Linear-Algebra-Done-Right · Unit 16 · Video 2 · Interactive Practice

Orthogonal Projection onto U and Why U⊥⊥ = U

IKey Formulas

Formula	Name	Hypotheses
$(U^\perp)^\perp = U$ , $\quad U^\perp = \{0\} \iff U = V$	Double complement; test for $U = V$	$U$ a finite-dimensional subspace of $V$
$P_U v = u$ where $v = u + w$ , $\ u \in U$ , $\ w \in U^\perp$	Orthogonal projection	$V = U \oplus U^\perp$ , so the $u$ is unique
$P_U v = \dfrac{\langle v, u \rangle}{\\|u\\|^2}\, u$	Projection onto a line	$U = \operatorname{span}(u)$ , $\ u \neq 0$
$P_U v = \langle v, e_1 \rangle e_1 + \cdots + \langle v, e_m \rangle e_m$	Property (9)	$e_1, \dots, e_m$ an orthonormal basis of $U$

Key Insight: $P_U$ is linear with $\operatorname{range} P_U = U$ and $\operatorname{null} P_U = U^\perp$ , it fixes $U$ and kills $U^\perp$ , $P_U^2 = P_U$ , and $\|P_U v\| \le \|v\|$ , all because $v = u + w$ splits in exactly one way.

IIProjecting onto a Line $U = \operatorname{span}(u)$

The foot $P_U v$ is the multiple of $u$ that leaves a remainder $v - P_U v$ orthogonal to $u$ .

IIIProperty (9) Does Not Depend on the Basis

Rotating an orthonormal basis of $U$ changes both terms $\langle v, e_k \rangle e_k$ but never their sum $P_U v$ .

Basis angle θ 30° Basis

IVRange, Null Space and $P_U^2 = P_U$

One application of $P_U$ moves every point onto $U$ ; a second application moves nothing.

Applications of P_U 0.00

VQuiz Questions

Problem 1 · Projection onto a Line

Given: In $\mathbf{R}^3$ with the Euclidean inner product, $u = (1, 2, 2)$ , $U = \operatorname{span}(u)$ and $v = (4, 1, 1)$ . Find $P_U v$ .

✅ Correct!

\langle v, u \rangle = 8

and

\|u\|^2 = 9

, so

P_U v = \tfrac{8}{9} u

; the remainder

v - P_U v = \tfrac{7}{9}(4, -1, -1)

is orthogonal to

u

❌ Not quite. You used

\langle v, u \rangle u

without dividing by

\|u\|^2 = 9

. That shortcut is right only when

\|u\| = 1

; here it even gives

\|P_U v\| = 24 > \|v\| = \sqrt{18}

, which property (8) forbids.

❌ Close, but… you divided by

\|u\| = 3

instead of

\|u\|^2 = 9

. Writing

e_1 = u / \|u\|

, the term

\langle v, e_1 \rangle e_1

divides by

\|u\|

twice.

❌ Not quite. That is

v - P_U v

, the piece in

U^\perp

; check that

\langle (28, -7, -7), (1, 2, 2) \rangle = 0

. The projection keeps the piece in

U

, a multiple of

u

❌ Not quite. Use

P_U v = \dfrac{\langle v, u \rangle}{\|u\|^2} u

Show solution

Compute $\langle v, u \rangle = 4 \cdot 1 + 1 \cdot 2 + 1 \cdot 2 = 8$ and $\|u\|^2 = 1 + 4 + 4 = 9$ . Then

P_U v = \frac{\langle v, u \rangle}{\|u\|^2}\, u = \frac{8}{9}(1, 2, 2) = \Big(\frac{8}{9}, \frac{16}{9}, \frac{16}{9}\Big).

Check: $v - P_U v = \big(\tfrac{28}{9}, -\tfrac{7}{9}, -\tfrac{7}{9}\big)$ and $\langle v - P_U v, u \rangle = \tfrac{28 - 14 - 14}{9} = 0$ , so $v - P_U v \in U^\perp$ , as property (6) requires.

Problem 2 · An Orthogonal Basis That Is Not Orthonormal

Given: In $\mathbf{R}^3$ with the Euclidean inner product, $U = \operatorname{span}(f_1, f_2)$ with $f_1 = (1, 1, 0)$ and $f_2 = (1, -1, 1)$ , and $v = (1, 2, 3)$ . Note $\langle f_1, f_2 \rangle = 0$ . Find $P_U v$ .

