Linear-Algebra-Done-Right · Unit 26 · Video 4 · Interactive Practice

The Trace Equals the Sum of the Eigenvalues

IKey Formulas

Formula	Name	Hypotheses
$\operatorname{tr} A = A_{1,1} + \cdots + A_{n,n}$	Trace of a square matrix	$A$ an $n \times n$ matrix
$\operatorname{tr}(AB) = \displaystyle\sum_{j=1}^{m}\sum_{k=1}^{n} A_{j,k}B_{k,j} = \operatorname{tr}(BA)$	Swap the order of the double sum	$A$ is $m \times n$ , $B$ is $n \times m$
$\operatorname{tr} T = \operatorname{tr}\mathcal{M}(T)$ , the same for every basis	Trace of an operator	$T \in \mathcal{L}(V)$ ; $\mathcal{M}(T, (u)) = C^{-1}\mathcal{M}(T, (v))\,C$
$\operatorname{tr} T = \lambda_1 + \cdots + \lambda_n$ ; $\ q(z) = z^n - (\operatorname{tr} T)z^{n-1} + \cdots$	Sum of eigenvalues; characteristic polynomial	$\mathbf{F} = \mathbf{C}$ , $n = \dim V$ , each eigenvalue repeated as often as its multiplicity

Key Insight: A change of basis replaces $\mathcal{M}(T)$ by $C^{-1}\mathcal{M}(T)C$ , and $\operatorname{tr}\bigl((C^{-1}A)C\bigr) = \operatorname{tr}\bigl(C(C^{-1}A)\bigr) = \operatorname{tr} A$ , so the trace belongs to the operator.

IIThe Diagonal Changes, Its Sum Does Not

For $T(x, y) = (4x - y,\ 2x + y)$ , a change of basis moves the diagonal entries of $\mathcal{M}(T)$ but keeps their sum $5 = 2 + 3$ .

s 1 t 1

III $\operatorname{tr}(AB) = \operatorname{tr}(BA)$ by Swapping a Double Sum

Every product $A_{j,k}B_{k,j}$ is counted once in $\operatorname{tr}(AB)$ , by rows, and once in $\operatorname{tr}(BA)$ , by columns.

💡 $AB$ is $2 \times 2$ and $BA$ is $3 \times 3$ , yet their traces agree; $AB$ and $BA$ themselves need not be equal even when both are square.

IVQuiz Questions

Problem 1 · Count the Multiplicity

Given: $T \in \mathcal{L}(\mathbf{C}^4)$ whose eigenvalues are $3$ with multiplicity $2$ , $-1$ with multiplicity $1$ , and $2i$ with multiplicity $1$ — find $\operatorname{tr} T$ .

✅ Correct!

3 + 3 + (-1) + 2i = 5 + 2i

❌ Each eigenvalue counts as often as its multiplicity.

3

appears twice:

3 + 3 - 1 + 2i

❌ The trace of a complex operator can be non-real. Keep the eigenvalue

2i

in the sum.

❌ Check the sign of $-1$ . The sum uses the eigenvalues themselves, not their absolute values.

Show solution

On a complex vector space, $\operatorname{tr} T$ is the sum of the eigenvalues, each repeated as often as its multiplicity (the multiplicities add up to $\dim \mathbf{C}^4 = 4$ ):

\operatorname{tr} T = 3 + 3 + (-1) + 2i = 5 + 2i.

Problem 2 · Trace from the Characteristic Polynomial

Given: $T \in \mathcal{L}(\mathbf{C}^3)$ with characteristic polynomial $z^3 - 4z^2 + z + 6$ — find $\operatorname{tr} T$ .

✅ Correct!

\operatorname{tr} T

is the negative of the coefficient of

z^2

; indeed

z^3 - 4z^2 + z + 6 = (z + 1)(z - 2)(z - 3)

and

-1 + 2 + 3 = 4

❌ Mind the sign.

