Linear-Algebra-Done-Right · Unit 2 · Video 4 · Interactive Practice

Direct Sums: Only One Way to Write Zero

IKey Formulas

Statement	Name	Meaning
$V_1 \oplus \cdots \oplus V_m$	Direct sum	$V_1 + \cdots + V_m$ when each of its elements is $v_1 + \cdots + v_m$ , with $v_k \in V_k$ , in only one way
$0 = v_1 + \cdots + v_m,\ v_k \in V_k \;\Rightarrow\; v_1 = \cdots = v_m = 0$	Condition for a direct sum	The sum is direct $\iff$ $0$ has only the trivial representation
$U + W$ direct $\iff U \cap W = \{0\}$	Two subspaces	A test for two subspaces only: for three or more, pairwise intersections $\{0\}$ do not make the sum direct
$\mathbf{F}^3 = \{(x, y, 0)\} \oplus \{(0, 0, z)\}$	Example	$(x, y, z) = (x, y, 0) + (0, 0, z)$ , and no other split

Key Insight: Uniqueness for every vector reduces to uniqueness for one vector: if $v = v_1 + \cdots + v_m = u_1 + \cdots + u_m$ , subtracting gives $0 = (v_1 - u_1) + \cdots + (v_m - u_m)$ with each $v_k - u_k \in V_k$ .

IITwo Lines Through the Origin

Two different lines through $0$ meet only at $0$ , so every $v \in \mathbf{R}^2$ splits along them in exactly one way.

IIITesting Only Zero

A sum is direct exactly when $0 = v_1 + v_2 + v_3$ forces every piece $v_k$ to be $0$ .

Sum v₁ = (a, b, 0): a 1 b 2 v₂ = (0, 0, c): c 1 v₃ = (0, d, d): d 0

IVPairwise Intersections Are Not Enough

Every pair among $V_1, V_2, V_3$ meets only at $0$ , yet $V_1 + V_2 + V_3$ is not direct.

Step 1

Step 1 — $V_1 \cap V_2$

(x, y, 0) = (0, 0, z) \Rightarrow x = y = z = 0

, so

V_1 \cap V_2 = \{0\}

✓

Step 2 — $V_1 \cap V_3$

(0, y, y)

with last coordinate

0

has

y = 0

, so

V_1 \cap V_3 = \{0\}

✓

Step 3 — $V_2 \cap V_3$

(0, y, y)

with middle coordinate

0

has

y = 0

, so

V_2 \cap V_3 = \{0\}

✓

Step 4 — an overlap with $V_1 + V_2$

(0, 1, 0) + (0, 0, 1) = (0, 1, 1) \in V_3

, a nonzero vector of

V_3 \cap (V_1 + V_2)

, so

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

is a second representation of

0

: not a direct sum.

VQuiz Questions

Problem 1 · Split a Vector Along Two Lines

Given: $U = \{(3t, t) : t \in \mathbf{R}\}$ and $W = \{(s, 2s) : s \in \mathbf{R}\}$ , so $\mathbf{R}^2 = U \oplus W$ . Write $(7, 9) = u + w$ with $u \in U$ and $w \in W$ . Find $u$ .

✅ Correct!

u = (3, 1)

and

w = (4, 8)

(3, 1) + (4, 8) = (7, 9)

, and since

U \cap W = \{0\}

this is the only split.

❌ Check which coefficient goes with which line.

(12, 4) = 4(3, 1)

uses the

W

-coefficient

s = 4

U

; then

w = 1 \cdot (1, 2)

and

(12, 4) + (1, 2) = (13, 6) \ne (7, 9)

❌ Not quite. That is the split along the coordinate axes.

(7, 0)

is not in

U

: a vector of

U

has first coordinate

3

times its second.

❌ Not quite.

(6, 2)

is in

U

, but then

w = (7, 9) - (6, 2) = (1, 7)

, and

(1, 7) \notin W

because

7 \ne 2 \cdot 1

❌ Not quite. Write

u = (3t, t)

w = (s, 2s)

and solve

(3t + s,\ t + 2s) = (7, 9)

Show solution

Write $u = (3t, t)$ and $w = (s, 2s)$ . Then $u + w = (7, 9)$ means

3t + s = 7, \qquad t + 2s = 9.

From the first equation $s = 7 - 3t$ ; substituting, $t + 14 - 6t = 9$ , so $t = 1$ and $s = 4$ . Hence

u = (3, 1), \qquad w = (4, 8), \qquad (3, 1) + (4, 8) = (7, 9).

The system has exactly one solution, matching the fact that $U \cap W = \{0\}$ makes the sum direct.

Problem 2 · Three Lines in a Plane

Given: in $\mathbf{R}^3$ , $U_1 = \{(x, 0, 0)\}$ , $U_2 = \{(0, y, 0)\}$ , $U_3 = \{(z, z, 0)\}$ , where $x, y, z \in \mathbf{R}$ . Each pairwise intersection is $\{0\}$ . Which statement is correct?

✅ Correct! The pieces lie in

U_1

U_2

U_3

(take

z = -1

) and are not all

0

, so

0

has a second representation.

❌ That is the pitfall. The intersection test

U \cap W = \{0\}

covers exactly two subspaces. Here

(1, 1, 0) \in U_3

also lies in

U_1 + U_2

, which gives a nontrivial way to write

0

❌ Not quite.