✅ Correct! Normalizing gives

e_k = f_k / \|f_k\|

, so

\langle v, e_k \rangle e_k = \dfrac{\langle v, f_k \rangle}{\|f_k\|^2} f_k

: here

\tfrac{3}{2} f_1 + \tfrac{2}{3} f_2

❌ Not quite.

3 f_1 + 2 f_2

applies formula (9) to

f_1, f_2

, which are orthogonal but not unit vectors (

\|f_1\|^2 = 2

\|f_2\|^2 = 3

). The remainder

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

has inner product

-3

with

f_1

, so it is not in

U^\perp

❌ Not quite. This vector is

v - P_U v

, the piece in

U^\perp

: it is orthogonal to both

f_1

and

f_2

. The projection is the other piece.

❌ Close, but…

\tfrac{3}{2} f_1

is only the projection onto

\operatorname{span}(f_1)

. Formula (9) sums over a whole orthonormal basis of

U

, so the term

\tfrac{2}{3} f_2

must be added.

❌ Not quite. Normalize

f_1, f_2

first, then use formula (9).

Show solution

Since $\langle f_1, f_2 \rangle = 1 - 1 + 0 = 0$ , the list $e_1 = f_1 / \sqrt{2}$ , $e_2 = f_2 / \sqrt{3}$ is an orthonormal basis of $U$ , and $\langle v, e_k \rangle e_k = \dfrac{\langle v, f_k \rangle}{\|f_k\|^2} f_k$ .

Here $\langle v, f_1 \rangle = 1 + 2 = 3$ and $\langle v, f_2 \rangle = 1 - 2 + 3 = 2$ , so

P_U v = \frac{3}{2}(1, 1, 0) + \frac{2}{3}(1, -1, 1) = \Big(\frac{13}{6}, \frac{5}{6}, \frac{2}{3}\Big).

Check: $v - P_U v = \big(-\tfrac{7}{6}, \tfrac{7}{6}, \tfrac{7}{3}\big) = \tfrac{7}{6}(-1, 1, 2)$ , and $\langle (-1, 1, 2), f_1 \rangle = 0$ , $\langle (-1, 1, 2), f_2 \rangle = -1 - 1 + 2 = 0$ .

Problem 3 · Matrix and Null Space of $P_U$

Given: In $\mathbf{R}^3$ with the Euclidean inner product, $U = \operatorname{span}\big((1, 1, 1)\big)$ . Find the matrix of $P_U$ with respect to the standard basis, and $\dim \operatorname{null} P_U$ .

What is $\mathcal{M}(P_U)$ ?

What is $\dim \operatorname{null} P_U$ ?

✅ Correct!

P_U(x, y, z) = \tfrac{x + y + z}{3}(1, 1, 1)

, and

\operatorname{null} P_U = U^\perp = \{x + y + z = 0\}

has dimension

3 - 1 = 2

❌ Not quite. This matrix sends

(1, 1, 1)

(3, 3, 3)

, but property (2) requires

P_U u = u

for

u \in U

. Divide by

\|(1, 1, 1)\|^2 = 3

❌ Close, but… the coefficient is

\langle v, u \rangle / \|u\|^2

, so divide by

3

, not

\sqrt{3}

. Check: this matrix squared is

\sqrt{3}

times itself, not itself, so it fails

P_U^2 = P_U

❌ Not quite. This is

I - \mathcal{M}(P_U)

, the matrix of

P_{U^\perp}

: it sends

(1, 1, 1)

0

. The projection onto

U

keeps the piece along

(1, 1, 1)

❌ Not quite.

P_U

is not injective unless

U = V

: by property (5),

\operatorname{null} P_U = U^\perp

, and here

(1, -1, 0) \in U^\perp

is sent to

0

❌ Not quite.

1 = \dim U = \dim \operatorname{range} P_U

. The null space is

U^\perp

, and

\dim U + \dim U^\perp = 3

❌ Not quite.

P_U

is not the zero map:

P_U(1, 1, 1) = (1, 1, 1) \neq 0

by property (2).

❌ Check the matrix. Column

k

P_U

of the

k

-th standard basis vector.

❌ Check the null space. Use property (5):

\operatorname{null} P_U = U^\perp

Show solution

With $u = (1, 1, 1)$ and $\|u\|^2 = 3$ ,

P_U(x, y, z) = \frac{x + y + z}{3}(1, 1, 1),

so each standard basis vector goes to $\tfrac{1}{3}(1, 1, 1)$ and every column of $\mathcal{M}(P_U)$ is $\big(\tfrac13, \tfrac13, \tfrac13\big)$ . Check: squaring the matrix gives entries $3 \cdot \tfrac19 = \tfrac13$ , so $\mathcal{M}(P_U)^2 = \mathcal{M}(P_U)$ , matching property (7).

