Linear-Algebra-Done-Right · Unit 19 · Video 2 · Interactive Practice

The Complex Spectral Theorem: Normal Is Enough

IKey Formulas

Formula	Name	When it applies
$T$ normal $\iff$ $\mathcal{M}(T)$ diagonal w.r.t. some orthonormal basis $\iff$ $V$ has an orthonormal basis of eigenvectors of $T$	Complex spectral theorem	$\mathbf{F} = \mathbf{C}$ , $V$ finite-dimensional, $T \in \mathcal{L}(V)$
$\\|Te_k\\|^2 = \|a_{k,k}\|^2$ , $\ \\|T^*e_k\\|^2 = \|a_{k,k}\|^2 + \|a_{k,k+1}\|^2 + \cdots + \|a_{k,n}\|^2$	Row $k$ of the proof	Schur's orthonormal basis, rows $1, \dots, k-1$ already cleared
$\mathcal{M}(TT^) = \operatorname{diag}(\|\lambda_1\|^2, \dots, \|\lambda_n\|^2) = \mathcal{M}(T^T)$	Diagonal implies normal	$\mathcal{M}(T) = \operatorname{diag}(\lambda_1, \dots, \lambda_n)$ w.r.t. an orthonormal basis
$V = E(\lambda_1, T) \oplus \cdots \oplus E(\lambda_m, T)$	Orthogonal eigenspace decomposition	$T$ normal, $\mathbf{F} = \mathbf{C}$ , $V$ finite-dimensional, distinct eigenvalues $\lambda_1, \dots, \lambda_m$

Key Insight: A diagonal matrix with real entries equals its conjugate transpose, so over $\mathbf{R}$ the condition is $T = T^*$ . Over $\mathbf{C}$ the diagonal entries can be non-real, so an operator with an orthonormal eigenbasis is always normal, and self-adjoint exactly when every eigenvalue is real.

IIClearing Schur's Matrix Row by Row

With $\mathcal{M}(T)$ upper triangular in an orthonormal basis, $\|Te_k\| = \|T^*e_k\|$ for each $k$ forces every entry above the diagonal to be $0$ .

a₁₂ a₁₃ a₂₃

IIIDiagonal Means Normal

If $\mathcal{M}(T) = \operatorname{diag}(\lambda_1, \lambda_2)$ in an orthonormal basis, $T$ is normal; it is self-adjoint exactly when both $\lambda_k$ are real.

IVThe Opening Example on $\mathbf{C}^2$

For $T(w, z) = (2w - 3z,\ 3w + 2z)$ , $v = (w, z)$ is an eigenvector exactly when $T$ multiplies both coordinates by one complex number.

💡 Challenge: find eigenvectors for both $2 + 3i$ and $2 - 3i$ . Each eigenspace is a line, so every eigenvector for $2 + 3i$ is orthogonal to every eigenvector for $2 - 3i$ , as $\langle (i, 1), (-i, 1) \rangle = i\,\overline{(-i)} + 1 = 0$ shows, so rescaling them to norm $1$ gives an orthonormal eigenbasis of $\mathbf{C}^2$ .

VQuiz Questions

Problem 1 · Normal or Self-Adjoint?

Given: $T \in \mathcal{L}(\mathbf{C}^2)$ has matrix $A = \begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix}$ with respect to the standard basis. Which statement is true?

✅ Correct!

T

is normal but not self-adjoint, and over

\mathbf{C}

normality is exactly the condition for an orthonormal eigenbasis.

❌ Not quite. Over

\mathbf{C}

the adjoint's matrix is the conjugate transpose:

A^* = \begin{pmatrix} 1 & -i \\ -i & 1 \end{pmatrix} \neq A

. A symmetric complex matrix need not be self-adjoint.

❌ Not quite. "Orthonormal eigenbasis

\iff T = T^*

" is the real spectral theorem. On a complex space the condition is

TT^* = T^*T

, which this

T

satisfies.

