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.

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Contents

1. Definition
2. Basic Linear Properties
3. Cyclic Property
4. Trace of $A{A}^T$
5. Trace and Eigenvalues
6. Summary & Caution

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Definition

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

$$\mathrm{tr}(A)=\sum _{i=1}^n{a}_{ii}.$$

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

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Basic Linear Properties

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

Transpose Invariance

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

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

$$\mathrm{tr}(A^T)=\sum _{i=1}^n{a}_{ii}=\mathrm{tr}(A).\square$$

Additivity

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

Proof.

$$\mathrm{tr}(A+B)=\sum _{i=1}^n(a_{ii}+b_{ii})=\sum _{i=1}^n{a}_{ii}+\sum _{i=1}^n{b}_{ii}=\mathrm{tr}(A)+\mathrm{tr}(B).\square$$

Homogeneity

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

Proof.

$$\mathrm{tr}(\lambda A)=\sum _{i=1}^n\lambda a_{ii}=\lambda \sum _{i=1}^n{a}_{ii}=\lambda \mathrm{tr}(A).\square$$

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

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

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Cyclic Property

Two-Factor Identity

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

Proof. The $i$-th diagonal entry of $AB$ is $(AB{)}_{ii}={\sum }_j{a}_{ij}{b}_{ji}$, so

$$\mathrm{tr}(AB)=\sum _i\sum _j{a}_{ij}{b}_{ji}.$$

Swapping the order of summation and recognising $(BA{)}_{jj}={\sum }_i{b}_{ji}{a}_{ij}$ yields

$$\mathrm{tr}(AB)=\sum _j\sum _i{b}_{ji}{a}_{ij}=\sum _j(BA{)}_{jj}=\mathrm{tr}(BA).\square$$

Transpose Variants

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

$$\mathrm{tr}(AB)=\mathrm{tr}(BA)=\mathrm{tr}(B^T{A}^T)=\mathrm{tr}(A^T{B}^T).$$

Similarity Invariance

If $P$ is invertible, then

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

Proof. Apply the two-factor identity twice:

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

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

$$\mathrm{tr}(A_1{A}_2\cdots A_k)=\mathrm{tr}(A_2\cdots A_k{A}_1).$$

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Trace of $A{A}^T$

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

$$\mathrm{tr}(A{A}^T)=\sum _{i=1}^n\sum _{j=1}^n{a}_{ij}^2.$$

Proof. The $i$-th diagonal entry of $A{A}^T$ is

$$(A{A}^T{)}_{ii}=\sum _j{a}_{ij}(A^T{)}_{ji}=\sum _j{a}_{ij}^2.$$

Summing over all $i$ gives

$$\mathrm{tr}(A{A}^T)=\sum _i\sum _j{a}_{ij}^2.\square$$

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

$$\mathrm{tr}(A{A}^{\ast })=\sum _{i,j}|a_{ij}{|}^2.$$

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Trace and Eigenvalues

If ${\lambda }_1,\dots ,{\lambda }_n$ are the eigenvalues of $A$ counted with algebraic multiplicity, then

$$\mathrm{tr}(A)=\sum _{i=1}^n{\lambda }_i.$$

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

$$\mathrm{tr}(J)=\sum _{i=1}^n{\lambda }_i.\square$$

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

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Summary & Caution

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

Identity	Formula
Definition	$\mathrm{tr}(A)={\sum }_i{a}_{ii}$
Transpose	$\mathrm{tr}(A^T)=\mathrm{tr}(A)$
Additivity	$\mathrm{tr}(A+B)=\mathrm{tr}(A)+\mathrm{tr}(B)$
Homogeneity	$\mathrm{tr}(\lambda A)=\lambda \mathrm{tr}(A)$
Cyclic	$\mathrm{tr}(AB)=\mathrm{tr}(BA)$
Similarity	$\mathrm{tr}(P^{-1}AP)=\mathrm{tr}(A)$
Frobenius norm	$\mathrm{tr}(A{A}^T)={\sum }_{i,j}{a}_{ij}^2$
Eigenvalues	$\mathrm{tr}(A)={\sum }_i{\lambda }_i$

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