Linear-Algebra-Done-Right · Unit 13 · Video 6 · Interactive Practice

Commuting Operators Over C: One Upper-Triangular Basis

IKey Formulas

Formula	Name	What it says
$ST = TS \Rightarrow Sv = \lambda v$ and $Tv = \mu v$ for some $v \neq 0$	Common eigenvector	For $V \neq \{0\}$ finite-dimensional complex: $T$ maps $E(\lambda, S)$ into itself, and $T\|_{E(\lambda, S)}$ has an eigenvector
$ST = TS \Rightarrow \mathcal{M}(S)$ and $\mathcal{M}(T)$ upper triangular for one basis	Common upper-triangular basis	Induction on $\dim V$ : $v_1$ a common eigenvector, $V = \operatorname{span}(v_1) \oplus W$ , $P(av_1 + w) = w$ , then $\hat{S}w = P(Sw)$ and $\hat{T}w = P(Tw)$ commute on $W$
$\mathcal{M}(S + T)_{k,k} = \mathcal{M}(S)_{k,k} + \mathcal{M}(T)_{k,k}$ , $\ \mathcal{M}(ST)_{k,k} = \mathcal{M}(S)_{k,k}\,\mathcal{M}(T)_{k,k}$	Eigenvalues of $S + T$ and $ST$	In the common basis every eigenvalue of $S + T$ is $\lambda + \mu$ , and every eigenvalue of $ST$ is $\lambda\mu$ , with $\lambda$ , $\mu$ in the same diagonal position $k$

Key Insight: The pairing is by diagonal position. With $\mathcal{M}(S) = \operatorname{diag}(1, 2)$ and $\mathcal{M}(T) = \operatorname{diag}(10, 20)$ , the eigenvalues of $S + T$ are $11$ and $22$ , while $1 + 20 = 21$ is not one.

IIWithout $ST = TS$ There Is No Rule

$S$ and $T$ each have only the eigenvalue $0$ , yet $S + T$ and $ST$ can have nonzero eigenvalues unless they commute.

Pair t 1

IIIBuilding the Common Upper-Triangular Basis on $\mathbf{C}^3$

For a commuting pair on $\mathbf{C}^3$ , the proof's steps produce one basis making both matrices upper triangular.

Step 1

Step 1 — An eigenspace of $S$ . The pair on $\mathbf{C}^3$ :

S = \begin{pmatrix} -1 & 3 & 3 \\ -3 & 5 & 3 \\ -3 & 3 & 5 \end{pmatrix}, \quad T = \begin{pmatrix} 0 & 1 & 2 \\ -1 & 2 & 2 \\ -2 & 2 & 3 \end{pmatrix}, \quad ST = TS = \begin{pmatrix} -9 & 11 & 13 \\ -11 & 13 & 13 \\ -13 & 13 & 15 \end{pmatrix}.

Every row of $S - 2I$ is $(-3, 3, 3)$ , so $E(2, S) = \{x : x_1 = x_2 + x_3\} = \operatorname{span}(a, b)$ with $a = (1, 1, 0)$ and $b = (1, 0, 1)$ .

Step 2 — $T$ maps $E(2, S)$ into itself. $Ta = (1, 1, 0) = a$ and $Tb = (2, 1, 1) = a + b$ , so with respect to $a, b$

\mathcal{M}\big(T|_{E(2, S)}\big) = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}.

Step 3 — An eigenvector of $T|_{E(2, S)}$ . $(T - I)(xa + yb) = ya$ , which is $0$ exactly when $y = 0$ . So $v_1 = a = (1, 1, 0)$ gives $Sv_1 = 2v_1$ and $Tv_1 = v_1$ .

$b$ is an eigenvector of $S$ but not of $T$ : $Tb = a + b \notin \operatorname{span}(b)$ .

Step 4 — Drop the $v_1$ component. $V = \operatorname{span}(v_1) \oplus W$ with $W = \operatorname{span}(e_2, e_3)$ , and $P(x_1, x_2, x_3) = (0,\ x_2 - x_1,\ x_3)$ .

$\hat{S}e_2 = P(3, 5, 3) = 2e_2 + 3e_3$ , $\ \hat{S}e_3 = P(3, 3, 5) = 5e_3$ , $\ \hat{T}e_2 = P(1, 2, 2) = e_2 + 2e_3$ , $\ \hat{T}e_3 = P(2, 2, 3) = 3e_3$ .

