Linear-Algebra-Done-Right · Unit 26 · Video 5 · Interactive Practice

Why ST − TS Can Never Be the Identity

IKey Formulas

Formula	Name	Hypotheses
$\operatorname{tr} T = \langle Te_1, e_1\rangle + \cdots + \langle Te_n, e_n\rangle$	Trace on an inner product space	$V$ a finite-dimensional inner product space, $e_1, \dots, e_n$ an orthonormal basis, $T \in \mathcal{L}(V)$
$\operatorname{tr}(\lambda T) = \lambda \operatorname{tr} T$ , $\ \operatorname{tr}(S + T) = \operatorname{tr} S + \operatorname{tr} T$	$\operatorname{tr}$ is a linear functional on $\mathcal{L}(V)$	$S, T \in \mathcal{L}(V)$ , $\lambda \in \mathbf{F}$
$\operatorname{tr}(ST) = \operatorname{tr}(TS)$	Order does not matter inside the trace	$S, T \in \mathcal{L}(V)$
$\operatorname{tr}(ST - TS) = 0 \neq \dim V = \operatorname{tr} I$	$ST - TS = I$ is impossible	$V$ finite-dimensional and nonzero

Key Insight: Every commutator $ST - TS$ has trace $0$ , while the identity has trace $\dim V$ ; in infinite dimensions there is no trace, and the equation can hold.

IIEvery Commutator Has Trace $0$

Whatever $S$ and $T$ are, the diagonal of $ST - TS$ adds up to $0$ , so it never matches $\operatorname{tr} I = 2$ .

S T₁₁ 0 T₁₂ 1 T₂₁ 2 T₂₂ 0

IIIDiagonal Entries $\langle Te_k, e_k\rangle$

Turning the orthonormal basis changes each $\langle Te_k, e_k\rangle$ but not their sum, $\operatorname{tr} T = 4$ for $T(x, y) = (x + 2y,\ 3y)$ .

angle of e₁ 30°

IVQuiz Questions

Problem 1 · Linearity of the Trace

Given: $S, T \in \mathcal{L}(V)$ with $\operatorname{tr} S = 4$ and $\operatorname{tr} T = -1$ — find $\operatorname{tr}(3S - 2T + ST - TS)$ .

✅ Correct!

3 \cdot 4 - 2 \cdot (-1) + \bigl(\operatorname{tr}(ST) - \operatorname{tr}(TS)\bigr) = 12 + 2 + 0 = 14

❌ Check the sign.

-2 \operatorname{tr} T = -2 \cdot (-1) = +2

❌ The commutator contributes nothing.

\operatorname{tr}(ST - TS) = \operatorname{tr}(ST) - \operatorname{tr}(TS) = 0

for all

S, T

❌ Keep the coefficients. Linearity gives

\operatorname{tr}(3S) = 3 \operatorname{tr} S = 12

and

\operatorname{tr}(-2T) = 2

Show solution

The trace is linear and $\operatorname{tr}(ST) = \operatorname{tr}(TS)$ :

\operatorname{tr}(3S - 2T + ST - TS) = 3\operatorname{tr} S - 2\operatorname{tr} T + \operatorname{tr}(ST) - \operatorname{tr}(TS) = 12 + 2 + 0 = 14.

Problem 2 · Which Is Not a Commutator?

Given: four $2 \times 2$ matrices — find the one that is NOT of the form $\mathcal{M}(ST - TS)$ for any $S, T \in \mathcal{L}(\mathbf{C}^2)$ (matrices with respect to the standard basis).

✅ Correct! Its trace is

1

, but every

ST - TS

has trace

0

❌ That one is a commutator. With

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

and

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

ST - TS = T - 0 = T

❌ That one is a commutator. With

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

and

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

ST - TS = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} - \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}

❌ That one is a commutator. Take

S = T

: then

ST - TS = 0

Show solution

Since $\operatorname{tr}(ST - TS) = \operatorname{tr}(ST) - \operatorname{tr}(TS) = 0$ , a matrix with nonzero trace is never a commutator. Option (d) has trace $1$ .