❌ Not quite. That product is

A \cdot A = A^2

: you transposed without conjugating. With

A^* = \begin{pmatrix} 1 & -i \\ -i & 1 \end{pmatrix}

, the

(1,1)

entry of

AA^*

1 \cdot 1 + i \cdot (-i) = 2

and the

(1,2)

entry is

1 \cdot (-i) + i \cdot 1 = 0

Show solution

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

A^* = \begin{pmatrix} 1 & -i \\ -i & 1 \end{pmatrix} \neq A,

so $T$ is not self-adjoint. Multiplying,

AA^* = \begin{pmatrix} 1 + i(-i) & -i + i \\ i - i & i(-i) + 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}, \qquad A^*A = \begin{pmatrix} 1 + (-i)i & i - i \\ -i + i & (-i)i + 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}.

So $TT^* = T^*T$ : $T$ is normal, and by the complex spectral theorem $\mathbf{C}^2$ has an orthonormal basis of eigenvectors of $T$ , even though $T \neq T^*$ .

Problem 2 · Distinct Eigenvalues Are Not Enough

Given: $T \in \mathcal{L}(\mathbf{C}^2)$ has matrix $\begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}$ with respect to the standard basis $e_1, e_2$ . Does $\mathbf{C}^2$ have an orthonormal basis of eigenvectors of $T$ ?

✅ Correct! An orthonormal eigenbasis would make

T

normal, and the first row of the proof already fails: the entry

2

above the diagonal cannot be cleared.

❌ Not quite. Those eigenvectors do form a basis, but

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

. Dividing by norms changes lengths, not angles, and each eigenspace is a line, so no other choice of eigenvectors is orthogonal either.

❌ Not quite. Schur's theorem gives an upper-triangular matrix, not a diagonal one; an orthonormal eigenbasis needs a diagonal matrix. Clearing the entry

2

requires normality, which fails here.

❌ Right answer, wrong reason. Over

\mathbf{C}

normal is enough:

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

is not self-adjoint yet has an orthonormal eigenbasis. Here the obstruction is that

T

is not normal.

Show solution

The standard basis is orthonormal, so $\|Te_1\|$ is the norm of column $1$ and $\|T^*e_1\|$ is the norm of row $1$ (conjugated):

\|Te_1\|^2 = |1|^2 + |0|^2 = 1, \qquad \|T^*e_1\|^2 = |1|^2 + |2|^2 = 5.

Since $\|Te_1\| \neq \|T^*e_1\|$ , $T$ is not normal. By the complex spectral theorem, $\mathbf{C}^2$ has no orthonormal basis of eigenvectors of $T$ .

Directly: the eigenvectors are the nonzero multiples of $(1, 0)$ (eigenvalue $1$ ) and of $(1, 1)$ (eigenvalue $3$ , from $\begin{pmatrix} -2 & 2 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = 0$ ), and $\langle a(1, 0), b(1, 1) \rangle = a\overline{b} \neq 0$ for nonzero $a, b$ . So there is an eigenbasis, but never an orthonormal one.

Problem 3 · Diagonalizing a Normal Matrix

Given: $T \in \mathcal{L}(\mathbf{C}^2)$ has matrix $A = \begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix}$ with respect to the standard basis, and $T(1, 1) = (1 + i)(1, 1)$ . Find the eigenvalue of $(1, -1)$ and the matrix of $T$ with respect to the orthonormal basis $\tfrac{1}{\sqrt{2}}(1, 1),\ \tfrac{1}{\sqrt{2}}(1, -1)$ .

Eigenvalue of $(1, -1)$ ?

Matrix of $T$ in the orthonormal basis?

✅ Correct! The diagonal entries

1 \pm i

are not real, which is why

T

is normal without being self-adjoint.

❌ Not quite.

1 + i

belongs to

(1, 1)

. Compute

A(1, -1) = (1 - i,\ i - 1)

and factor out the scalar that multiplies

(1, -1)

❌ Not quite.

i - 1

is the second coordinate of

A(1, -1)

, not the eigenvalue:

(1 - i,\ i - 1) = (1 - i)(1, -1)

, since the second coordinate is

(1 - i)(-1)

❌ Not quite. Reading eigenvalues off the diagonal works for an upper-triangular matrix, and

A

is not upper triangular. Compute

A(1, -1)

instead.

❌ Not quite. Dividing an eigenvector by

\sqrt{2}

does not change its eigenvalue:

T\bigl(\tfrac{1}{\sqrt{2}}(1, 1)\bigr) = (1 + i)\,\tfrac{1}{\sqrt{2}}(1, 1)

❌ Not quite. A matrix depends on the basis. Each new basis vector is an eigenvector, so column

k

has only the eigenvalue on the diagonal.

