Linear-Algebra-Done-Right · Unit 18 · Video 4 · Interactive Practice

Normal Operators: $\|Tv\| = \|T^*v\|$ and Orthogonal Eigenvectors

IKey Formulas

Formula	Name	What it says
$TT^* = T^*T$	Normal operator	$T$ commutes with its adjoint; every self-adjoint operator is normal
$\\|Tv\\| = \\|T^*v\\|$ for every $v \in V$	Norm characterization	Equivalent to $T$ being normal
$Tv = \lambda v \iff T^*v = \overline{\lambda} v$	Shared eigenvectors	Holds when $T$ is normal; then $\operatorname{null} T = \operatorname{null} T^$ and $\operatorname{range} T = \operatorname{range} T^$
$A = \frac{T + T^}{2}$ , $B = \frac{T - T^}{2i}$ , $T = A + iB$	Real and imaginary parts ( $\mathbf{F} = \mathbf{C}$ )	$A$ , $B$ are self-adjoint; $T$ is normal $\iff AB = BA$

Key Insight: If $T$ is normal, $Tu = \alpha u$ , $Tv = \beta v$ and $\alpha \ne \beta$ , then $(\alpha - \beta)\langle u, v \rangle = \langle Tu, v \rangle - \langle u, T^*v \rangle = 0$ , so $u \perp v$ .

IIComparing $\|Tv\|$ With $\|T^*v\|$

$T$ is normal exactly when the circles of radius $\|Tv\|$ and $\|T^*v\|$ coincide for every $v$ .

Operator on

\mathbf{R}^2

💡 For $N$ , the vector $e_1$ has $\|Ne_1\| = 0$ but $\|N^*e_1\| = 1$ : $e_1 \in \operatorname{null} N$ yet $e_1 \notin \operatorname{null} N^*$ , which a normal operator never allows.

IIIOrthogonal Eigenvectors

Here $Tu = 2u$ and $Tw = -w$ ; this $T$ is normal exactly when $u \perp w$ .

IVReal and Imaginary Parts: $T = A + iB$

Every operator on $\mathbf{C}^2$ splits as $A + iB$ with $A$ , $B$ self-adjoint; normality is exactly $AB = BA$ .

Matrix of

T

Re t₁₁ 1 Im t₁₁ 0 Re t₁₂ 0 Im t₁₂ 2 Re t₂₁ 0 Im t₂₁ 2 Re t₂₂ 1 Im t₂₂ 0

VQuiz Questions

Problem 1 · Norms of $Tv$ and $T^*v$

Given: $T \in \mathcal{L}(\mathbf{F}^2)$ has standard matrix $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ , and $v = (1, 1)$ .

Compute $\|Tv\|$ and $\|T^*v\|$ .

Is $T$ normal?

✅ Correct! Equal norms at one vector prove nothing; a single vector with

\|Tv\| \ne \|T^*v\|

, here

e_2

, proves

T

is not normal.

❌ Not quite.

(1, 1) \mapsto (0, 1)

is what

T^{-1}

does. The adjoint's matrix is the conjugate transpose

\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}

, so

T^*v = (1, 2)

❌ Not quite.

3 = 2 + 1

adds the coordinates of

Tv = (2, 1)

. The norm is

\sqrt{2^2 + 1^2} = \sqrt{5}

❌ Not quite. The characterization needs

\|Tv\| = \|T^*v\|

for every

v

. Try

v = e_2

Te_2 = (1, 1)

but

T^*e_2 = (0, 1)

❌ Not quite. Invertibility and the eigenvalues say nothing about whether

T

commutes with

T^*

. Here

TT^* = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}

while

T^*T = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}

Show solution

The standard basis is orthonormal, so $\mathcal{M}(T^*)$ is the conjugate transpose:

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

For $v = (1, 1)$ : $Tv = (2, 1)$ and $T^*v = (1, 2)$ , so $\|Tv\| = \|T^*v\| = \sqrt{5}$ .

For $e_2 = (0, 1)$ : $Te_2 = (1, 1)$ and $T^*e_2 = (0, 1)$ , so

\|Te_2\| = \sqrt{2} \ne 1 = \|T^*e_2\|.

Since $T$ is normal if and only if $\|Tv\| = \|T^*v\|$ for every $v$ , the vector $e_2$ shows $T$ is not normal. Directly: $TT^* = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \ne \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} = T^*T$ .

Problem 2 · Normal or Not?

Given: operators on $\mathbf{C}^2$ described by their matrices with respect to the standard basis.

Which of these operators is normal?