\mathcal{M}(\hat{S})\mathcal{M}(\hat{T}) = \begin{pmatrix} 2 & 0 \\ 3 & 5 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 13 & 15 \end{pmatrix} = \mathcal{M}(\hat{T})\mathcal{M}(\hat{S})

Step 5 — Finish inside $W$ . $\hat{S}e_3 = 5e_3$ and $\hat{T}e_3 = 3e_3$ , so take $v_2 = e_3$ and $v_3 = e_2$ . With respect to $v_1, v_2, v_3$ :

\mathcal{M}(S) = \begin{pmatrix} 2 & 3 & 3 \\ 0 & 5 & 3 \\ 0 & 0 & 2 \end{pmatrix}, \quad \mathcal{M}(T) = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 3 & 2 \\ 0 & 0 & 1 \end{pmatrix}.

So $S + T$ has eigenvalues $2 + 1 = 3$ and $5 + 3 = 8$ , and $ST$ has eigenvalues $2 \cdot 1 = 2$ and $5 \cdot 3 = 15$ .

💡 $S$ is diagonalizable but $T$ is not: $(T - I)x = 0$ only for $x \in \operatorname{span}(1, 1, 0)$ and $(T - 3I)x = 0$ only for $x \in \operatorname{span}(1, 1, 1)$ , so no diagonalizing basis serves both.

IVWhich Sums Are Eigenvalues?

Is every $\lambda + \mu$ an eigenvalue of $S + T$ , or only the pairs sharing a diagonal position?

Pair Operator

VQuiz Questions

Problem 1 · A Common Eigenvector on $\mathbf{C}^2$

Given: $S, T \in \mathcal{L}(\mathbf{C}^2)$ with $S = \begin{pmatrix} 3 & 1 \\ -1 & 1 \end{pmatrix}$ and $T = \begin{pmatrix} 2 & -1 \\ 1 & 4 \end{pmatrix}$ , which satisfy $ST = TS = \begin{pmatrix} 7 & 1 \\ -1 & 5 \end{pmatrix}$ — find a common eigenvector, then the eigenvalues of $ST$ .

Which vector is an eigenvector of both S and T?

What are the eigenvalues of ST?

✅ Correct!

S(1, -1) = 2(1, -1)

and

T(1, -1) = 3(1, -1)

, and

ST = 6I + N

with

N^2 = 0

has only the eigenvalue

2 \cdot 3 = 6

❌ Check $S$ first.

S(1, 1) = (4, 0)

, which is not a multiple of

(1, 1)

❌ Check $S$ first.

S(1, 0) = (3, -1)

, which is not a multiple of

(1, 0)

❌ Diagonalizability is not needed.

ST = TS

on a complex space already guarantees a common eigenvector; neither operator has to be diagonalizable.

❌ Not quite. Look inside the eigenspace of

S

❌ That is $S + T$ . Here

S + T = 5I

; the question asks for

ST

, whose eigenvalues are products

\lambda\mu

❌ Those are the eigenvalues of $S$ and of $T$ separately. The eigenvalue of

ST

is the product of the diagonal entries in one position of the common upper-triangular basis.

❌ The diagonal only counts in an upper-triangular matrix. In the standard basis

ST = \begin{pmatrix} 7 & 1 \\ -1 & 5 \end{pmatrix}

is not upper triangular, and

ST - 7I = \begin{pmatrix} 0 & 1 \\ -1 & -2 \end{pmatrix}

is invertible.

❌ Not quite. Write

S

and

T

in terms of

N = S - 2I

Show solution

Let $N = S - 2I = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$ . Then $N^2 = 0$ , $S = 2I + N$ and $T = 3I - N$ .

Eigenvalues of $S$ and $T$ . If $Sv = \lambda v$ with $v \neq 0$ , then $Nv = (\lambda - 2)v$ , so $0 = N^2 v = (\lambda - 2)^2 v$ and $\lambda = 2$ . Likewise $T$ has only the eigenvalue $3$ .

Common eigenvector. $E(2, S) = \operatorname{null} N = \operatorname{span}\big((1, -1)\big)$ , and

S(1, -1) = (2, -2) = 2(1, -1), \qquad T(1, -1) = (3, -3) = 3(1, -1).

Neither operator is diagonalizable (each eigenspace is one-dimensional), yet $(1, -1)$ is a common eigenvector.

