Linear-Algebra-Done-Right · Unit 18 · Video 1 · Interactive Practice

The Adjoint $T^*$ : Moving $T$ into the Second Slot of $\langle Tv, w \rangle$

IKey Formulas

Formula	Name	What it says
$\langle Tv, w \rangle = \langle v, T^*w \rangle$ for all $v \in V$ , $w \in W$	Definition of $T^*: W \to V$	For $T \in \mathcal{L}(V, W)$ with $V$ , $W$ finite-dimensional inner product spaces over $\mathbf{F}$ : $T^*w$ is the Riesz vector of $v \mapsto \langle Tv, w \rangle$ , so it exists and is unique
$T^(w_1 + w_2) = T^w_1 + T^w_2$ , $\;T^(\lambda w) = \lambda\, T^*w$	$T^* \in \mathcal{L}(W, V)$	The two conjugations in the homogeneity proof cancel
$(S + T)^* = S^* + T^$ , $\;(\lambda T)^ = \overline{\lambda}\, T^$ , $\;(T^)^* = T$	Rules (a)–(c)	$\lambda$ enters through the first slot, so it is conjugated once
$(ST)^* = T^S^$ , $\;I^* = I$ , $\;(T^)^{-1} = (T^{-1})^$	Rules (d)–(f)	Products reverse order; $S \in \mathcal{L}(W, U)$ in (d); in (f), $T$ is invertible, and then so is $T^*$

Key Insight: To compute $T^*w$ , start from $\langle Tv, w \rangle$ , rearrange until $v$ stands alone in the first slot, and read off whatever sits in the second slot.

IIReading Off $T^*w$ from $\langle Tv, w \rangle$

For $T(x, y) = (x - y, 2x, y)$ , the vector $T^*w$ reproduces $\langle Tv, w \rangle$ as $\langle v, T^*w \rangle$ for every $v$ .

w₁ 1 w₂ 1 w₃ 2

IIIA Scalar Through the First Slot Comes Out Conjugated

$\lambda$ leaves the first slot unchanged but enters the second slot conjugated, so $(\lambda T)^* = \overline{\lambda}\, T^*$ , not $\lambda T^*$ .

Re λ 1 Im λ 1

IVWhy $(ST)^* = T^S^$ : The Adjoints Run Back in Reverse

With $T(x, y) = (x - y, 2x, y)$ as above and $S(a, b, c) = a + b - c$ from $\mathbf{R}^3$ to $\mathbf{R}$ , moving $ST$ across the inner product strips off $S$ first, then $T$ .

Step 1 v₁ 2 v₂ 1 t 1

VQuiz Questions

Problem 1 · Compute an Adjoint by Regrouping

Given: $T \in \mathcal{L}(\mathbf{R}^2, \mathbf{R}^3)$ , $T(x, y) = (x + 2y,\ 3y,\ -x)$ , with the Euclidean inner products. Find $T^*(w_1, w_2, w_3)$ .

✅ Correct!

\langle T(x, y), w \rangle = x(w_1 - w_3) + y(2w_1 + 3w_2) = \langle (x, y), (w_1 - w_3, 2w_1 + 3w_2) \rangle

❌ Not quite. These are the right coefficients in the wrong slots:

2w_1 + 3w_2

multiplies

y

, so it is the second coordinate of

T^*w

, and

w_1 - w_3

multiplies

x

, so it is the first.

❌ Not quite. The third coordinate of

T(x, y)

-x

, so it contributes

-x\,w_3

; the coefficient of

x

w_1 - w_3

, not

w_1 + w_3

❌ Not quite.

T

goes from

\mathbf{R}^2

\mathbf{R}^3

, so

T^*

comes back from

\mathbf{R}^3

\mathbf{R}^2

T^*w

must have exactly two coordinates, one for each of

x

and

y

Show solution

Start from $\langle Tv, w \rangle$ with $v = (x, y)$ and $w = (w_1, w_2, w_3)$ :

\langle T(x, y), w \rangle = (x + 2y)w_1 + 3y\,w_2 - x\,w_3.

Regroup around $x$ and $y$ so that $(x, y)$ stands alone in the first slot:

= x(w_1 - w_3) + y(2w_1 + 3w_2) = \big\langle (x, y),\ (w_1 - w_3,\ 2w_1 + 3w_2) \big\rangle.

The second slot is $T^*w$ , so $T^*(w_1, w_2, w_3) = (w_1 - w_3,\ 2w_1 + 3w_2)$ .

