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# Comprehensive Guide on the Trace of Matrices

schedule May 20, 2023
Last updated
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Linear Algebra
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Definition.

# Trace of a square matrix

The trace of an $n\times{n}$ matrix is defined as the sum of its diagonals:

$$\mathrm{tr}(\boldsymbol{A})=\sum^n_{i=1}a_{ii}$$

Where $a_{ii}$ is the value of the matrix $\boldsymbol{A}$ at row $i$ column $i$.

Example.

## Computing the trace of a 2x2 matrix

Compute the trace of the following matrix:

$$\boldsymbol{A}= \begin{pmatrix} 1&2\\ 5&7\\ \end{pmatrix}$$

Solution. The trace of $\boldsymbol{A}$ is simply the sum of its diagonals:

\begin{align*} \mathrm{tr}(\boldsymbol{A}) &=1+7\\ &=8 \end{align*}
Theorem.

# Matrix and its transpose have the same trace

If $\boldsymbol{A}$ is a square matrix, then:

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

Proof. Let $\boldsymbol{A}$ be a square matrix. The transpose $\boldsymbol{A}^T$ has the same diagonal entries as $\boldsymbol{A}$ by theorem, which means that the sum of the diagonals of $\boldsymbol{A}^T$ and $\boldsymbol{A}$ are the same. We therefore conclude that $\mathrm{tr}(\boldsymbol{A}^T)= \mathrm{tr}(A)$. This completes the proof.

Theorem.

# Trace of cA where c is a scalar

If $\boldsymbol{A}$ is any matrix and $c$ is any scalar, then:

$$\mathrm{tr}(c\boldsymbol{A})= c\cdot\mathrm{tr}(\boldsymbol{A})$$

Proof. Let $\boldsymbol{A}$ be the following square matrix:

$$\boldsymbol{A}=\begin{pmatrix} a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\smash\ddots&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn}\\ \end{pmatrix}$$

The trace of $\boldsymbol{A}$ is:

$$\begin{equation}\label{eq:o9IsYjUgzEbeq8g84Fn} \mathrm{tr}(\boldsymbol{A})= a_{11}+a_{22}+\cdots+a_{nn} \end{equation}$$

If $c$ is any scalar, then $c\boldsymbol{A}$ is:

$$c\boldsymbol{A}=\begin{pmatrix} ca_{11}&ca_{12}&\cdots&ca_{1n}\\ ca_{21}&ca_{22}&\cdots&ca_{2n}\\ \vdots&\vdots&\smash\ddots&\vdots\\ ca_{n1}&ca_{n2}&\cdots&ca_{nn}\\ \end{pmatrix}$$

The trace of $c\boldsymbol{A}$ is:

\begin{align*} \mathrm{tr}(c\boldsymbol{A}) &=ca_{11}+ ca_{22}+ \cdots+ ca_{nn}\\ &=c(a_{11}+ a_{22}+ \cdots+ a_{nn})\\ &=c\cdot\mathrm{tr}(\boldsymbol{A}) \end{align*}

This completes the proof.

Theorem.

# Trace of the sum of two matrices

If $\boldsymbol{A}$ and $\boldsymbol{B}$ are square matrices of the same shape, then:

\begin{align*} \mathrm{tr}(\boldsymbol{A}+\boldsymbol{B})= \mathrm{tr}(\boldsymbol{A})+ \mathrm{tr}(\boldsymbol{B}) \end{align*}

Solution. Let A and B be the following square matrices:

$$\boldsymbol{A}= \begin{pmatrix} a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\smash\ddots&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn}\\ \end{pmatrix},\;\;\;\; \boldsymbol{B}= \begin{pmatrix} b_{11}&b_{12}&\cdots&b_{1n}\\ b_{21}&b_{22}&\cdots&b_{2n}\\ \vdots&\vdots&\smash\ddots&\vdots\\ b_{n1}&b_{n2}&\cdots&b_{nn}\\ \end{pmatrix}$$

The sum $\boldsymbol{A}+\boldsymbol{B}$ is:

$$\boldsymbol{A}+\boldsymbol{B}= \begin{pmatrix} a_{11}+b_{11}&a_{12}+b_{12}&\cdots&a_{1n}+b_{1n}\\ a_{21}+b_{21}&a_{22}+b_{22}&\cdots&a_{2n}+b_{2n}\\ \vdots&\vdots&\smash\ddots&\vdots\\ a_{n1}+b_{n1}&a_{n2}+b_{n2}&\cdots&a_{nn}+b_{nn}\\ \end{pmatrix}$$

