Linear-Algebra-Done-Right · Unit 2 · Video 3 · Interactive Practice

Sums of Subspaces: The Smallest Subspace Containing Them

IKey Formulas

Formula	Name	Key fact
$V_1 + \cdots + V_m = \{v_1 + \cdots + v_m : v_1 \in V_1, \dots, v_m \in V_m\}$	Sum of subspaces	All possible sums, one vector from each $V_k$
$V_k \subseteq V_1 + \cdots + V_m \subseteq U$	Smallest subspace containing $V_1, \dots, V_m$	Holds for every subspace $U$ of $V$ that contains $V_1, \dots, V_m$
$\{(x_1, x_2, x_3, x_4) \in \mathbf{F}^4 : x_3 = 5x_4 + b\}$	Subspace test with a constant	A subspace $\iff b = 0$ , because $0$ must be in it
$\{0\}$ , lines through $0$ , $\mathbf{R}^2$	Subspaces of $\mathbf{R}^2$	A line missing the origin is not a subspace

Key Insight: Sums of subspaces play the role that unions play for subsets: the smallest subspace containing $U$ and $W$ is $U + W$ , while $U \cup W$ is rarely a subspace at all.

IIThe Subspace Test with a Condition $f'(2) = b$

The differentiable $f$ on $(0, 3)$ with $f'(2) = b$ pass all three subspace conditions only when $b = 0$ .

b 1/2 Condition λ 2

IIIUnions Fail, Sums Work

The $x$ -axis $U$ and the $y$ -axis $W$ are subspaces of $\mathbf{R}^2$ , yet $U \cup W$ is not closed under addition.

W =

IVComputing $U + W$ in $\mathbf{F}^4$

With $U = \{(x, x, y, y)\}$ and $W = \{(x, x, x, y)\}$ , a list lies in $U + W$ exactly when its first two coordinates agree.

shift t 0

💡 The split is not unique: $(t, t, t, t)$ lies in both $U$ and $W$ , so it can move from one summand to the other. Sums in which every split is unique are the next video's direct sums.

VQuiz Questions

Problem 1 · A Sum in F³

Given: $U = \{(x, 0, 0) \in \mathbf{F}^3 : x \in \mathbf{F}\}$ and $W = \{(0, y, y) \in \mathbf{F}^3 : y \in \mathbf{F}\}$ — find $U + W$ .

✅ Correct!

(x, 0, 0) + (0, y, y) = (x, y, y)

, and every

(x, y, y)

splits this way.

❌ Not quite. That is the video's example, where

W

was the second axis. Here

W

's elements have equal second and third coordinates, so

(x, 0, 0) + (0, y, y) = (x, y, y)

❌ Not quite.

(1, 0, 0) \in U

and

(0, 1, 1) \in W

, but

(1, 1, 1)

lies in neither, so

U \cup W

misses elements of

U + W

❌ Not quite. Every element of

U + W

has the form

(x, y, y)

, so

(0, 0, 1)

is not in

U + W

❌ Not quite. Add a typical element of each:

(x, 0, 0) + (0, y, y)

Show solution

$\subseteq$ : a typical element of $U + W$ is

(x, 0, 0) + (0, y, y) = (x, y, y),

whose second and third coordinates are equal.

$\supseteq$ : any $(x, y, y)$ splits as $(x, 0, 0) + (0, y, y)$ with $(x, 0, 0) \in U$ and $(0, y, y) \in W$ .

So $U + W = \{(x, y, y) : x, y \in \mathbf{F}\}$ . It is not all of $\mathbf{F}^3$ : for instance $(0, 0, 1)$ has unequal second and third coordinates.

Problem 2 · Which Subset Is a Subspace?

Given: four subsets of $\mathbf{R}^2$ — which one is a subspace of $\mathbf{R}^2$ ?

✅ Correct!

y = -3x

is a line through the origin, one of the three kinds of subspace of

\mathbf{R}^2

❌ Not quite. This is the union of the two axes:

(1, 0)

and

(0, 1)

are in it, but their sum

(1, 1)

is not.

❌ Not quite. At

x = 0

the line has

y = 1

, so

(0, 0)

is not on it and the first condition fails.

❌ Not quite. This set is closed under scalar multiplication, but

(2, 1)

and

(-1, -2)

are in it while their sum

(1, -1)

has

xy = -1 < 0

❌ Not quite. Check

0

, sums and scalar multiples for each set.

Show solution

For $\{(x, y) : y = -3x\}$ : $(0, 0)$ satisfies $0 = -3 \cdot 0$ . If $y_1 = -3x_1$ and $y_2 = -3x_2$ , then $y_1 + y_2 = -3(x_1 + x_2)$ ; and $\lambda y_1 = -3(\lambda x_1)$ . All three conditions hold.

The others fail:

Union of the axes: $(1, 0) + (0, 1) = (1, 1)$ lies on neither axis.
$y = 2x + 1$ : the zero vector would need $0 = 2 \cdot 0 + 1$ .
$xy \ge 0$ : $(2, 1) + (-1, -2) = (1, -1)$ , and $1 \cdot (-1) = -1 < 0$ .