(z - \lambda_1)(z - \lambda_2)(z - \lambda_3) = z^3 - (\lambda_1 + \lambda_2 + \lambda_3)z^2 + \cdots

, so the trace is minus the coefficient.

❌ The constant term is $(-1)^3\lambda_1\lambda_2\lambda_3$ . It records the product of the eigenvalues, not their sum.

❌ That is the coefficient of $z$ . The trace sits in the coefficient of

z^{n-1} = z^2

Show solution

With $n = 3$ , the characteristic polynomial is $z^3 - (\lambda_1 + \lambda_2 + \lambda_3)z^2 + \cdots$ , so $\operatorname{tr} T = -(-4) = 4$ .

Check: $z^3 - 4z^2 + z + 6 = (z + 1)(z - 2)(z - 3)$ , so the eigenvalues are $-1$ , $2$ , $3$ , and $-1 + 2 + 3 = 4$ .

Problem 3 · Traces of Products of Different Sizes

Given: a $2 \times 3$ matrix $A$ and a $3 \times 2$ matrix $B$ with $AB = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ — find $\operatorname{tr}(BA)$ .

✅ Correct!

\operatorname{tr}(BA) = \operatorname{tr}(AB) = 1 + 4 = 5

❌ The identity holds for an $m \times n$ $A$ and an $n \times m$ $B$ .

BA

is unknown, but its trace is not.

❌ That adds every entry of $AB$ . The trace adds only the diagonal entries.

❌ That is only one diagonal entry of $AB$ . The trace adds both:

1 + 4

Show solution

For $A$ of size $m \times n$ and $B$ of size $n \times m$ ,

\operatorname{tr}(AB) = \sum_{j=1}^{m}\sum_{k=1}^{n} A_{j,k}B_{k,j} = \sum_{k=1}^{n}\sum_{j=1}^{m} B_{k,j}A_{j,k} = \operatorname{tr}(BA).

Here $m = 2$ , $n = 3$ , and $\operatorname{tr}(AB) = 1 + 4 = 5$ , so $\operatorname{tr}(BA) = 5$ although $BA$ itself is not determined by $AB$ .

Problem 4 · Trace of an Operator on Polynomials

Given: $T \in \mathcal{L}(\mathcal{P}_2(\mathbf{R}))$ defined by $Tp = xp' + p(1)$ , where $p(1)$ is the constant polynomial — find $\operatorname{tr} T$ .

✅ Correct! In the basis

1, x, x^2

T1 = 1

Tx = 1 + x

Tx^2 = 1 + 2x^2

, so the diagonal is

1, 1, 2

❌ The term $p(1)$ also reaches the diagonal.

T1 = 0 + 1 = 1

, so the first diagonal entry is

1

, not

0

❌ That adds every entry of $\mathcal{M}(T)$ . The constants

1

Tx

and

Tx^2

lie above the diagonal.

❌ Not quite. Write

T1

Tx

Tx^2

in the basis

1, x, x^2

and add the coefficient of

1

T1

, of

x

Tx

, of

x^2

Tx^2

Show solution

Use the basis $1, x, x^2$ :

T1 = 0 + 1 = 1, \qquad Tx = x + 1, \qquad Tx^2 = 2x^2 + 1.

\mathcal{M}(T) = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}, \qquad \operatorname{tr} T = 1 + 1 + 2 = 4.

The matrix is upper triangular, so the eigenvalues are $1, 1, 2$ , and their sum is $4$ as well.

Solved: 0 / 4

The Trace Equals the Sum of the Eigenvalues

IKey Formulas

IIThe Diagonal Changes, Its Sum Does Not

IIItr⁡(AB)=tr⁡(BA)\operatorname{tr}(AB) = \operatorname{tr}(BA)tr(AB)=tr(BA) by Swapping a Double Sum

IVQuiz Questions

III $\operatorname{tr}(AB) = \operatorname{tr}(BA)$ by Swapping a Double Sum