Linear-Algebra-Done-Right · Unit 13 · Video 5 · Interactive Practice

Commuting Operators Share a Diagonalizing Basis

IKey Formulas

Formula	Statement	Hypotheses
$ST = TS \iff \mathcal{M}(S)\mathcal{M}(T) = \mathcal{M}(T)\mathcal{M}(S)$	Operators commute exactly when their matrices commute	$S, T \in \mathcal{L}(V)$ , matrices with respect to one basis $v_1, \ldots, v_n$ of $V$
$p(T)\,q(T) = (pq)(T) = (qp)(T) = q(T)\,p(T)$	Polynomials in one operator commute; in particular $(\lambda I)T = \lambda T = T(\lambda I)$	$T \in \mathcal{L}(V)$ , $p, q \in \mathcal{P}(\mathbf{F})$
$T\big(E(\lambda, S)\big) \subseteq E(\lambda, S)$	Every eigenspace of $S$ is invariant under $T$	$ST = TS$ , $\lambda \in \mathbf{F}$
one basis diagonalizes both $\iff ST = TS$	Simultaneous diagonalizability	$S, T \in \mathcal{L}(V)$ both diagonalizable

Key Insight: On $E(\lambda, S)$ the operator $S$ is just $\lambda I$ , so every basis of that eigenspace works for $S$ . Commuting makes $T$ map $E(\lambda, S)$ into itself, and there $T$ is diagonalizable, so choose its eigenvectors. Putting these bases together over $V = E(\lambda_1, S) \oplus \cdots \oplus E(\lambda_m, S)$ gives one basis for both.

IIOrder Usually Matters

Two $2 \times 2$ matrices commute exactly when $AB - BA$ is the zero matrix.

pair entry A₁₁ 1

IIIEigenspaces of S Are Invariant Under T

For $S = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ , does $T$ keep each eigenline $E(3, S)$ and $E(1, S)$ inside itself?

eigenspace T t 1

IVChoosing One Basis for Both

$S = \operatorname{diag}(1, 1, 2)$ and $T(x, y, z) = (y, x, 3z)$ commute; which bases of $\mathbf{R}^3$ make both matrices diagonal?

basis u₁, u₂, u₃

💡 Any basis of the plane $E(1, S)$ makes $\mathcal{M}(S)$ diagonal, but $\mathcal{M}(T)$ is diagonal too only when that plane's basis consists of eigenvectors of $T|_{E(1, S)}$ , nonzero multiples of $(1, 1, 0)$ and $(1, -1, 0)$ .

VQuiz Questions

Problem 1 · Which One Commutes?

Given: $A = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}$ . Find the matrix $B$ with $AB = BA$ .

✅ Correct!

B = A + 2I

is a polynomial in

A

, so it commutes with

A

; indeed

AB = BA = \begin{pmatrix} 3 & 5 \\ 0 & 8 \end{pmatrix}

❌ The products differ.

AB = \begin{pmatrix} 1 & 1 \\ 2 & 0 \end{pmatrix}

but

BA = \begin{pmatrix} 0 & 2 \\ 1 & 1 \end{pmatrix}

❌ The products differ.

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

but

BA = \begin{pmatrix} 1 & 1 \\ 1 & 3 \end{pmatrix}

❌ Diagonal does not mean it commutes with everything. Two diagonal matrices commute, but

A

is not diagonal:

AB = \begin{pmatrix} 1 & 2 \\ 0 & 4 \end{pmatrix}

while

BA = \begin{pmatrix} 1 & 1 \\ 0 & 4 \end{pmatrix}

. Only scalar multiples of

I

commute with every matrix.

❌ Not quite. Compute

AB

and

BA

for each candidate and compare all four entries.

Show solution

Notice that $\begin{pmatrix} 3 & 1 \\ 0 & 4 \end{pmatrix} = A + 2I$ . With $p(z) = z$ and $q(z) = z + 2$ , the multiplicative property gives $p(A)q(A) = (pq)(A) = (qp)(A) = q(A)p(A)$ , so $A$ and $A + 2I$ commute. Directly:

A(A + 2I) = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 3 & 1 \\ 0 & 4 \end{pmatrix} = \begin{pmatrix} 3 & 5 \\ 0 & 8 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ 0 & 4 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}.