Problem 3 · A Sum in F⁴, Both Inclusions

Given: $U = \{(x, y, y, y) \in \mathbf{F}^4 : x, y \in \mathbf{F}\}$ and $W = \{(x, x, x, y) \in \mathbf{F}^4 : x, y \in \mathbf{F}\}$ — find $U + W$ , and split $(3, 1, 1, 5)$ as $u + w$ with $u \in U$ , $w \in W$ .

What is U + W?

Which is a valid split of (3, 1, 1, 5)?

✅ Correct! Every sum

(a + c, b + c, b + c, b + d)

has equal second and third coordinates, and

(3, 1, 1, 5) = (3, 1, 1, 1) + (0, 0, 0, 4)

❌ Check U + W. That is the video's answer for different subspaces. Here a typical sum is

(a, b, b, b) + (c, c, c, d) = (a + c, b + c, b + c, b + d)

: the second and third coordinates agree.

❌ Check U + W.

(1, 0, 1, 0)

is not a sum: every element of

U + W

has equal second and third coordinates.

❌ Check U + W.

W

is not inside

U

(1, 1, 1, 0) \in W

but its last three coordinates are not equal, so

U + W

is larger than

U

❌ Check U + W. Add a typical element of each and compare coordinates.

❌ Check the split.

(3, 1, 1, 5)

is not in

U

: its last three coordinates

1, 1, 5

are not all equal.

❌ Check the split.

(0, 1, 1, 5)

is not in

W

: its first three coordinates

0, 1, 1

are not all equal.

❌ Check the split. Both summands have the right shape, but

(2, 1, 1, 1) + (1, 1, 1, 4) = (3, 2, 2, 5)

❌ Check the split. The first summand needs equal last three coordinates; the second needs equal first three.

Show solution

$\subseteq$ : take $(a, b, b, b) \in U$ and $(c, c, c, d) \in W$ :

(a, b, b, b) + (c, c, c, d) = (a + c,\ b + c,\ b + c,\ b + d).

The second and third coordinates are equal, so $U + W \subseteq \{(x, y, y, z) : x, y, z \in \mathbf{F}\}$ .

$\supseteq$ : for any $x, y, z \in \mathbf{F}$ ,

(x, y, y, z) = (x, y, y, y) + (0, 0, 0, z - y).

The first summand is in $U$ (last three coordinates equal) and the second is in $W$ (first three coordinates equal, all $0$ ). So $U + W = \{(x, y, y, z) : x, y, z \in \mathbf{F}\}$ .

The split: with $x = 3$ , $y = 1$ , $z = 5$ the formula gives

(3, 1, 1, 5) = (3, 1, 1, 1) + (0, 0, 0, 4).

Problem 4 · The Smallest Subspace Containing Two Lines

Given: in $\mathbf{R}^3$ , $V_1 = \{(t, -t, 0) : t \in \mathbf{R}\}$ , $V_2 = \{(0, s, -s) : s \in \mathbf{R}\}$ , and the plane $P = \{(x, y, z) : x + y + z = 0\}$ , a subspace containing both $V_1$ and $V_2$ — which statement is true?

✅ Correct! The theorem gives

V_1 + V_2 \subseteq P

, and every point of

P

splits as

(x, -x, 0) + (0, -z, z)

, so the two sets are equal.

❌ Not quite.

(1, -1, 0) + (0, 1, -1) = (1, 0, -1)

lies on neither line, so

V_1 + V_2

is bigger than the union.

❌ Not quite.

P

is a subspace containing

V_1

and

V_2

, so the smallest such subspace satisfies

V_1 + V_2 \subseteq P

; and

(1, 0, 0) \notin P

❌ Not quite. The inclusion

V_1 + V_2 \subseteq P

is right, but it is not strict: any

(x, y, z)

with

x + y + z = 0

equals

(x, -x, 0) + (0, -z, z)

❌ Not quite. Compute a typical sum

(t, -t, 0) + (0, s, -s)

Show solution

$\subseteq$ : $P$ is a subspace of $\mathbf{R}^3$ containing $V_1$ and $V_2$ (their elements have coordinate sum $0$ ). Since $V_1 + V_2$ is the smallest subspace containing $V_1$ and $V_2$ , $V_1 + V_2 \subseteq P$ . Directly:

(t, -t, 0) + (0, s, -s) = (t,\ s - t,\ -s), \qquad t + (s - t) + (-s) = 0.

$\supseteq$ : let $x + y + z = 0$ , so $y = -x - z$ . Take $t = x$ and $s = -z$ :

(x, -x, 0) + (0, -z, z) = (x,\ -x - z,\ z) = (x, y, z).

So $V_1 + V_2 = P$ : two distinct lines through the origin add up to the plane containing them, not to all of $\mathbf{R}^3$ (for example $(1, 0, 0) \notin P$ ).

Solved: 0 / 4

Sums of Subspaces: The Smallest Subspace Containing Them

IKey Formulas

IIThe Subspace Test with a Condition f′(2)=bf'(2) = bf′(2)=b

IIIUnions Fail, Sums Work

IVComputing U+WU + WU+W in F4\mathbf{F}^4F4

VQuiz Questions

IIThe Subspace Test with a Condition $f'(2) = b$

IVComputing $U + W$ in $\mathbf{F}^4$