The operator with matrix $\begin{pmatrix} i & 1 \\ -1 & i \end{pmatrix}$ is …

✅ Correct!

\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}

has

TT^* = T^*T = 2I

, and

\begin{pmatrix} i & 1 \\ -1 & i \end{pmatrix}

satisfies

T^* = -T

, which commutes with

T

❌ Not quite. Distinct eigenvalues

1

2

do not make an operator normal. Its eigenvectors

(1, 0)

and

(1, 1)

are not orthogonal, which is impossible for a normal operator; indeed

TT^* = \begin{pmatrix} 2 & 2 \\ 2 & 4 \end{pmatrix} \ne \begin{pmatrix} 1 & 1 \\ 1 & 5 \end{pmatrix} = T^*T

❌ Not quite. The rotation-like pattern is misleading: the off-diagonal entries have different sizes.

TT^* = \begin{pmatrix} 4 & 0 \\ 0 & 1 \end{pmatrix}

but

T^*T = \begin{pmatrix} 1 & 0 \\ 0 & 4 \end{pmatrix}

, and

\|Te_1\| = 1 \ne 2 = \|T^*e_1\|

❌ Not quite. A scalar diagonal does not rescue the off-diagonal

1

T = iI + E

with

E = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}

, and

T

is normal only if

E

is (part 4 with

\lambda = i

). But

EE^* \ne E^*E

; numerically

TT^* = \begin{pmatrix} 2 & -i \\ i & 1 \end{pmatrix} \ne \begin{pmatrix} 1 & -i \\ i & 2 \end{pmatrix} = T^*T

❌ Not quite. Self-adjoint requires each diagonal entry to equal its own conjugate, but

\overline{i} = -i \ne i

. The conjugate transpose is

\begin{pmatrix} -i & -1 \\ 1 & -i \end{pmatrix} = -T

❌ Not quite. Real eigenvalues are a property of self-adjoint operators; normal operators may have any complex eigenvalues. Here

T^* = -T

, so

TT^* = -T^2 = T^*T

and

T

is normal.

Show solution

Compute $TT^*$ and $T^*T$ with $\mathcal{M}(T^*) = \overline{\mathcal{M}(T)}^{\,t}$ .

\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix},

so (d) is normal. The others fail: (a) gives $\begin{pmatrix} 2 & 2 \\ 2 & 4 \end{pmatrix}$ versus $\begin{pmatrix} 1 & 1 \\ 1 & 5 \end{pmatrix}$ , (b) gives $\begin{pmatrix} 4 & 0 \\ 0 & 1 \end{pmatrix}$ versus $\begin{pmatrix} 1 & 0 \\ 0 & 4 \end{pmatrix}$ , (c) gives $\begin{pmatrix} 2 & -i \\ i & 1 \end{pmatrix}$ versus $\begin{pmatrix} 1 & -i \\ i & 2 \end{pmatrix}$ .

For $T = \begin{pmatrix} i & 1 \\ -1 & i \end{pmatrix}$ : conjugating and transposing gives $T^* = \begin{pmatrix} -i & -1 \\ 1 & -i \end{pmatrix} = -T$ . So $T^* \ne T$ (not self-adjoint), but $TT^* = -T^2 = T^*T$ (normal).

Problem 3 · Splitting $T = A + iB$

Given: $T \in \mathcal{L}(\mathbf{C}^2)$ with standard matrix $\begin{pmatrix} 1 & 2i \\ 0 & 3 \end{pmatrix}$ . Find $A = \frac{T + T^*}{2}$ , $B = \frac{T - T^*}{2i}$ , and decide whether $T$ is normal.

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

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

Is $T$ normal?

✅ Correct!

A

and

B

are self-adjoint with

T = A + iB

, but

AB = \begin{pmatrix} i & 1 \\ 3 & -i \end{pmatrix} \ne \begin{pmatrix} -i & 3 \\ 1 & i \end{pmatrix} = BA

, so

T

is not normal.

❌ Not quite. That is the entrywise real part of the matrix.