Eigenvalues of $ST$ . $ST = (2I + N)(3I - N) = 6I + N - N^2 = 6I + N$ . If $STv = \lambda v$ with $v \neq 0$ , then $Nv = (\lambda - 6)v$ and $(\lambda - 6)^2 v = 0$ , so $\lambda = 6 = 2 \cdot 3$ . In the basis $(1, -1), (1, 0)$ both matrices are upper triangular, with diagonals $2, 2$ and $3, 3$ , and the diagonal of $\mathcal{M}(ST)$ is $6, 6$ .

Problem 2 · A Pair That Does Not Commute

Given: $S = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $T = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$ on $\mathbf{C}^2$ ; each has $1$ as its only eigenvalue — find the eigenvalues of $S + T$ .

✅ Correct!

(S + T)(1, 1) = 3(1, 1)

and

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

; neither

3

nor

1

equals

1 + 1

, which is allowed because

ST \neq TS

❌ The theorem needs $ST = TS$ . Here

ST = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}

and

TS = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}

differ, so nothing ties the eigenvalues of

S + T

1 + 1

❌ Those belong to $ST$ .

ST = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}

is the product; the question asks for the sum

S + T = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}

❌ Test them.

S + T - 2I = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}

and

S + T

are both invertible, so neither

2

nor

0

is an eigenvalue.

Show solution

$S$ and $T$ . $(S - I)^2 = 0$ , so $Sv = \lambda v$ with $v \neq 0$ gives $(\lambda - 1)^2 v = 0$ and $\lambda = 1$ ; the same holds for $T$ .

The sum. $S + T = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ , and

(S + T)(1, 1) = (3, 3) = 3(1, 1), \qquad (S + T)(1, -1) = (1, -1).

With respect to the basis $(1, 1), (1, -1)$ the matrix of $S + T$ is $\operatorname{diag}(3, 1)$ , so its eigenvalues are exactly $3$ and $1$ .

Why no contradiction. $ST = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \neq \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} = TS$ . The rule "every eigenvalue of $S + T$ is $\lambda + \mu$ " is proved only for commuting operators, and this pair shows it can fail without that hypothesis.

Problem 3 · One Basis for Both, Then $ST$

Given: commuting operators $S, T \in \mathcal{L}(\mathbf{C}^3)$ — find a basis giving both upper-triangular matrices, then the eigenvalues of $ST$ .

S = \begin{pmatrix} 2 & 0 & 0 \\ 1 & 2 & 0 \\ 0 & 1 & 3 \end{pmatrix}, \qquad T = \begin{pmatrix} -4 & 0 & 0 \\ 0 & -4 & 0 \\ 1 & 1 & -3 \end{pmatrix}

Which basis makes both matrices upper triangular?

What are the eigenvalues of ST?

✅ Correct! Reversing the basis turns both lower-triangular matrices upper triangular, with diagonals

3, 2, 2

and

-3, -4, -4

, so

ST

has eigenvalues

-9

and

-8

❌ Those matrices are lower triangular.

Se_1 = 2e_1 + e_2

has a component along

e_2

, so

Se_1 \notin \operatorname{span}(e_1)

❌ Check the second vector.

Se_1 = 2e_1 + e_2

, and

e_2

is not in

\operatorname{span}(e_3, e_1)

❌ Check the first vector.

Se_2 = 2e_2 + e_3

is not a multiple of

e_2

, so

e_2

cannot come first.

❌ Not quite. The first basis vector must be a common eigenvector.

❌ Not every product counts. Only

\lambda\mu

with

\lambda

and

\mu

in the same diagonal position:

-6 = 2 \cdot (-3)

and

-12 = 3 \cdot (-4)

pair different positions.

❌ Those are the eigenvalues of $S + T$ . The diagonal of

\mathcal{M}(S + T)

0, -2, -2

; for

ST

multiply the diagonal entries instead.

❌ Those pair the wrong positions.

2 \cdot (-3)

and

3 \cdot (-4)

mix positions; the diagonal pairs are

(3, -3)

and

(2, -4)

❌ Not quite. Multiply the diagonals position by position.

Show solution

Commuting. $T = S^2 - 4S$ , so $ST = TS = \begin{pmatrix} -8 & 0 & 0 \\ -4 & -8 & 0 \\ 3 & -1 & -9 \end{pmatrix}$ . $S$ is not diagonalizable: $(S - 2I)x = 0$ forces $x_1 = 0$ and $x_2 = -x_3$ , a one-dimensional space.

