Which changes leave the diagonal sum alone?

Source scope: §3.7. Worked numerical examples, experiments, and practice are teaching supplements unless identified as source results.

Add the diagonal

For $A=\begin{bmatrix}2&7\\0&-1\end{bmatrix}$, the trace is $2+(-1)=1$. The off-diagonal seven contributes nothing directly:

The source lists four rules for compatible square sums and square cyclic products:

The first three follow by summing diagonal entries. The last is cyclic invariance; it applies to any number of factors, not only four. A teaching derivation of the basic step is

This also works for rectangular $A\in\mathbb R^{m\times n}$ and $B\in\mathbb R^{n\times m}$ because both products are square, though their sizes differ. Group a longer product into two factors to move its first factor to the end.

A cycle is not a shuffle

Let $A=\begin{bmatrix}1&0\\0&0\end{bmatrix}$, $B=\begin{bmatrix}0&1\\0&0\end{bmatrix}$, and $C=\begin{bmatrix}0&0\\1&0\end{bmatrix}$. Then $BC=A$, so $\operatorname{tr}(ABC)=1$. But $BA=0$, so $\operatorname{tr}(BAC)=0$. Arbitrary swaps fail. Taking a fourth factor $D=I$ gives the source’s four-factor warning.

The trace also equals the sum of eigenvalues, counted with algebraic multiplicity:

The source states this identity without proof. A general real matrix can have complex eigenvalues; the sum still equals its real trace. For symmetric matrices all eigenvalues are real, as the spectral theorem will establish.

Transfer

Must $\operatorname{tr}(ABC)$ equal $\operatorname{tr}(CAB)$? Must it equal $\operatorname{tr}(ACB)$?