Linear-Algebra-Done-Right · Unit 16 · Video 1 · Interactive Practice

Orthogonal Complements and the Direct Sum V = U ⊕ U⊥

IKey Formulas

Formula	Statement	Hypotheses
$U^\perp = \{ v \in V : \langle u, v \rangle = 0 \text{ for every } u \in U \}$	Orthogonal complement; it is always a subspace of $V$	$U$ any subset of $V$ (one vector, a list, a subspace)
$\{0\}^\perp = V$ , $\ V^\perp = \{0\}$ , $\ U \cap U^\perp \subseteq \{0\}$	The two extremes, and $U$ and $U^\perp$ share at most the zero vector	$U \subseteq V$
$G \subseteq H \Rightarrow H^\perp \subseteq G^\perp$	A larger set imposes more conditions, so its complement is smaller	$G, H \subseteq V$
$V = U \oplus U^\perp$ , $\ v = \underbrace{\textstyle\sum_{k=1}^{m} \langle v, e_k \rangle e_k}_{u \in U} + \underbrace{(v - u)}_{w \in U^\perp}$	Direct sum theorem; when $V$ is finite-dimensional, $\dim U^\perp = \dim V - \dim U$	$U$ a finite-dimensional subspace with orthonormal basis $e_1, \dots, e_m$

Key Insight: In $\mathbf{R}^3$ , $\{(2, 3, 5)\}^\perp$ is the plane $2x + 3y + 5z = 0$ , and the plane's complement is the line $\{(2t, 3t, 5t)\}$ : the line lies in it, and $\dim = 3 - 2 = 1$ forces equality.

IIA Point, a Plane, a Line in $\mathbf{R}^3$

Every $v \in \mathbf{R}^3$ is, in exactly one way, a vector in the plane $2x + 3y + 5z = 0$ plus a multiple of $(2, 3, 5)$ .

x 3 y 1 z 2 view −120°

IIIThe Complement of Any Subset

For any set $U$ of points, $U^\perp$ is a subspace, and $U \cap U^\perp$ is $\{0\}$ or empty.

set

IVComplements in $\mathbf{F}^5$ : Direct Sum and Reversed Inclusions

In $\mathbf{F}^5$ , the complement of a span of standard basis vectors is the span of the remaining ones.

💡 The direct sum $V = U \oplus U^\perp$ needs $U$ finite-dimensional; $V$ itself may be infinite-dimensional, but for an infinite-dimensional $U$ the theorem can fail.

VQuiz Questions

Problem 1 · Membership in a Complement

Given: $V = \mathbf{R}^3$ with the Euclidean inner product and $U = \{(1, -2, 4)\}$ — which vector lies in $U^\perp$ ?

✅ Correct!

\langle (1, -2, 4), (2, 1, 0) \rangle = 2 - 2 + 0 = 0

, so

(2, 1, 0)

lies in the plane

x - 2y + 4z = 0

, which is

U^\perp

❌ Not quite. That is the vector in

U

itself:

\langle (1, -2, 4), (1, -2, 4) \rangle = 1 + 4 + 16 = 21 \ne 0

. A nonzero vector is never orthogonal to itself.

❌ Not quite.

(-1, 2, -4)

points along the same line as

(1, -2, 4)

; its inner product with it is

-1 - 4 - 16 = -21 \ne 0

. That line is the orthogonal complement of

U^\perp

, not part of

U^\perp

❌ Check the sign.

\langle (1, -2, 4), (0, 2, -1) \rangle = 0 - 4 - 4 = -8 \ne 0

. The vector

(0, 2, 1)

would work:

0 - 4 + 4 = 0

Show solution

A vector $(x, y, z)$ is in $U^\perp$ exactly when $\langle (1, -2, 4), (x, y, z) \rangle = x - 2y + 4z = 0$ . Test each option:

(1, -2, 4): \ 1 + 4 + 16 = 21, \qquad (-1, 2, -4): \ -1 - 4 - 16 = -21,

(2, 1, 0): \ 2 - 2 + 0 = 0, \qquad (0, 2, -1): \ 0 - 4 - 4 = -8.

Only $(2, 1, 0)$ satisfies the equation, so it is the vector in $U^\perp$ , the plane through the origin perpendicular to $(1, -2, 4)$ .

