 1, 0, 0) + c(0, 0, 1, 1)$ gives $\dim V_2 = 2$ .

Intersection. $(0, b, c, c) \in V_1 \iff 0 + b = 0$ , so

V_1 \cap V_2 = \{(0, 0, c, c)\} = \operatorname{span}(v_1), \quad v_1 = (0, 0, 1, 1), \quad m = 1.

Follow the proof. Extend $v_1$ into $V_1$ with $u_1 = (1, -1, 0, 0)$ , $u_2 = (0, 0, 1, 0)$ (an independent list of length $3 = \dim V_1$ , so a basis), and into $V_2$ with $w_1 = (0, 1, 0, 0)$ . The combined list $v_1, u_1, u_2, w_1$ is a basis of $V_1 + V_2$ , so

\dim(V_1 + V_2) = m + j + k = 1 + 2 + 1 = 4 = 3 + 2 - 1.

A subspace of $\mathbf{F}^4$ of dimension $4$ is $\mathbf{F}^4$ itself, so $V_1 + V_2 = \mathbf{F}^4$ .

Problem 4 · Where the Analogy Breaks

Given: each statement below is the vector-space translation of a true fact about finite sets ( $\cup \mapsto +$ , $\# \mapsto \dim$ , $\emptyset \mapsto \{0\}$ ), for subspaces $V_1, V_2, V_3$ of a finite-dimensional $V$ . Which translation is false?

✅ Correct! In

\mathbf{R}^2

the lines

\operatorname{span}((1, 0))

\operatorname{span}((0, 1))

\operatorname{span}((1, 1))

meet pairwise only in

0

, yet

1 + 1 + 1 = 3 > 2 = \dim \mathbf{R}^2

❌ This one is true. It is the unit 2 result that a sum of subspaces is the smallest subspace containing them, the analogue of

S_1 \cup S_2

being the smallest subset containing

S_1

and

S_2

❌ This one is true. It is the theorem of this video, the analogue of

\#(S_1 \cup S_2) = \#S_1 + \#S_2 - \#(S_1 \cap S_2)

❌ This one is true. By the formula, the dimensions add exactly when

\dim(V_1 \cap V_2) = 0

, and the only subspace of dimension

0

\{0\}

Show solution

Statements (a), (c) and (d) hold: (a) is the unit 2 result on sums, (c) is the dimension formula, and (d) follows from (c) because a subspace has dimension $0$ exactly when it is $\{0\}$ .

Statement (b) fails. Take $V_1 = \operatorname{span}((1, 0))$ , $V_2 = \operatorname{span}((0, 1))$ , $V_3 = \operatorname{span}((1, 1))$ in $\mathbf{R}^2$ . Two distinct lines through the origin meet only in $0$ , so all three pairwise intersections are $\{0\}$ . But $V_1 + V_2 + V_3 \subseteq \mathbf{R}^2$ , so

\dim(V_1 + V_2 + V_3) \leq 2 < 3 = \dim V_1 + \dim V_2 + \dim V_3.

The sum is not direct either: $(1, 0) + (0, 1) + (-1, -1) = 0$ writes $0$ with nonzero pieces. For sets, pairwise disjoint pieces always have counts that add; for three or more subspaces, pairwise intersections $\{0\}$ are not enough. The right analogue, stated in the video: a sum $V_1 + \cdots + V_m$ is direct exactly when the dimensions add, a result proved later.

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