Linear-Algebra-Done-Right · Unit 20 · Video 2 · Interactive Practice

Unitary Operators: $S^{-1} = S^*$ , Eigenvalues on $|\lambda| = 1$

IKey Formulas

Formula	Name	Hypotheses
$S$ unitary $\iff S^S = SS^ = I \iff S^{-1} = S^*$	Characterizations of unitary operators	$S \in \mathcal{L}(V)$ , $V$ a finite-dimensional inner product space
$S$ unitary $\iff$ the rows (equivalently, the columns) of $\mathcal{M}(S, (e_1, \dots, e_n))$ form an orthonormal basis of $\mathbf{F}^n$	Row and column tests	$e_1, \dots, e_n$ an orthonormal basis of $V$ ; Euclidean inner product on $\mathbf{F}^n$
$Sv = \lambda v$ , $v \ne 0 \implies \|\lambda\| = 1$	Eigenvalues of a unitary operator	$S$ unitary
$S$ unitary $\iff V$ has an orthonormal basis of eigenvectors of $S$ with every $\|\lambda_j\| = 1$	Unitary operators on complex spaces	$\mathbf{F} = \mathbf{C}$ , $S \in \mathcal{L}(V)$

Key Insight: under the analogy $z \leftrightarrow S$ , $\overline{z} \leftrightarrow S^*$ , the unit circle $\overline{z}z = 1$ becomes $S^*S = I$ : the unitary operators play the role of the complex numbers with $|z| = 1$ .

IIThe Unit Circle in $\mathbf{C}$

Multiplying by $z$ scales every absolute value by $|z|$ , so it keeps all lengths exactly when $\overline{z}z = 1$ .

IIIColumns, Rows and Eigenvalues of a $2 \times 2$ Operator

Unitary forces orthonormal columns, orthonormal rows and every $|\lambda| = 1$ ; eigenvalues on the circle alone do not force unitary.

Matrix

💡 The right panel shows the eigenvalues of the matrix acting on $\mathbf{C}^2$ ; on $\mathbf{R}^2$ a unitary operator can only have the eigenvalues $1$ and $-1$ .

IVQuiz Questions

Problem 1 · The Inverse Is the Adjoint

Given: $S \in \mathcal{L}(\mathbf{C}^2)$ has matrix $\mathcal{M}(S) = \tfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix}$ with respect to the standard basis. Its columns are orthonormal, so $S$ is unitary. Find $\mathcal{M}(S^{-1})$ .

✅ Correct!

S^{-1} = S^*

, and with respect to an orthonormal basis

\mathcal{M}(S^*)

is the conjugate transpose of

\mathcal{M}(S)

❌ Not quite. This is the transpose without conjugation, which is

\mathcal{M}(S)

itself because the matrix is symmetric. Its square is

\begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix} \ne I

: complex entries must be conjugated.

❌ Not quite. The scalar

\tfrac{1}{\sqrt{2}}

is real, so conjugation leaves it alone; it is not inverted. With

\sqrt{2}

the product with

\mathcal{M}(S)

2I

❌ Not quite. Both off-diagonal entries are

i

, so both conjugate to

-i

; this matrix conjugated only one. Multiplying it on the right by

\mathcal{M}(S)

gives

\begin{pmatrix} 1 & 0 \\ i & 0 \end{pmatrix} \ne I

Show solution

A unitary operator satisfies $S^{-1} = S^*$ . The standard basis is orthonormal, so $\mathcal{M}(S^*)$ is the conjugate transpose of $\mathcal{M}(S)$ . Transposing the symmetric matrix changes nothing; conjugating replaces each $i$ by $-i$ and keeps the real factor $\tfrac{1}{\sqrt{2}}$ :

\mathcal{M}(S^{-1}) = \mathcal{M}(S^*) = \tfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & -i \\ -i & 1 \end{pmatrix}.

Check: $\tfrac12\begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix}\begin{pmatrix} 1 & -i \\ -i & 1 \end{pmatrix} = \tfrac12\begin{pmatrix} 1 + 1 & -i + i \\ i - i & 1 + 1 \end{pmatrix} = I$ .