Problem 2 · A Set Containing Zero

Given: $V = \mathbf{R}^2$ with the Euclidean inner product and the two-point set $U = \{(1, 0), (0, 0)\}$ — find $U^\perp$ and $U \cap U^\perp$ .

What is U⊥?

What is U ∩ U⊥?

✅ Correct!

U^\perp

is the

y

-axis, and since

0

belongs to

U

and to every complement,

U \cap U^\perp = \{0\}

: the inclusion

U \cap U^\perp \subseteq \{0\}

is an equality here.

❌ Not quite. Two vectors do not give two conditions here:

\langle (0, 0), v \rangle = 0

for every

v

, so the zero vector imposes no condition. Only

v_1 = 0

remains.

❌ Not quite. "Complement" here is not the set complement.

(1, 1) \in \mathbf{R}^2 \setminus U

, but

\langle (1, 0), (1, 1) \rangle = 1 \ne 0

❌ Not quite. That is the span of

U

, the

x

-axis. Its vectors are parallel to

(1, 0)

, not orthogonal:

\langle (1, 0), (t, 0) \rangle = t

❌ Not quite. The intersection is empty only when

0 \notin U

. Here

(0, 0) \in U

, and

0

lies in every

U^\perp

❌ Not quite.

\langle (1, 0), (1, 0) \rangle = 1 \ne 0

, so

(1, 0) \notin U^\perp

. A nonzero vector is never orthogonal to itself.

❌ Not quite. That is

U^\perp

itself. The intersection keeps only the points of

U

that also lie in

U^\perp

, and

U

has just two points.

❌ Check U⊥. Write out

\langle u, v \rangle = 0

for each

u \in U

❌ Check the intersection. Test each of the two points of

U

for membership in

U^\perp

Show solution

For $v = (v_1, v_2)$ the two conditions are

\langle (1, 0), v \rangle = v_1 = 0, \qquad \langle (0, 0), v \rangle = 0.

The second holds for every $v$ , so $U^\perp = \{(0, t) : t \in \mathbf{R}\}$ , the $y$ -axis, a subspace even though $U$ is just two points.

For the intersection, test the two points of $U$ : $(1, 0)$ has $v_1 = 1 \ne 0$ , so it is not in $U^\perp$ ; $(0, 0)$ is in $U^\perp$ . Hence $U \cap U^\perp = \{0\}$ .

Problem 3 · Split a Vector in R⁴

Given: $V = \mathbf{R}^4$ with the Euclidean inner product, $U = \operatorname{span}\big((1, 1, 1, 1)\big)$ and $v = (4, 0, 2, 6)$ — find $\dim U^\perp$ and the vector $u \in U$ with $v - u \in U^\perp$ .

What is dim U⊥?

What is u?

✅ Correct!

\dim U^\perp = 4 - 1 = 3

, and

v = (3, 3, 3, 3) + (1, -3, -1, 3)

with the second piece orthogonal to

(1, 1, 1, 1)

❌ Not quite.

1

\dim U

. The dimension formula gives

\dim U^\perp = \dim V - \dim U

❌ Not quite.

4

\dim V

; that would make

U^\perp = V

, which happens only for

U = \{0\}

❌ Not quite.

U^\perp = \{0\}

only when

U = V

. Here

U

is a line in

\mathbf{R}^4

, and one linear condition

x_1 + x_2 + x_3 + x_4 = 0

leaves three dimensions.

❌ Normalize first.

\langle v, e \rangle e

needs a unit vector

e

. Using

(1, 1, 1, 1)

, whose norm is

2

, multiplies the answer by

\|(1,1,1,1)\|^2 = 4

❌ Normalize in both places. With

e = \tfrac12 (1, 1, 1, 1)

\langle v, e \rangle = 6

, and

u = 6e

, not

6 \cdot (1, 1, 1, 1)

❌ Not quite.

(1, -3, -1, 3)

w = v - u

, the piece in

U^\perp

: its entries sum to

0

. The question asks for the piece in

U

❌ Check the dimension. Use

\dim U^\perp = \dim V - \dim U

❌ Check u. Use

u = \langle v, e \rangle e

with

e

a unit vector spanning

U

Show solution

Dimension. $\dim U = 1$ , so $\dim U^\perp = 4 - 1 = 3$ .

