Matrix Trace

Shilong Wang

397人浏览 · 2026-04-03 10:52:04

Shilong Wang · 2026-04-03 10:52:04 发布

The Matrix Trace: Identities and Proofs

A compact, self-contained reference covering every major trace identity — from basic linearity to the eigenvalue connection — with complete proofs.

Definition
Basic Linear Properties
Cyclic Property
Trace of $AA^T$
Trace and Eigenvalues
Summary & Caution

Definition

Let $(a_{ij}) \in F^{n \times n}$ , where $F$ is a field. The trace of $A$ is the sum of its diagonal entries:

$\operatorname{tr}(A) = \sum_{i=1}^{n} a_{ii}.$

Simple as it looks, this single number encodes surprising depth — it is invariant under similarity, additive, and equals the sum of all eigenvalues.

Basic Linear Properties

Trace satisfies three elementary properties that together make it a linear map $F^{n\times n} \to F$ .

Transpose Invariance

$\operatorname{tr}(A^T) = \operatorname{tr}(A).$

Proof. Since $A^T)_{ii} = a_{ii}$ for every $i$ , summing gives

$\operatorname{tr}(A^T) = \sum_{i=1}^n a_{ii} = \operatorname{tr}(A). \qquad \square$

Additivity

$\operatorname{tr}(A + B) = \operatorname{tr}(A) + \operatorname{tr}(B).$

Proof.

$\operatorname{tr}(A+B) = \sum_{i=1}^n (a_{ii}+b_{ii}) = \sum_{i=1}^n a_{ii} + \sum_{i=1}^n b_{ii} = \operatorname{tr}(A)+\operatorname{tr}(B). \qquad \square$

Homogeneity

$\operatorname{tr}(\lambda A) = \lambda\,\operatorname{tr}(A), \qquad \lambda \in F.$

Proof.

$\operatorname{tr}(\lambda A) = \sum_{i=1}^n \lambda a_{ii} = \lambda\sum_{i=1}^n a_{ii} = \lambda\operatorname{tr}(A). \qquad \square$

Combining additivity and homogeneity: for any $\alpha, \beta \in F$ ,

$\operatorname{tr}(\alpha A + \beta B) = \alpha\operatorname{tr}(A) + \beta\operatorname{tr}(B).$

Cyclic Property

Two-Factor Identity

$\operatorname{tr}(AB) = \operatorname{tr}(BA).$

Proof. The $i$ -th diagonal entry of $A B$ is $(AB)_{ii} = \sum_{j} a_{ij}b_{ji}$ , so

$\operatorname{tr}(AB) = \sum_i\sum_j a_{ij}b_{ji}.$

Swapping the order of summation and recognising $(BA)_{jj} = \sum_i b_{ji}a_{ij}$ yields

$\operatorname{tr}(AB) = \sum_j\sum_i b_{ji}a_{ij} = \sum_j (BA)_{jj} = \operatorname{tr}(BA). \qquad \square$

Transpose Variants

Because $AB)^T = B^T A^T$ and trace is transpose-invariant, one obtains the chain

$\operatorname{tr}(AB) = \operatorname{tr}(BA) = \operatorname{tr}(B^TA^T) = \operatorname{tr}(A^TB^T).$

Similarity Invariance

If $P$ is invertible, then

$\operatorname{tr}(P^{-1}AP) = \operatorname{tr}(A).$

Proof. Apply the two-factor identity twice:

$\operatorname{tr}(P^{-1}AP) = \operatorname{tr}(APP^{-1}) = \operatorname{tr}(A). \qquad \square$

More generally, trace is fully cyclic over any number of factors:

$\operatorname{tr}(A_1 A_2 \cdots A_k) = \operatorname{tr}(A_2 \cdots A_k A_1).$

Trace of $AA^T$

For real matrices $(a_{ij}) \in \mathbb{R}^{n\times n}$ ,

$\operatorname{tr}(AA^T) = \sum_{i=1}^n\sum_{j=1}^n a_{ij}^2.$

Proof. The $i$ -th diagonal entry of $AA^T$ is

$(AA^T)_{ii} = \sum_j a_{ij}(A^T)_{ji} = \sum_j a_{ij}^2.$

Summing over all $i$ gives

$\operatorname{tr}(AA^T) = \sum_i\sum_j a_{ij}^2. \qquad \square$

In other words, $\operatorname{tr}(AA^T)$ is exactly the sum of squares of all entries of $A$ — it is the squared Frobenius norm $A\|_F^2$ . For complex matrices the analogue replaces squaring with the modulus:

$\operatorname{tr}(AA^*) = \sum_{i,j}|a_{ij}|^2.$

Trace and Eigenvalues

If $\lambda_1, \dots, \lambda_n$ are the eigenvalues of $A$ counted with algebraic multiplicity, then

$\operatorname{tr}(A) = \sum_{i=1}^n \lambda_i.$

Proof. Over an algebraic closure, write $A = PJP^{-1}$ where $J$ is the Jordan normal form of $A$ . By similarity invariance, $\operatorname{tr}(A) = \operatorname{tr}(J)$ . The diagonal entries of $J$ are precisely the eigenvalues of $A$ , so

$\operatorname{tr}(J) = \sum_{i=1}^n \lambda_i. \qquad \square$

This identity shows that the trace is a basis-free invariant of the linear operator — not just a convenient coordinate-dependent sum.

Summary & Caution

For $\in F^{n\times n}$ and $\lambda \in F$ , the complete set of identities:

Identity	Formula
Definition	$\operatorname{tr}(A)=\sum_i a_{ii}$
Transpose	$\operatorname{tr}(A^T)=\operatorname{tr}(A)$
Additivity	$\operatorname{tr}(A+B)=\operatorname{tr}(A)+\operatorname{tr}(B)$
Homogeneity	$\operatorname{tr}(\lambda A)=\lambda\operatorname{tr}(A)$
Cyclic	$\operatorname{tr}(AB)=\operatorname{tr}(BA)$
Similarity	$\operatorname{tr}(P^{-1}AP)=\operatorname{tr}(A)$
Frobenius norm	$\operatorname{tr}(AA^T)=\sum_{i,j}a_{ij}^2$
Eigenvalues	$\operatorname{tr}(A)=\sum_i\lambda_i$

Caution — cyclic ≠ commutative. Trace is cyclic, not fully commutative. While
$\operatorname{tr}(ABC) = \operatorname{tr}(BCA) = \operatorname{tr}(CAB),$
in general $\operatorname{tr}(ABC) \neq \operatorname{tr}(ACB)$ . Cycling preserves the trace; arbitrary reordering does not.