Check: $T(1, 1) = (3, 3, -1)$ and $\langle (3, 3, -1), (1, 1, 1) \rangle = 5$ ; $T^*(1, 1, 1) = (0, 5)$ and $\langle (1, 1), (0, 5) \rangle = 5$ .

Problem 2 · Conjugates in the Second Slot

Given: $T \in \mathcal{L}(\mathbf{C}^2, \mathbf{C}^3)$ , $T(z_1, z_2) = (i z_1,\ z_1 + z_2,\ 2z_2)$ , with the Euclidean inner products $\langle a, b \rangle = \sum_k a_k \overline{b_k}$ . Find $T^*(w_1, w_2, w_3)$ .

✅ Correct! The coefficient of

z_1

i\overline{w_1} + \overline{w_2}

, and moving it into the second slot conjugates it to

-i w_1 + w_2

❌ Not quite. You conjugated the constant

i

but left the

\overline{w_k}

alone; the whole coefficient

i\overline{w_1} + \overline{w_2}

must be conjugated, which turns each

\overline{w_k}

back into

w_k

. As written, this map is not even linear in

w

❌ Not quite. You undid the bars on the

w_k

but forgot that the constant

i

is conjugated too:

\overline{i\overline{w_1}} = -i w_1

. With

+i

\langle z, T^*w \rangle

would contain

-i z_1 \overline{w_1}

instead of

i z_1 \overline{w_1}

❌ Not quite. These are the coefficients of

z_1

and

z_2

read off directly, but the second slot conjugates whatever sits in it:

\langle z, u \rangle = z_1 \overline{u_1} + z_2 \overline{u_2}

, so

u_1

must be the conjugate of

i\overline{w_1} + \overline{w_2}

Show solution

Start from $\langle Tz, w \rangle$ with $z = (z_1, z_2)$ :

\langle T(z_1, z_2), w \rangle = i z_1 \overline{w_1} + (z_1 + z_2)\overline{w_2} + 2z_2 \overline{w_3} = z_1\big(i\overline{w_1} + \overline{w_2}\big) + z_2\big(\overline{w_2} + 2\overline{w_3}\big).

To write $z_1 c_1$ as $z_1 \overline{u_1}$ we need $u_1 = \overline{c_1}$ ; a coefficient entering the second slot is conjugated:

\overline{i\overline{w_1} + \overline{w_2}} = -i w_1 + w_2, \qquad \overline{\overline{w_2} + 2\overline{w_3}} = w_2 + 2w_3.

So $\langle Tz, w \rangle = \big\langle (z_1, z_2),\ (-i w_1 + w_2,\ w_2 + 2w_3) \big\rangle$ and $T^*(w_1, w_2, w_3) = (-i w_1 + w_2,\ w_2 + 2w_3)$ , which is linear in $w$ , as the theorem promises.

Check: $z = (1, 0)$ , $w = (1, 0, 0)$ : $\langle Tz, w \rangle = \langle (i, 1, 0), (1, 0, 0) \rangle = i$ , and $\langle (1, 0), T^*w \rangle = \langle (1, 0), (-i, 0) \rangle = \overline{-i} = i$ .

Problem 3 · Combining the Rules

Given: $T \in \mathcal{L}(V, W)$ and $S \in \mathcal{L}(W, U)$ , where $V$ , $W$ , $U$ are finite-dimensional inner product spaces over $\mathbf{C}$ . Simplify each adjoint.

What is $\big((2i)\,ST\big)^*$ ?

What is $\big(((3 - i)\,T)^*\big)^*$ ?

✅ Correct! Rule (b) conjugates the scalar and rule (d) reverses the product; applying the adjoint twice undoes everything, scalar included.

❌ Not quite. The order

T^*S^*

is right, but by rule (b) the scalar comes out conjugated:

\overline{2i} = -2i

❌ Not quite. The scalar

-2i

is right, but

(ST)^* = T^*S^*

(ST)^*

maps

U

back to

V

, so

S^*: U \to W

must act first and

T^*: W \to V

second.

S^*T^*

would apply

T^*

to a vector of

U

❌ Not quite. Two errors: the scalar must be conjugated,

\overline{2i} = -2i

, and the product must reverse,

(ST)^* = T^*S^*

❌ Not quite. You conjugated only once. The first adjoint gives

(3 + i)\,T^*

, and the second adjoint conjugates the scalar again:

\overline{3 + i} = 3 - i

❌ Not quite.

