Linear-Algebra-Done-Right · Unit 18 · Video 2 · Interactive Practice

$\operatorname{null} T^* = (\operatorname{range} T)^\perp$ and $\mathcal{M}(T^*)$ as the Conjugate Transpose

IKey Formulas

Formula	Name	Where it lives
$\operatorname{null} T^* = (\operatorname{range} T)^\perp$ , $\ \operatorname{range} T = (\operatorname{null} T^*)^\perp$	Formulas (1) and (4)	Subspaces of $W$
$\operatorname{range} T^* = (\operatorname{null} T)^\perp$ , $\ \operatorname{null} T = (\operatorname{range} T^*)^\perp$	Formulas (2) and (3)	Subspaces of $V$
$(A^*)_{j,k} = \overline{A_{k,j}}$	Conjugate transpose	$A$ is $m$ -by- $n$ , $A^*$ is $n$ -by- $m$
$\mathcal{M}(T^) = (\mathcal{M}(T))^$	Matrix of the adjoint	Orthonormal bases of $V$ and $W$ only

Key Insight: $T^*$ is defined by $\langle Tv, w\rangle = \langle v, T^*w\rangle$ with no basis in sight; the matrix identity $\mathcal{M}(T^*) = (\mathcal{M}(T))^*$ is where a basis enters, and the proof uses $\mathcal{M}(T)_{j,k} = \langle Te_k, f_j\rangle$ , which needs orthonormality.

IIWhere $\operatorname{null} T^*$ Lives

For $T(x, y) = (x - y,\ 2x - 2y)$ , $T^*w = 0$ exactly when $w$ is orthogonal to $\operatorname{range} T$ .

IIIBuilding a Conjugate Transpose

Row $k$ of $A$ becomes column $k$ of $A^*$ , and every entry is conjugated on the way.

Matrix Entries moved 0

IVWhy Orthonormal Bases Matter

For $T(x, y) = (x + 2y,\ -y)$ and one basis $u_1, u_2$ of $\mathbf{R}^2$ , compare $\mathcal{M}(T^*)$ with $(\mathcal{M}(T))^*$ .

💡 $T^*(p, q) = (p,\ 2p - q)$ does not depend on the basis; only its matrix does. Orthonormal bases are sufficient for $\mathcal{M}(T^*) = (\mathcal{M}(T))^*$ , and a basis that is a common rescaling of an orthonormal one works too.

VQuiz Questions

Problem 1 · Conjugate Transpose of a 3-by-2 Matrix

Given: $B = \begin{pmatrix} 2 - i & 0 \\ 4 & 1 + 3i \\ -5i & 6 \end{pmatrix}$ — find $B^*$ .

✅ Correct! Each row of

B

became a column of

B^*

, real parts unchanged and imaginary parts negated.

❌ That is only the transpose. The shape is right, but

B^*

also conjugates every entry:

2 - i

must become

2 + i

, and

-5i

must become

5i

❌ That is only the conjugate. The entries are conjugated, but rows and columns were not interchanged:

B

3

-by-

2

, so

B^*

must be

2

-by-

3

❌ Conjugation is not negation. This negated whole entries, real parts included:

\overline{2 - i} = 2 + i

, not

-2 + i

, and real entries such as

4

and

6

stay as they are.

Show solution

$B$ is $3$ -by- $2$ , so $B^*$ is $2$ -by- $3$ with $(B^*)_{j,k} = \overline{B_{k,j}}$ . Row $k$ of $B$ , conjugated, becomes column $k$ of $B^*$ :

Row 1: $(2 - i,\ 0) \to (2 + i,\ 0)$ . Row 2: $(4,\ 1 + 3i) \to (4,\ 1 - 3i)$ . Row 3: $(-5i,\ 6) \to (5i,\ 6)$ .

B^* = \begin{pmatrix} 2 + i & 4 & 5i \\ 0 & 1 - 3i & 6 \end{pmatrix}

Problem 2 · The Adjoint of a Linear Functional

Given: $T \in \mathcal{L}(\mathbf{C}^3, \mathbf{C})$ , $T(z_1, z_2, z_3) = z_1 + i z_2 - 2z_3$ , with the standard inner products — find $T^*(1)$ .