By property (5), $\operatorname{null} P_U = U^\perp = \{(x, y, z) : x + y + z = 0\}$ , spanned by $(1, -1, 0)$ and $(1, 0, -1)$ . Its dimension is $2$ , consistent with $\dim \operatorname{null} P_U + \dim \operatorname{range} P_U = 2 + 1 = 3$ .

Problem 4 · Projection in a Function Space

Given: $V = \mathcal{P}_2(\mathbf{R})$ with $\langle p, q \rangle = \int_{0}^{1} pq$ , and $U = \operatorname{span}(x)$ , which has orthonormal basis $e_1 = \sqrt{3}\, x$ . Find $P_U(x^2)$ and $\|P_U(x^2)\|^2$ .

What is $P_U(x^2)$ ?

What is $\|P_U(x^2)\|^2$ ?

✅ Correct!

P_U(x^2) = \tfrac{3}{4}x

and

\|P_U(x^2)\|^2 = \tfrac{9}{16}\cdot\tfrac{1}{3} = \tfrac{3}{16} \le \|x^2\|^2 = \tfrac{1}{5}

, as property (8) promises.

❌ Not quite.

\tfrac{1}{4}x = \langle x^2, x \rangle\, x

, but

x

is not a unit vector:

\|x\|^2 = \int_0^1 x^2 = \tfrac13

. Divide by it, or use

e_1 = \sqrt{3}\,x

: the term is

\dfrac{1/4}{1/3}\,x = \tfrac34 x

❌ Not quite. Property (2) fixes only vectors of

U

, and

x^2 \notin \operatorname{span}(x)

❌ Not quite.

\tfrac{1}{3} = \langle x^2, 1 \rangle

is the projection onto

\operatorname{span}(1)

. Here

U = \operatorname{span}(x)

, so

P_U(x^2)

must be a multiple of

x

❌ Close, but…

\tfrac{9}{16}

is the square of the coefficient

\tfrac34

, but

\|\tfrac34 x\|^2 = \tfrac{9}{16}\|x\|^2

and

\|x\|^2 = \int_0^1 x^2 = \tfrac13

. Note

\tfrac{9}{16} > \|x^2\|^2 = \tfrac15

, which property (8) rules out.

❌ Not quite.

\tfrac{1}{5} = \int_0^1 x^4

\|x^2\|^2

, the norm of

v

itself, not of its projection.

❌ Not quite.

\tfrac{1}{48} = \|\tfrac14 x\|^2

comes from the unnormalized projection

\langle x^2, x \rangle\, x

. With

e_1 = \sqrt3\, x

the projection is

\tfrac34 x

❌ Check the projection. Use property (9) with

e_1 = \sqrt{3}\,x

❌ Check the norm. Integrate the square of

P_U(x^2)

over

[0, 1]

Show solution

First, $\|x\|^2 = \int_0^1 x^2\,dx = \tfrac13$ , so $e_1 = \sqrt{3}\,x$ is a unit vector and an orthonormal basis of $U$ . By property (9), $P_U(x^2) = \langle x^2, e_1 \rangle e_1$ , where

\langle x^2, e_1 \rangle = \sqrt{3}\int_{0}^{1} x^3\,dx = \frac{\sqrt 3}{4}, \qquad P_U(x^2) = \frac{\sqrt 3}{4}\cdot\sqrt{3}\,x = \frac{3}{4}x.

Check: $\big\langle x^2 - \tfrac34 x,\ x \big\rangle = \int_0^1 \big(x^3 - \tfrac34 x^2\big)\,dx = \tfrac14 - \tfrac14 = 0$ , so $x^2 - \tfrac34 x \in U^\perp$ , as property (6) requires.

Then $\|P_U(x^2)\|^2 = \int_0^1 \tfrac{9}{16}x^2\,dx = \tfrac{9}{16}\cdot\tfrac13 = \tfrac{3}{16}$ , while $\|x^2\|^2 = \int_0^1 x^4\,dx = \tfrac15$ ; indeed $\tfrac{3}{16} \le \tfrac15$ .

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Orthogonal Projection onto U and Why U⊥⊥ = U

IKey Formulas

IIProjecting onto a Line U=span⁡(u)U = \operatorname{span}(u)U=span(u)

IIIProperty (9) Does Not Depend on the Basis

IVRange, Null Space and PU2=PUP_U^2 = P_UPU2​=PU​

VQuiz Questions

IIProjecting onto a Line $U = \operatorname{span}(u)$

IVRange, Null Space and $P_U^2 = P_U$