The trace of $\boldsymbol{A}+\boldsymbol{B}$ is:

\begin{align*} \mathrm{tr}(\boldsymbol{A}+\boldsymbol{B}) &=(a_{11}+b_{11})+ (a_{22}+b_{22})+ \cdots+ (a_{nn}+b_{nn})\\ &=(a_{11}+a_{22}+\cdots+a_{nn})+ (b_{11}+b_{22}+\cdots+b_{nn})\\ &=\mathrm{tr}(\boldsymbol{A})+ \mathrm{tr}(\boldsymbol{B}) \end{align*}

This completes the proof.

Theorem.

# Trace of AB is equal to the trace of BA

If $\boldsymbol{A}$ and $\boldsymbol{B}$ are matrices, then the trace of $\boldsymbol{AB}$ is equal to the trace of $\boldsymbol{BA}$, that is:

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

Proof. Suppose we have the following matrices:

$$\boldsymbol{A}= \begin{pmatrix} a_{11}&a_{12}&\cdots&a_{1m}\\ a_{21}&a_{22}&\cdots&a_{2m}\\ \vdots&\vdots&\smash\ddots&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nm}\\ \end{pmatrix},\;\;\;\; \boldsymbol{B}= \begin{pmatrix} b_{11}&b_{12}&\cdots&b_{1n}\\ b_{21}&b_{22}&\cdots&b_{2n}\\ \vdots&\vdots&\smash\ddots&\vdots\\ b_{m1}&b_{m2}&\cdots&b_{mn}\\ \end{pmatrix}$$

Since $\boldsymbol{A}$ is an $n\times{m}$ matrix and $\boldsymbol{B}$ is an $m\times{n}$ matrix, their product $\boldsymbol{AB}$ will be an $n\times{n}$ matrix.

By definition of traces, we have that:

\begin{align*} \mathrm{tr}(\boldsymbol{AB})&= \sum^n_{i=1}(\boldsymbol{AB})_{ii}\\ &=(\boldsymbol{AB})_{11}+(\boldsymbol{AB})_{22}+\cdots+(\boldsymbol{AB})_{nn}\\ \end{align*}

Here, $(\boldsymbol{AB})_{11}$ represents the value at row $1$ column $1$ in matrix $\boldsymbol{AB}$. By matrix multiplication, we can compute each of these terms like so:

$$\begin{gather*} (\boldsymbol{AB})_{11}= a_{11}b_{11}+ a_{12}b_{21}+ \cdots+ a_{1m}b_{m1}\\ (\boldsymbol{AB})_{22}= a_{21}b_{12}+ a_{22}b_{22}+ \cdots+ a_{2m}b_{m2}\\ \vdots\\ (\boldsymbol{AB})_{nn}= a_{n1}b_{1n}+ a_{n2}b_{2n}+ \cdots+ a_{nm}b_{mn} \end{gather*}$$

The trick here is to take the sum column-wise instead of row-wise:

$$\begin{gather*} a_{11}b_{11}+a_{21}b_{12}+\cdots+a_{n1}b_{1n}=(\boldsymbol{BA})_{11}\\ a_{12}b_{21}+a_{22}b_{22}+\cdots+a_{n2}b_{2n}=(\boldsymbol{BA})_{22}\\ \cdots\\ a_{1m}b_{m1}+a_{2m}b_{m2}+\cdots+a_{nm}b_{mn}=(\boldsymbol{BA})_{mm}\\ \end{gather*}$$

Therefore, we can say that:

\begin{align*} \mathrm{tr}(\boldsymbol{AB}) &=(\boldsymbol{AB})_{11}+(\boldsymbol{AB})_{22}+\cdots+(\boldsymbol{AB})_{nn}\\ &=(\boldsymbol{BA})_{11}+(\boldsymbol{BA})_{22}+\cdots+(\boldsymbol{BA})_{mm}\\ &=\mathrm{tr}(\boldsymbol{BA}) \end{align*}

This completes the proof.

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