The other three have trace $0$ and are commutators: (a) $= \bigl[\begin{smallmatrix} 1 & 0 \\ 0 & 0 \end{smallmatrix}\bigr]\bigl[\begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix}\bigr] - \bigl[\begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix}\bigr]\bigl[\begin{smallmatrix} 1 & 0 \\ 0 & 0 \end{smallmatrix}\bigr]$ ; (b) $= \bigl[\begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix}\bigr]\bigl[\begin{smallmatrix} 0 & 0 \\ 1 & 0 \end{smallmatrix}\bigr] - \bigl[\begin{smallmatrix} 0 & 0 \\ 1 & 0 \end{smallmatrix}\bigr]\bigl[\begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix}\bigr]$ ; (c) $= SS - SS$ for any $S$ .

Problem 3 · Diagonal Entries in an Orthonormal Basis

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ defined by $T(x, y) = (2x + y,\ -x + 4y)$ , the Euclidean inner product, and the orthonormal basis $e_1 = \tfrac15(3, 4)$ , $e_2 = \tfrac15(-4, 3)$ — find $\langle Te_1, e_1\rangle$ and $\langle Te_2, e_2\rangle$ .

$\langle Te_1, e_1\rangle$

$\langle Te_2, e_2\rangle$

✅ Correct!

\tfrac{82}{25} + \tfrac{68}{25} = 6 = 2 + 4 = \operatorname{tr} T

❌ $e_1$ has length $1$ only with the factor $\tfrac15$ .

82 = \langle T(3, 4), (3, 4)\rangle

must be divided by

5^2

❌ $2$ is the entry for the standard basis vector $(1, 0)$ . Diagonal entries depend on the orthonormal basis; only their sum does not.

❌ $4$ is the entry for $(0, 1)$ . Compute

\langle Te_2, e_2\rangle

for

e_2 = \tfrac15(-4, 3)

, or subtract from

\operatorname{tr} T = 6

❌ The two entries need not be equal. They must add up to

\operatorname{tr} T = 6

❌ Not quite.

Te_1 = \tfrac15(10, 13)

; take its inner product with

\tfrac15(3, 4)

❌ Not quite. The two diagonal entries add up to

\operatorname{tr} T = 2 + 4 = 6

Show solution

Te_1 = \tfrac15(6 + 4,\ -3 + 16) = \tfrac15(10, 13), \qquad \langle Te_1, e_1\rangle = \tfrac1{25}(30 + 52) = \tfrac{82}{25}.

Te_2 = \tfrac15(-8 + 3,\ 4 + 12) = \tfrac15(-5, 16), \qquad \langle Te_2, e_2\rangle = \tfrac1{25}(20 + 48) = \tfrac{68}{25}.

The sum is $\tfrac{150}{25} = 6$ , which is $\operatorname{tr} T = 2 + 4$ computed from the standard basis.

Problem 4 · A Scaled Identity as a Commutator

Given: $V$ with $\dim V = 5$ , and $\lambda \in \mathbf{F}$ such that $ST - TS = \lambda I$ for some $S, T \in \mathcal{L}(V)$ — find which values of $\lambda$ are possible.

✅ Correct! Taking traces gives

0 = 5\lambda

, so

\lambda = 0

, and

\lambda = 0

does occur (take

S = T

❌ Take the trace of both sides.

\operatorname{tr}(ST - TS) = 0

while

\operatorname{tr}(\lambda I) = 5\lambda

❌ $5$ is $\operatorname{tr} I$ , not a possible $\lambda$ . The trace equation is

0 = 5\lambda

❌ $\lambda = 0$ is possible. Any operator commutes with itself, so

SS - SS = 0 = 0 \cdot I

Show solution

Apply the trace, which is linear and satisfies $\operatorname{tr}(ST) = \operatorname{tr}(TS)$ :

0 = \operatorname{tr}(ST) - \operatorname{tr}(TS) = \operatorname{tr}(ST - TS) = \operatorname{tr}(\lambda I) = \lambda \dim V = 5\lambda.

So $\lambda = 0$ . Conversely $\lambda = 0$ is attained, for instance by $S = T = I$ . The case $\lambda = 1$ is the video's statement that $ST - TS = I$ is impossible.

Solved: 0 / 4

Why ST − TS Can Never Be the Identity

IKey Formulas

IIEvery Commutator Has Trace 000

IIIDiagonal Entries ⟨Tek,ek⟩\langle Te_k, e_k\rangle⟨Tek​,ek​⟩

IVQuiz Questions

IIEvery Commutator Has Trace $0$

IIIDiagonal Entries $\langle Te_k, e_k\rangle$