The basis. Use the span condition with $v_1, v_2, v_3 = e_3, e_2, e_1$ :

$Se_3 = 3e_3$ and $Te_3 = -3e_3$ : both in $\operatorname{span}(e_3)$ .
$Se_2 = 2e_2 + e_3$ and $Te_2 = -4e_2 + e_3$ : both in $\operatorname{span}(e_3, e_2)$ .
$Se_1$ and $Te_1$ lie in $\operatorname{span}(e_3, e_2, e_1) = \mathbf{C}^3$ .

\mathcal{M}(S) = \begin{pmatrix} 3 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}, \qquad \mathcal{M}(T) = \begin{pmatrix} -3 & 1 & 1 \\ 0 & -4 & 0 \\ 0 & 0 & -4 \end{pmatrix}.

The order $e_1, e_2, e_3$ fails because $Se_1 = 2e_1 + e_2$ ; the orders $e_3, e_1, e_2$ and $e_2, e_1, e_3$ fail at $Se_1$ and $Se_2$ respectively.

Eigenvalues of $ST$ . $\mathcal{M}(ST) = \mathcal{M}(S)\mathcal{M}(T)$ is upper triangular with diagonal

3 \cdot (-3) = -9, \qquad 2 \cdot (-4) = -8, \qquad 2 \cdot (-4) = -8,

so the eigenvalues of $ST$ are $-9$ and $-8$ .

Problem 4 · Differentiation and a Shift on $\mathcal{P}_2(\mathbf{C})$

Given: on $\mathcal{P}_2(\mathbf{C})$ , $Dp = p'$ and $(Tp)(x) = p(x + 1)$ . Both are linear and $DT = TD$ , since $\frac{d}{dx}\,p(x + 1) = p'(x + 1)$ — find the common eigenvectors, then the eigenvalues of $D + T$ .

Which polynomials are common eigenvectors of D and T?

What are the eigenvalues of D + T?

✅ Correct! In the basis

1, x, x^2

both matrices are upper triangular with diagonals

0, 0, 0

and

1, 1, 1

, so

D + T

has only the eigenvalue

0 + 1 = 1

❌ $x$ is an eigenvector of neither.

Dx = 1 \notin \operatorname{span}(x)

and

Tx = x + 1 \notin \operatorname{span}(x)

❌ Check $T$ .

T(1 + x) = 2 + x

, which is not a multiple of

1 + x

❌ Check $D$ .

Dx^2 = 2x

, which is not a multiple of

x^2

❌ Not quite. Find the eigenvectors of

D

first.

❌ That is the eigenvalue of $D$ alone.

T

contributes its eigenvalue

1

in every diagonal position.

❌ That lists the eigenvalues of $D$ and $T$ . The eigenvalues of

D + T

are diagonal sums

0 + 1

;

0

itself is not one, since

(D + T)p = 0

forces

p = 0

❌ Over $\mathbf{C}$ every operator has an eigenvalue.

\mathcal{P}_2(\mathbf{C})

is finite-dimensional and nonzero, so

D + T

has one.

❌ Not quite. Add the diagonals in the basis

1, x, x^2

Show solution

Matrices. With respect to $1, x, x^2$ : $D1 = 0$ , $Dx = 1$ , $Dx^2 = 2x$ , and $T1 = 1$ , $Tx = 1 + x$ , $Tx^2 = 1 + 2x + x^2$ , so

\mathcal{M}(D) = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{pmatrix}, \qquad \mathcal{M}(T) = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix}.

Common eigenvectors. $D^3 = 0$ , so $Dp = \lambda p$ with $p \neq 0$ gives $\lambda^3 p = 0$ and $\lambda = 0$ ; then $p' = 0$ and $p$ is a nonzero constant. Every nonzero constant $c$ satisfies $Dc = 0$ and $Tc = c$ , so the common eigenvectors are exactly the nonzero constants. Neither operator is diagonalizable.

Eigenvalues of $D + T$ . Both matrices are upper triangular in the same basis, so $\mathcal{M}(D + T)$ is upper triangular with diagonal $0 + 1, 0 + 1, 0 + 1$ . Its diagonal entries are its eigenvalues, so $1$ is the only eigenvalue of $D + T$ .

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Commuting Operators Over C: One Upper-Triangular Basis

IKey Formulas

IIWithout ST=TSST = TSST=TS There Is No Rule

IIIBuilding the Common Upper-Triangular Basis on C3\mathbf{C}^3C3

IVWhich Sums Are Eigenvalues?

VQuiz Questions

IIWithout $ST = TS$ There Is No Rule

IIIBuilding the Common Upper-Triangular Basis on $\mathbf{C}^3$