Each other candidate gives $AB \neq BA$ :

(a) $AB = \begin{pmatrix} 1 & 1 \\ 2 & 0 \end{pmatrix}$ , $BA = \begin{pmatrix} 0 & 2 \\ 1 & 1 \end{pmatrix}$ ;
(b) $AB = \begin{pmatrix} 2 & 1 \\ 2 & 2 \end{pmatrix}$ , $BA = \begin{pmatrix} 1 & 1 \\ 1 & 3 \end{pmatrix}$ ;
(d) $AB = \begin{pmatrix} 1 & 2 \\ 0 & 4 \end{pmatrix}$ , $BA = \begin{pmatrix} 1 & 1 \\ 0 & 4 \end{pmatrix}$ .

Problem 2 · The Standard Basis Is Not Enough

Given: $S, T \in \mathcal{L}(\mathbf{R}^2)$ with $\mathcal{M}(S) = 3I$ and $\mathcal{M}(T) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ in the standard basis. They commute, since $S = 3I$ . Find a basis of $\mathbf{R}^2$ with respect to which both $\mathcal{M}(S)$ and $\mathcal{M}(T)$ are diagonal.

✅ Correct!

E(3, S) = \mathbf{R}^2

, so any basis makes

\mathcal{M}(S) = \operatorname{diag}(3, 3)

; the eigenvectors

(1, 1)

and

(1, -1)

T

also give

\mathcal{M}(T) = \operatorname{diag}(1, -1)

❌ That diagonalizes $S$ only. In the standard basis

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

, which is not diagonal:

Te_1 = e_2

is not a multiple of

e_1

❌ $(1, 0)$ is not an eigenvector of $T$ .

T(1, 0) = (0, 1) = -(1, 0) + (1, 1)

, so in this basis

\mathcal{M}(T) = \begin{pmatrix} -1 & 0 \\ 1 & 1 \end{pmatrix}

❌ One eigenvalue makes it easier, not harder.

S

and

T

are diagonalizable and commute, so the theorem guarantees a common diagonalizing basis. Here

E(3, S)

is all of

\mathbf{R}^2

, so any eigenvector basis of

T

works.

❌ Not quite. Every basis diagonalizes

S = 3I

; look for eigenvectors of

T

Show solution

The only eigenvalue of $S = 3I$ is $3$ , and $E(3, S) = \mathbf{R}^2$ . Every nonzero vector is an eigenvector of $S$ , so $\mathcal{M}(S) = \operatorname{diag}(3, 3)$ with respect to every basis.

So the question is only which bases diagonalize $T$ . Since $T(x, y) = (y, x)$ ,

T(1, 1) = (1, 1), \qquad T(1, -1) = (-1, 1) = -(1, -1),

so $(1, 1), (1, -1)$ is a basis of eigenvectors of $T$ and $\mathcal{M}(T) = \operatorname{diag}(1, -1)$ . This is the proof of the theorem in miniature: inside the eigenspace $E(3, S)$ , choose a basis of eigenvectors of $T$ .

The standard basis fails because $Te_1 = e_2$ , and $(1, 0), (1, 1)$ fails because $T(1, 0) = (0, 1)$ is not a multiple of $(1, 0)$ .

Problem 3 · T on an Eigenspace of S

Given: $S = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$ and $T = \begin{pmatrix} 1 & 4 \\ 0 & 5 \end{pmatrix}$ , which commute. Find $E(3, S)$ and the scalar by which $T$ acts on it.

What is E(3, S)?

T restricted to E(3, S) equals which multiple of I?

✅ Correct!

S(1, 1) = (3, 3)

, and

T(1, 1) = (5, 5) = 5(1, 1)

stays in

E(3, S)

, as the invariance theorem promises.

❌ $(0, 1)$ is not an eigenvector of $S$ .

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

, which is not a multiple of

(0, 1)

. Solve

(S - 3I)v = 0

instead.

❌ That eigenspace belongs to $2$ .

S(1, 0) = (2, 0) = 2(1, 0)

, so

\operatorname{span}\big((1, 0)\big) = E(2, S)

❌ Check the eigenvector.