(3 + i)\,T^*

((3 - i)\,T)^*

, the result after ONE adjoint; the question takes the adjoint twice, and

(T^*)^* = T

❌ Not quite. The scalar is right after two conjugations, but the map is not:

(T^*)^* = T

, and

T^*

goes from

W

V

, the wrong direction for a map built from

T \in \mathcal{L}(V, W)

Show solution

First part. By rule (b) with $\lambda = 2i$ , then rule (d):

\big((2i)\,ST\big)^* = \overline{2i}\,(ST)^* = -2i\,(ST)^* = -2i\,T^*S^*.

Second part. Apply rule (b) twice and rule (c):

\big((3 - i)\,T\big)^* = \overline{3 - i}\,T^* = (3 + i)\,T^*, \qquad \big((3 + i)\,T^*\big)^* = \overline{3 + i}\,(T^*)^* = (3 - i)\,T.

This is rule (c) applied to the single map $(3 - i)\,T$ : the adjoint of an adjoint returns the original map.

Problem 4 · The Adjoint of an Inverse

Given: With the Euclidean inner products, $T(x, y) = (x + 2y,\ 3y,\ -x)$ from $\mathbf{R}^2$ to $\mathbf{R}^3$ and $S(a, b, c) = (a,\ b + c)$ from $\mathbf{R}^3$ to $\mathbf{R}^2$ , so $ST(x, y) = (x + 2y,\ 3y - x)$ is invertible on $\mathbf{R}^2$ . Find $\big((ST)^{-1}\big)^*(5, 0)$ .

✅ Correct!

(ST)^*(u_1, u_2) = (u_1 - u_2,\ 2u_1 + 3u_2)

, and by rule (f) the answer solves

(ST)^*u = (5, 0)

, giving

u = (3, -2)

❌ Not quite.

(3, 1) = (ST)^{-1}(5, 0)

: you inverted

ST

but never took the adjoint.

ST

is not its own adjoint here, since

(ST)^*(u_1, u_2) = (u_1 - u_2,\ 2u_1 + 3u_2) \neq ST(u_1, u_2)

❌ Not quite.

(5, 10) = (ST)^*(5, 0)

: you took the adjoint but not the inverse. By rule (f),

\big((ST)^{-1}\big)^* = \big((ST)^*\big)^{-1}

, so solve

(ST)^*u = (5, 0)

instead.

❌ Not quite. Only the operator

ST \in \mathcal{L}(\mathbf{R}^2)

needs to be invertible, and it is;

(ST)^{-1}

is never split into

T^{-1}S^{-1}

, and the adjoint of any linear map exists.

Show solution

Step 1: $(ST)^*$ by regrouping. For $v = (x, y)$ and $u = (u_1, u_2)$ :

\langle ST(x, y), u \rangle = (x + 2y)u_1 + (3y - x)u_2 = x(u_1 - u_2) + y(2u_1 + 3u_2),

so $(ST)^*(u_1, u_2) = (u_1 - u_2,\ 2u_1 + 3u_2)$ . Rule (d) agrees: $S^*(u_1, u_2) = (u_1, u_2, u_2)$ and $T^*(w_1, w_2, w_3) = (w_1 - w_3,\ 2w_1 + 3w_2)$ , so $T^*S^*(u_1, u_2) = (u_1 - u_2,\ 2u_1 + 3u_2)$ .

Step 2: rule (f). $\big((ST)^{-1}\big)^* = \big((ST)^*\big)^{-1}$ , so the answer is the $u$ with $(ST)^*u = (5, 0)$ :

u_1 - u_2 = 5, \qquad 2u_1 + 3u_2 = 0 \quad\Longrightarrow\quad u_2 = -2,\ u_1 = 3.

Check against the definition: $(ST)^{-1}(1, 0) = \big(\tfrac{3}{5}, \tfrac{1}{5}\big)$ , and $\big\langle (\tfrac{3}{5}, \tfrac{1}{5}), (5, 0) \big\rangle = 3 = \langle (1, 0), (3, -2) \rangle$ .

Solved: 0 / 4

The Adjoint T∗T^*T∗: Moving TTT into the Second Slot of ⟨Tv,w⟩\langle Tv, w \rangle⟨Tv,w⟩

IKey Formulas

IIReading Off T∗wT^*wT∗w from ⟨Tv,w⟩\langle Tv, w \rangle⟨Tv,w⟩

IIIA Scalar Through the First Slot Comes Out Conjugated

IVWhy (ST)∗=T∗S∗(ST)^* = T^*S^*(ST)∗=T∗S∗: The Adjoints Run Back in Reverse

VQuiz Questions

The Adjoint $T^*$ : Moving $T$ into the Second Slot of $\langle Tv, w \rangle$

IIReading Off $T^*w$ from $\langle Tv, w \rangle$

IVWhy $(ST)^* = T^S^$ : The Adjoints Run Back in Reverse