Problem 2 · Eigenvalues Must Land on the Circle

Given: the same unitary $S$ with $\mathcal{M}(S) = \tfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix}$ . Find its eigenvalues (try the vectors $(1, 1)$ and $(1, -1)$ ).

✅ Correct! Both have

|\lambda|^2 = \tfrac{1 + 1}{2} = 1

, as every eigenvalue of a unitary operator must.

❌ Not quite. These have

|\lambda| = \sqrt{2}

, off the unit circle, which is impossible for a unitary operator: the factor

\tfrac{1}{\sqrt{2}}

was dropped.

❌ Not quite.

S(1, 1) = \tfrac{1 + i}{\sqrt{2}}(1, 1)

and

S(1, -1) = \tfrac{1 - i}{\sqrt{2}}(1, -1)

, and an operator on a

2

-dimensional space has at most two eigenvalues, so neither

1

nor

-1

is one. The restriction to

\pm 1

applies to real inner product spaces; this operator acts on

\mathbf{C}^2

❌ Not quite. These lie on the unit circle, but they are not eigenvalues:

S(1, 1) = \tfrac{1 + i}{\sqrt{2}}(1, 1)

, not

i(1, 1)

Show solution

S(1, 1) = \tfrac{1}{\sqrt{2}}(1 + i,\ i + 1) = \tfrac{1 + i}{\sqrt{2}}\,(1, 1), \qquad S(1, -1) = \tfrac{1}{\sqrt{2}}(1 - i,\ i - 1) = \tfrac{1 - i}{\sqrt{2}}\,(1, -1).

The vectors $(1, 1)$ and $(1, -1)$ form a basis of $\mathbf{C}^2$ , so these are all the eigenvalues. Each has $|\lambda|^2 = \tfrac{1^2 + 1^2}{2} = 1$ . The eigenvectors are orthogonal, $\langle (1, 1), (1, -1) \rangle = 1 - 1 = 0$ , so after dividing by $\sqrt{2}$ they give the orthonormal eigenbasis the complex characterization promises.

Problem 3 · A Reflection on $\mathbf{R}^2$

Given: $R \in \mathcal{L}(\mathbf{R}^2)$ with $\mathcal{M}(R) = \begin{pmatrix} \tfrac12 & \tfrac{\sqrt{3}}{2} \\ \tfrac{\sqrt{3}}{2} & -\tfrac12 \end{pmatrix}$ (standard basis). Its rows are orthonormal, so $R$ is unitary. Find its eigenvalues and an eigenvector for the eigenvalue $-1$ .

What are the eigenvalues?

Which vector is an eigenvector for $\lambda = -1$ ?

✅ Correct! On a real inner product space the eigenvalues of a unitary operator lie in

\{1, -1\}

, and here both occur, with orthogonal eigenvectors

\big(\tfrac{\sqrt{3}}{2}, \tfrac12\big)

and

\big(-\tfrac12, \tfrac{\sqrt{3}}{2}\big)

❌ Not quite. These are the eigenvalues of the rotation by

60\degree

. This matrix is symmetric, hence self-adjoint, so its eigenvalues are real.

❌ Not quite.

1

is an eigenvalue, but so is

-1

: the vector

\big(-\tfrac12, \tfrac{\sqrt{3}}{2}\big)

is sent to its negative.

❌ Not quite. The diagonal entries are guaranteed to be the eigenvalues only when the matrix is upper triangular, and this one is not. Also

\big|\tfrac12\big| \ne 1

, impossible for a unitary operator.

❌ Check the eigenvalues. A unitary operator on a real space can only have eigenvalues

\pm 1

❌ Not quite.

R\big(\tfrac{\sqrt{3}}{2}, \tfrac12\big) = \big(\tfrac{\sqrt{3}}{2}, \tfrac12\big)

: this is an eigenvector for

\lambda = 1

, the line of reflection.

❌ Not quite.

R\big(\tfrac12, \tfrac{\sqrt{3}}{2}\big) = (1, 0)

, which is not a multiple of

\big(\tfrac12, \tfrac{\sqrt{3}}{2}\big)

❌ Not quite.