Orthonormal basis of $U$ . $\|(1, 1, 1, 1)\| = 2$ , so $e = \tfrac12 (1, 1, 1, 1)$ .

\langle v, e \rangle = \tfrac12 (4 + 0 + 2 + 6) = 6, \qquad u = 6e = (3, 3, 3, 3).

Check. $w = v - u = (1, -3, -1, 3)$ and $\langle w, (1, 1, 1, 1) \rangle = 1 - 3 - 1 + 3 = 0$ , so $w \in U^\perp$ and $v = u + w$ is the split $V = U \oplus U^\perp$ guarantees.

Problem 4 · A Complement of Polynomials

Given: $V = \mathcal{P}_2(\mathbf{R})$ with $\langle p, q \rangle = \int_{-1}^{1} p(x) q(x)\, dx$ and $U = \operatorname{span}(1, x)$ — find $\dim U^\perp$ and a polynomial spanning $U^\perp$ .

What is dim U⊥?

U⊥ is the span of …

✅ Correct!

\dim U^\perp = 3 - 2 = 1

, and

x^2 - \tfrac13

is orthogonal to both

1

and

x

, so

U^\perp = \operatorname{span}\big(x^2 - \tfrac13\big)

❌ Not quite.

2

\dim U

, or the dimension of

\{1\}^\perp

. Orthogonality to

x

as well cuts one more dimension:

3 - 2 = 1

❌ Not quite.

3 = \dim \mathcal{P}_2(\mathbf{R})

, which would make

U^\perp = V

; but

\langle 1, 1 \rangle = 2 \ne 0

❌ Not quite.

U^\perp = \{0\}

would need

U = V

, but

\dim U = 2 < 3

❌ Not quite.

x^2

is orthogonal to

x

(

\int_{-1}^{1} x^3\,dx = 0

) but not to

1

\int_{-1}^{1} x^2\,dx = \tfrac23 \ne 0

❌ Not quite.

x

lies in

\{1\}^\perp

, but

x \in U

and

\langle x, x \rangle = \tfrac23 \ne 0

. Since

\{1\} \subseteq U

U^\perp

is the smaller set, and it excludes

x

❌ Check the constant.

\langle x^2 - \tfrac12, 1 \rangle = \tfrac23 - 1 = -\tfrac13 \ne 0

. The constant must make

\int_{-1}^{1} (x^2 - c)\,dx = \tfrac23 - 2c

vanish.

❌ Check the dimension. Use

\dim U^\perp = \dim V - \dim U

with

\dim \mathcal{P}_2(\mathbf{R}) = 3

❌ Check orthogonality. The polynomial must be orthogonal to both

1

and

x

Show solution

Dimension. $1, x$ is linearly independent, so $\dim U = 2$ and $\dim U^\perp = 3 - 2 = 1$ .

Conditions. Write $p = a + bx + cx^2$ . Orthogonality to $1$ and to $x$ gives

\langle p, 1 \rangle = 2a + \tfrac{2}{3}c = 0, \qquad \langle p, x \rangle = \tfrac{2}{3}b = 0.

So $b = 0$ and $a = -\tfrac{c}{3}$ , i.e. $p = c\big(x^2 - \tfrac13\big)$ .

Check. $\int_{-1}^{1} \big(x^2 - \tfrac13\big)\,dx = \tfrac23 - \tfrac23 = 0$ and $\int_{-1}^{1} \big(x^3 - \tfrac{x}{3}\big)\,dx = 0$ . Hence $U^\perp = \operatorname{span}\big(x^2 - \tfrac13\big)$ , matching the dimension count.

Solved: 0 / 4

Orthogonal Complements and the Direct Sum V = U ⊕ U⊥

IKey Formulas

IIA Point, a Plane, a Line in R3\mathbf{R}^3R3

IIIThe Complement of Any Subset

IVComplements in F5\mathbf{F}^5F5: Direct Sum and Reversed Inclusions

VQuiz Questions

IIA Point, a Plane, a Line in $\mathbf{R}^3$

IVComplements in $\mathbf{F}^5$ : Direct Sum and Reversed Inclusions