✅ Correct!

\mathcal{M}(T) = \begin{pmatrix} 1 & i & -2 \end{pmatrix}

, and its conjugate transpose is the column with entries

1, -i, -2

❌ Only imaginary parts change sign.

-2

is real, so

\overline{-2} = -2

; conjugation left it alone.

❌ You transposed without conjugating. Check with

z = (0, 1, 0)

\langle Tz, 1\rangle = i

, while

\langle (0,1,0), (1, i, -2)\rangle = \overline{i} = -i

. The middle entry must be

-i

❌ That is a number, but $T^*$ maps $\mathbf{C}$ to $\mathbf{C}^3$ .

-1 + i = T(1, 1, 1)

; the adjoint goes the other way, so

T^*(1)

is a vector in

\mathbf{C}^3

Show solution

The standard bases of $\mathbf{C}^3$ and $\mathbf{C}$ are orthonormal, so $\mathcal{M}(T^*) = (\mathcal{M}(T))^*$ . The $1$ -by- $3$ matrix of $T$ is $\begin{pmatrix} 1 & i & -2 \end{pmatrix}$ , and its conjugate transpose is the $3$ -by- $1$ column

\mathcal{M}(T^*) = \begin{pmatrix} 1 \\ -i \\ -2 \end{pmatrix}, \qquad T^*(\lambda) = (\lambda,\ -i\lambda,\ -2\lambda).

Check against the definition: $\langle z, T^*\lambda\rangle = z_1\overline{\lambda} + z_2\overline{(-i\lambda)} + z_3\overline{(-2\lambda)} = (z_1 + iz_2 - 2z_3)\overline{\lambda} = \langle Tz, \lambda\rangle$ . So $T^*(1) = (1, -i, -2)$ .

Problem 3 · Null Space and Range of $T^*$

Given: $T \in \mathcal{L}(\mathbf{R}^3, \mathbf{R}^2)$ , $T(x_1, x_2, x_3) = (x_1 + x_2 - x_3,\ 2x_1 + 2x_2 - 2x_3)$ , with the standard inner products — find $\operatorname{null} T^*$ and $\operatorname{range} T^*$ .

What is null T*?

What is range T*?

✅ Correct!

\operatorname{null} T^* = (\operatorname{range} T)^\perp = \operatorname{span}\big((2,-1)\big)

W = \mathbf{R}^2

, and

\operatorname{range} T^* = (\operatorname{null} T)^\perp = \operatorname{span}\big((1,1,-1)\big)

V = \mathbf{R}^3

❌ $T$ is not surjective. Every output

(s, 2s)

lies on the line

\operatorname{span}\big((1,2)\big)

, so

\operatorname{range} T \ne \mathbf{R}^2

and its orthogonal complement is a line, not

\{0\}

❌ That is $\operatorname{range} T$ itself. Formula (1) takes the orthogonal complement:

\operatorname{null} T^*

is the set of vectors in

\mathbf{R}^2

orthogonal to

(1, 2)

❌ Wrong space.

T^*

starts in

W = \mathbf{R}^2

, so

\operatorname{null} T^*

is a subspace of

\mathbf{R}^2

; a vector in

\mathbf{R}^3

cannot belong to it.

❌ That is $\operatorname{null} T$ . Formula (2) says

\operatorname{range} T^*

is the orthogonal complement of this plane, which is the line through its normal vector

(1, 1, -1)

❌ Wrong space.

T^*

lands in

V = \mathbf{R}^3

, so

\operatorname{range} T^*

is a subspace of

\mathbf{R}^3

;

\operatorname{span}\big((1,2)\big)

\operatorname{range} T

, in

\mathbf{R}^2

❌ Too big.