R(1, -1) = \big(\tfrac{1 - \sqrt{3}}{2}, \tfrac{\sqrt{3} + 1}{2}\big)

, which is not a multiple of

(1, -1)

❌ Check the eigenvector. Apply

\mathcal{M}(R)

and compare with

-1

times the vector.

Show solution

Apply $\mathcal{M}(R)$ to the two candidate directions:

R\big(\tfrac{\sqrt{3}}{2}, \tfrac12\big) = \big(\tfrac{\sqrt{3}}{4} + \tfrac{\sqrt{3}}{4},\ \tfrac34 - \tfrac14\big) = \big(\tfrac{\sqrt{3}}{2}, \tfrac12\big),

R\big(-\tfrac12, \tfrac{\sqrt{3}}{2}\big) = \big(-\tfrac14 + \tfrac34,\ -\tfrac{\sqrt{3}}{4} - \tfrac{\sqrt{3}}{4}\big) = \big(\tfrac12, -\tfrac{\sqrt{3}}{2}\big) = -\big(-\tfrac12, \tfrac{\sqrt{3}}{2}\big).

So the eigenvalues are $1$ and $-1$ (two eigenvalues on a $2$ -dimensional space, so there are no others), both with $|\lambda| = 1$ . The two eigenvectors are orthonormal, so they form an orthonormal eigenbasis of $\mathbf{R}^2$ .

Problem 4 · Circle Eigenvalues Are Not Enough

Given: four operators on $\mathbf{C}^2$ , by their matrices in the standard basis. Every eigenvalue of each lies on the unit circle. Which one is unitary?

✅ Correct! Its columns are orthonormal, and it has the orthonormal eigenbasis

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

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

with eigenvalues

1

and

i

❌ Not quite. The eigenvalues

1

and

-1

are on the circle, but the second column

(1, -1)

has norm

\sqrt{2}

. The eigenvectors

(1, 0)

and

(1, -2)

are not orthogonal.

❌ Not quite. The only eigenvalue is

i

, on the circle, but the second column

(1, i)

has norm

\sqrt{2}

, and the eigenvectors are only the multiples of

(1, 0)

: there is no eigenbasis.

❌ Not quite. The eigenvalues are

\pm 1

(with eigenvectors

(2, 1)

and

(2, -1)

, not orthogonal), but the columns

(0, \tfrac12)

and

(2, 0)

have norms

\tfrac12

and

2

;

S(1, 0)

has norm

\tfrac12

Show solution

For (c), let $c_1 = \tfrac12(1 + i, 1 - i)$ and $c_2 = \tfrac12(1 - i, 1 + i)$ . Then

\|c_1\|^2 = \tfrac{|1 + i|^2 + |1 - i|^2}{4} = \tfrac{2 + 2}{4} = 1, \qquad \|c_2\|^2 = 1,

\langle c_1, c_2 \rangle = \tfrac14\big[(1 + i)\overline{(1 - i)} + (1 - i)\overline{(1 + i)}\big] = \tfrac14\big[(1 + i)^2 + (1 - i)^2\big] = \tfrac14(2i - 2i) = 0.

The columns form an orthonormal basis of $\mathbf{C}^2$ , so the operator is unitary. Its eigenvectors: $(1, 1) \mapsto \tfrac12(2, 2) = (1, 1)$ , eigenvalue $1$ ; $(1, -1) \mapsto \tfrac12(2i, -2i) = i(1, -1)$ , eigenvalue $i$ .

Each of (a), (b), (d) has a column of norm $\ne 1$ , so none is an isometry, even though all their eigenvalues lie on the unit circle: eigenvalues on the circle are necessary for unitary, not sufficient.

Solved: 0 / 4

Unitary Operators: S−1=S∗S^{-1} = S^*S−1=S∗, Eigenvalues on ∣λ∣=1|\lambda| = 1∣λ∣=1

IKey Formulas

IIThe Unit Circle in C\mathbf{C}C

IIIColumns, Rows and Eigenvalues of a 2×22 \times 22×2 Operator

IVQuiz Questions

Unitary Operators: $S^{-1} = S^*$ , Eigenvalues on $|\lambda| = 1$

IIThe Unit Circle in $\mathbf{C}$

IIIColumns, Rows and Eigenvalues of a $2 \times 2$ Operator