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

is not a multiple of

(1, -1)

; solve

(S - 3I)(x, y) = (-x + y, 0) = (0, 0)

❌ $3$ is the eigenvalue of $S$ , not of $T$ . Apply

T

itself:

T(1, 1) = (1 + 4, 5) = (5, 5)

❌ $1$ is the entry in the top-left corner. That entry records

Te_1 = e_1

, on the other eigenspace

E(2, S)

. On

E(3, S)

T(1, 1) = (5, 5)

❌ Not quite. Compute

T(1, 1)

and compare it with

(1, 1)

Show solution

Commuting: $ST = \begin{pmatrix} 2 & 13 \\ 0 & 15 \end{pmatrix} = TS$ . (In fact $T = 4S - 7I$ , a polynomial in $S$ .)

Eigenspace: $(S - 3I)(x, y) = (-x + y, 0)$ , which is $0$ exactly when $y = x$ . So $E(3, S) = \operatorname{span}\big((1, 1)\big)$ .

T on it: by the invariance theorem $T$ maps $E(3, S)$ into itself, so $T(1, 1)$ is a multiple of $(1, 1)$ :

T(1, 1) = (1 \cdot 1 + 4 \cdot 1,\ 5 \cdot 1) = (5, 5) = 5(1, 1).

Since $E(3, S)$ is one-dimensional, $T|_{E(3, S)} = 5I$ .

Problem 4 · Diagonalize Both at Once

Given: $S, T \in \mathcal{L}(\mathbf{R}^2)$ with $\mathcal{M}(S) = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$ and $\mathcal{M}(T) = \begin{pmatrix} 1 & 1 \\ 2 & 0 \end{pmatrix}$ in the standard basis. Find a basis that diagonalizes both, then $\mathcal{M}(T)$ in it.

Which basis diagonalizes both?

What is M(T) in that basis (in that order)?

✅ Correct!

ST = TS

, and

(1, 1)

(1, -2)

are eigenvectors of both:

\mathcal{M}(S) = \operatorname{diag}(5, 2)

and

\mathcal{M}(T) = \operatorname{diag}(2, -1)

❌ Neither matrix is diagonal there. In the standard basis the matrices are the given ones, and both have nonzero entries off the diagonal.

❌ $(1, -1)$ is not an eigenvector of $S$ .

S(1, -1) = (3, -1) = (1, 1) + 2(1, -1)

, so in this basis

\mathcal{M}(S) = \begin{pmatrix} 5 & 1 \\ 0 & 2 \end{pmatrix}

, upper triangular but not diagonal.

❌ $(1, 2)$ is not an eigenvector of $S$ .

S(1, 2) = (6, 8)

, which is not a multiple of

(1, 2)

❌ Check the basis. Test each vector: is

Sv

a multiple of

v

, and is

Tv

❌ That is $\mathcal{M}(S)$ .

S(1, 1) = 5(1, 1)

and

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

; for

T

compute

T(1, 1)

and

T(1, -2)

❌ The order is swapped. Column

k

\mathcal{M}(T)

records

Tu_k

T(1, 1) = 2(1, 1)

comes first, then

T(1, -2) = -(1, -2)

❌ Those are the diagonal entries in the standard basis. A change of basis changes the matrix; read the new diagonal from

T(1, 1)

and

T(1, -2)

❌ Not quite. Compute

T(1, 1)

and

T(1, -2)

Show solution

Step 1, commuting:

\begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 2 & 0 \end{pmatrix} = \begin{pmatrix} 6 & 4 \\ 8 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 2 & 0 \end{pmatrix}\begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix},

so $ST = TS$ (in fact $T = S - 3I$ ).

Step 2, eigenvectors of $S$ : $S(1, 1) = (5, 5) = 5(1, 1)$ and $S(1, -2) = (2, -4) = 2(1, -2)$ . Two eigenvalues on $\mathbf{R}^2$ make $S$ diagonalizable with $E(5, S) = \operatorname{span}\big((1, 1)\big)$ and $E(2, S) = \operatorname{span}\big((1, -2)\big)$ .

Step 3, $T$ on each eigenspace: each eigenspace is invariant under $T$ and one-dimensional, so its basis vector is an eigenvector of $T$ :

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

With respect to $(1, 1), (1, -2)$ : $\mathcal{M}(S) = \operatorname{diag}(5, 2)$ and $\mathcal{M}(T) = \operatorname{diag}(2, -1)$ .

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