\operatorname{range} T^* = (\operatorname{null} T)^\perp

, and

\operatorname{null} T

is a plane, so its complement in

\mathbf{R}^3

has dimension

3 - 2 = 1

Show solution

In $W = \mathbf{R}^2$ . $T(x) = s\,(1, 2)$ with $s = x_1 + x_2 - x_3$ , and $s$ takes every real value, so $\operatorname{range} T = \operatorname{span}\big((1,2)\big)$ . By formula (1),

\operatorname{null} T^* = (\operatorname{range} T)^\perp = \{(p, q) : p + 2q = 0\} = \operatorname{span}\big((2, -1)\big).

In $V = \mathbf{R}^3$ . $\operatorname{null} T = \{x : x_1 + x_2 - x_3 = 0\} = \{x : \langle x, (1,1,-1)\rangle = 0\}$ , a plane with normal $(1,1,-1)$ . By formula (2),

\operatorname{range} T^* = (\operatorname{null} T)^\perp = \operatorname{span}\big((1, 1, -1)\big).

Check directly. $\mathcal{M}(T) = \begin{pmatrix} 1 & 1 & -1 \\ 2 & 2 & -2 \end{pmatrix}$ in the orthonormal standard bases, so $T^*(p, q) = (p + 2q)(1, 1, -1)$ : it vanishes exactly when $p + 2q = 0$ , and its values fill the line $\operatorname{span}\big((1,1,-1)\big)$ .

Problem 4 · A Basis That Is Not Orthonormal

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ , $T(x, y) = (x + y,\ 0)$ , the usual inner product, and the basis $u_1 = (1, 0)$ , $u_2 = (1, 1)$ used for both domain and target. Here $\mathcal{M}(T) = \begin{pmatrix} 1 & 2 \\ 0 & 0 \end{pmatrix}$ — find $\mathcal{M}(T^*)$ with respect to $u_1, u_2$ .

✅ Correct!

T^*u_1 = T^*u_2 = (1, 1) = u_2

, so both columns are

(0, 1)

— and the result is not

(\mathcal{M}(T))^*

❌ That is $(\mathcal{M}(T))^*$ . The theorem

\mathcal{M}(T^*) = (\mathcal{M}(T))^*

needs orthonormal bases, and

\langle u_1, u_2\rangle = 1 \ne 0

; compute

T^*u_1

and

T^*u_2

directly.

❌ That is the matrix of $T^*$ in the standard basis. Its columns are

T^*(1,0)

and

T^*(0,1)

in standard coordinates; the question asks for coordinates of

T^*u_1

and

T^*u_2

u_1, u_2

❌ Coordinates go in columns, not rows. Column

k

holds the coordinates of

T^*u_k

; you wrote each coordinate vector

(0, 1)

as a row.

Show solution

The adjoint. $\langle T(x, y), (p, q)\rangle = (x + y)p = \langle (x, y), (p, p)\rangle$ , so $T^*(p, q) = (p, p)$ .

Columns of $\mathcal{M}(T^*)$ . $T^*u_1 = T^*(1, 0) = (1, 1) = 0u_1 + 1u_2$ and $T^*u_2 = T^*(1, 1) = (1, 1) = 0u_1 + 1u_2$ . So

\mathcal{M}(T^*) = \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix} \ne \begin{pmatrix} 1 & 0 \\ 2 & 0 \end{pmatrix} = (\mathcal{M}(T))^*.

(For the given $\mathcal{M}(T)$ : $Tu_1 = (1, 0) = u_1$ and $Tu_2 = (2, 0) = 2u_1$ .) The basis is not orthonormal, so the conjugate-transpose theorem does not apply.

Solved: 0 / 4

null⁡T∗=(range⁡T)⊥\operatorname{null} T^* = (\operatorname{range} T)^\perpnullT∗=(rangeT)⊥ and M(T∗)\mathcal{M}(T^*)M(T∗) as the Conjugate Transpose

IKey Formulas

IIWhere null⁡T∗\operatorname{null} T^*nullT∗ Lives

IIIBuilding a Conjugate Transpose

IVWhy Orthonormal Bases Matter

VQuiz Questions

$\operatorname{null} T^* = (\operatorname{range} T)^\perp$ and $\mathcal{M}(T^*)$ as the Conjugate Transpose

IIWhere $\operatorname{null} T^*$ Lives