❌ Not quite. Diagonal entry

k

is the eigenvalue of basis vector

k

; the basis lists

\tfrac{1}{\sqrt{2}}(1, 1)

, with eigenvalue

1 + i

, first.

Show solution

Eigenvalue. $A(1, -1) = (1 \cdot 1 + i(-1),\ i \cdot 1 + 1 \cdot (-1)) = (1 - i,\ i - 1) = (1 - i)(1, -1)$ , so the eigenvalue is $1 - i$ .

Orthonormal basis. $\langle (1, 1), (1, -1) \rangle = 1 \cdot 1 + 1 \cdot (-1) = 0$ and $\|(1, 1)\|^2 = \|(1, -1)\|^2 = 2$ , so $f_1 = \tfrac{1}{\sqrt{2}}(1, 1)$ , $f_2 = \tfrac{1}{\sqrt{2}}(1, -1)$ is orthonormal.

Matrix. $Tf_1 = (1 + i)f_1$ and $Tf_2 = (1 - i)f_2$ , so

\mathcal{M}(T, (f_1, f_2)) = \begin{pmatrix} 1 + i & 0 \\ 0 & 1 - i \end{pmatrix}.

Problem 4 · An Orthonormal Eigenbasis for a Self-Adjoint Matrix

Given: $T \in \mathcal{L}(\mathbf{C}^2)$ has matrix $B = \begin{pmatrix} 2 & 1 + i \\ 1 - i & 3 \end{pmatrix}$ with respect to the standard basis; $T$ is self-adjoint, hence normal, with eigenvalues $1$ and $4$ . Which list is an orthonormal basis of $\mathbf{C}^2$ consisting of eigenvectors of $T$ ?

✅ Correct! Here the eigenvalues

1

and

4

are real, matching

T = T^*

; in Problem 3 the eigenvalues

1 \pm i

were not, and that

T

was only normal.

❌ Not quite. The directions are right but the norms are not:

|1 + i|^2 = 2

, so

\|(1 + i, -1)\|^2 = 2 + 1 = 3

and

\|(1 + i, 2)\|^2 = 2 + 4 = 6

❌ Not quite.

B(1 - i, 2) = (4,\ 6 - 2i)

is not a multiple of

(1 - i, 2)

. The pair only looks orthogonal if the inner product skips the conjugate:

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

❌ Not quite. Self-adjoint does not mean diagonal in the standard basis:

Te_1 = (2,\ 1 - i)

is not a multiple of

e_1

. The spectral theorem promises some orthonormal eigenbasis, not this one.

Show solution

Eigenvectors. For $\lambda = 1$ : $B - I = \begin{pmatrix} 1 & 1 + i \\ 1 - i & 2 \end{pmatrix}$ , and $(1 + i, -1)$ satisfies $(1 + i) - (1 + i) = 0$ and $(1 - i)(1 + i) - 2 = 0$ . For $\lambda = 4$ : $B - 4I = \begin{pmatrix} -2 & 1 + i \\ 1 - i & -1 \end{pmatrix}$ , and $(1 + i, 2)$ satisfies $-2(1 + i) + 2(1 + i) = 0$ and $(1 - i)(1 + i) - 2 = 0$ .

Orthogonality. $\langle (1 + i, -1), (1 + i, 2) \rangle = (1 + i)\overline{(1 + i)} + (-1) \cdot 2 = 2 - 2 = 0$ .

Norms. $\|(1 + i, -1)\|^2 = 2 + 1 = 3$ and $\|(1 + i, 2)\|^2 = 2 + 4 = 6$ .

So $\tfrac{1}{\sqrt{3}}(1 + i, -1)$ , $\tfrac{1}{\sqrt{6}}(1 + i, 2)$ is an orthonormal eigenbasis, and with respect to it $\mathcal{M}(T) = \begin{pmatrix} 1 & 0 \\ 0 & 4 \end{pmatrix}$ .

Solved: 0 / 4

The Complex Spectral Theorem: Normal Is Enough

IKey Formulas

IIClearing Schur's Matrix Row by Row

IIIDiagonal Means Normal

IVThe Opening Example on C2\mathbf{C}^2C2

VQuiz Questions

IVThe Opening Example on $\mathbf{C}^2$