A = \frac{T + T^*}{2}

also mixes the off-diagonal entries: its

(1,2)

entry is

\frac{2i + \overline{0}}{2} = i

❌ Not quite. You used the transpose without conjugating. With

\mathcal{M}(T^*) = \begin{pmatrix} 1 & 0 \\ -2i & 3 \end{pmatrix}

the

(2,1)

entry of

A

\frac{0 + (-2i)}{2} = -i

; a self-adjoint matrix needs

a_{21} = \overline{a_{12}}

❌ Not quite. That is

T + T^*

; divide by

2

❌ Not quite. That is

\frac{T - T^*}{2} = iB

; dividing it by

i

gives

B

❌ Not quite. That is the entrywise imaginary part, which is not even self-adjoint. Compute

T - T^* = \begin{pmatrix} 0 & 2i \\ 2i & 0 \end{pmatrix}

and divide by

2i

❌ Not quite. You used the transpose without conjugating:

T - T^{t} = \begin{pmatrix} 0 & 2i \\ -2i & 0 \end{pmatrix}

. With

T^*

the

(2,1)

entry of

T - T^*

0 - (-2i) = 2i

❌ Not quite.

A

and

B

are self-adjoint for every operator on a complex space; the theorem needs them to commute as well.

❌ Not quite.

\operatorname{tr}(AB) = \operatorname{tr}(BA)

holds for any two square matrices, so it cannot detect normality. The products themselves differ.

Show solution

The adjoint's matrix is the conjugate transpose: $\mathcal{M}(T^*) = \begin{pmatrix} 1 & 0 \\ -2i & 3 \end{pmatrix}$ .

A = \frac{1}{2}\begin{pmatrix} 2 & 2i \\ -2i & 6 \end{pmatrix} = \begin{pmatrix} 1 & i \\ -i & 3 \end{pmatrix}, \qquad B = \frac{1}{2i}\begin{pmatrix} 0 & 2i \\ 2i & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.

Check: $A + iB = \begin{pmatrix} 1 & i + i \\ -i + i & 3 \end{pmatrix} = \begin{pmatrix} 1 & 2i \\ 0 & 3 \end{pmatrix} = T$ .

AB = \begin{pmatrix} i & 1 \\ 3 & -i \end{pmatrix}, \qquad BA = \begin{pmatrix} -i & 3 \\ 1 & i \end{pmatrix}.

Since $AB \ne BA$ , $T$ is not normal. (Consistently, $AB - BA = \frac{T^*T - TT^*}{2i} \ne 0$ .)

Problem 4 · The Cyclic Permutation

Given: $P \in \mathcal{L}(\mathbf{C}^3)$ , $P(x, y, z) = (z, x, y)$ , which is normal since $PP^* = P^*P = I$ . Let $\omega = e^{2\pi i/3}$ and $w = (1, \omega^2, \omega)$ , so that $Pw = \omega w$ . Find $P^*w$ .

✅ Correct! For a normal operator

Tv = \lambda v \iff T^*v = \overline{\lambda} v

, and

\overline{\omega} = \omega^2

because

|\omega| = 1

and

\omega^3 = 1

❌ Not quite. The eigenvalue must be conjugated:

T^*v = \overline{\lambda} v

. Since

\omega

is not real,

\overline{\omega} = \omega^2 \ne \omega

❌ Not quite.

P^*

is the inverse permutation

(x, y, z) \mapsto (y, z, x)

, not the identity; it scales

w

\overline{\omega}

, not by

1

❌ Not quite. An operator and its adjoint can have different eigenvectors in general, but not when the operator is normal: every eigenvector of

P

is an eigenvector of

P^*

Show solution

The matrix of $P$ is a permutation matrix, so its conjugate transpose is its transpose and $P^*(x, y, z) = (y, z, x)$ . Then $PP^* = P^*P = I$ , so $P$ is normal.

By part 5 of the theorem, $Pw = \omega w$ implies $P^*w = \overline{\omega} w$ . Since $\omega\overline{\omega} = |\omega|^2 = 1$ and $\omega^3 = 1$ , $\overline{\omega} = \omega^{-1} = \omega^2$ .

Direct check: $P^*w = (\omega^2, \omega, 1)$ , and $\omega^2 w = (\omega^2, \omega^4, \omega^3) = (\omega^2, \omega, 1)$ . So $P^*w = \omega^2 w$ .

Solved: 0 / 4

Normal Operators: ∥Tv∥=∥T∗v∥\|Tv\| = \|T^*v\|∥Tv∥=∥T∗v∥ and Orthogonal Eigenvectors

IKey Formulas

IIComparing ∥Tv∥\|Tv\|∥Tv∥ With ∥T∗v∥\|T^*v\|∥T∗v∥

IIIOrthogonal Eigenvectors

IVReal and Imaginary Parts: T=A+iBT = A + iBT=A+iB

VQuiz Questions

Normal Operators: $\|Tv\| = \|T^*v\|$ and Orthogonal Eigenvectors

IIComparing $\|Tv\|$ With $\|T^*v\|$

IVReal and Imaginary Parts: $T = A + iB$