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

forces

x = y = 0

, so

U_1 \cap U_2 = \{0\}

. The failure needs all three subspaces at once.

❌ Not quite. A direct sum need not fill the whole space:

U_1 \oplus U_2 = \{(x, y, 0)\} \ne \mathbf{R}^3

is direct. Directness is about uniqueness, so test the zero vector.

❌ Not quite. Look for a way to write

0

with one piece from each

U_k

, not all pieces

0

Show solution

Pairs. $U_1 \cap U_2$ : $(x, 0, 0) = (0, y, 0)$ gives $x = y = 0$ . $U_1 \cap U_3$ : $(z, z, 0)$ with second coordinate $0$ has $z = 0$ . $U_2 \cap U_3$ : $(z, z, 0)$ with first coordinate $0$ has $z = 0$ .

All three. $(1, 0, 0) \in U_1$ , $(0, 1, 0) \in U_2$ and $(-1, -1, 0) \in U_3$ (with $z = -1$ ), and

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

Together with $0 = 0 + 0 + 0$ this gives two representations of $0$ , so by the condition for a direct sum, $U_1 + U_2 + U_3$ is not direct.

Problem 3 · When Does a Line Overlap a Plane?

Given: in $\mathbf{F}^3$ , $U = \{(x, y, z) : x + y + z = 0\}$ and $W = \{(t, 2t, kt) : t \in \mathbf{F}\}$ for a fixed $k \in \mathbf{F}$ .

For which k is U + W not a direct sum?

For that k, a second representation of 0

✅ Correct! For

k = -3

the line

W

lies inside the plane

U

, so

U \cap W = W \ne \{0\}

, and

(1, 2, -3) \in U

with

(-1, -2, 3) \in W

add to

0

❌ Check the sign.

(t, 2t, kt) \in U

means

t + 2t + kt = 0

, i.e.

(3 + k)t = 0

, which has a solution

t \ne 0

when

k = -3

❌ Not quite. A line can lie inside a plane. Find when

(t, 2t, kt)

satisfies

x + y + z = 0

for some

t \ne 0

❌ Check the value of k.

U + W

fails to be direct exactly when

U \cap W \ne \{0\}

: solve

t + 2t + kt = 0

with

t \ne 0

❌ That is the trivial representation. Every sum has it; a failure of directness needs a representation with a nonzero piece.

❌ Not quite.

(2, 1, -3)

is in

U

, but

(-2, -1, 3)

is not in

W

: a vector of

W

has second coordinate twice its first, and

-1 \ne 2 \cdot (-2)

❌ Not quite. Use a nonzero vector of

U \cap W

and its negative.

Show solution

By the two-subspace test, $U + W$ is direct $\iff U \cap W = \{0\}$ . A vector $(t, 2t, kt)$ of $W$ lies in $U$ exactly when

t + 2t + kt = (3 + k)\,t = 0.

If $k \ne -3$ this forces $t = 0$ , so $U \cap W = \{0\}$ and the sum is direct. If $k = -3$ , every $t$ works: $(1, 2, -3) \in U \cap W$ .

Then $(1, 2, -3) \in U$ and $-(1, 2, -3) = (-1, -2, 3) \in W$ (take $t = -1$ ), and

(1, 2, -3) + (-1, -2, 3) = (0, 0, 0),

a representation of $0$ with nonzero pieces, so $U + W$ is not direct.

Problem 4 · Even Plus Odd Polynomials

Given: in $\mathcal{P}(\mathbf{R})$ , $U = \{p : p(-x) = p(x)\}$ (even) and $W = \{p : p(-x) = -p(x)\}$ (odd), and $\mathcal{P}(\mathbf{R}) = U \oplus W$ . Let $p(x) = 2 + 3x - x^2 + 5x^3$ .

The piece of p in U

Why is this split the only one?

✅ Correct!

p = (2 - x^2) + (3x + 5x^3)

, an even plus an odd polynomial, and

U \cap W = \{0\}

makes this the only such split.

❌ That is the other piece.

3x + 5x^3

changes sign when

x \mapsto -x

, so it lies in

W

❌ Not quite. That piece contains

3x

, and

3(-x) \ne 3x

, so it is not even. Collect the even powers of

x

❌ Check the even piece. It is

\tfrac{1}{2}\big(p(x) + p(-x)\big)

❌ Not quite. That shows the split exists. Uniqueness is the separate claim, and for two subspaces it is

U \cap W = \{0\}

❌ Two subspaces are never disjoint. Both contain the zero polynomial; for subspaces, disjointness is replaced by

U \cap W = \{0\}

❌ Not quite. For two subspaces, uniqueness is equivalent to

U \cap W = \{0\}

Show solution

The split. Here $p(-x) = 2 - 3x - x^2 - 5x^3$ , so

\tfrac{1}{2}\big(p(x) + p(-x)\big) = 2 - x^2 \in U, \qquad \tfrac{1}{2}\big(p(x) - p(-x)\big) = 3x + 5x^3 \in W,

and these two add to $p$ .

Uniqueness. If $q \in U \cap W$ , then $q(x) = q(-x) = -q(x)$ , so $2q(x) = 0$ and $q = 0$ . Thus $U \cap W = \{0\}$ , and by the two-subspace test $U + W$ is direct: $2 - x^2$ is the only even piece of $p$ .

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