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# AB and BA have the same nonzero eigenvalues

## Problem

Let $A$ and $B$ be $m\times n$ and $n\times m$ matrices, respectively. Show that $AB$ and $BA$ have the same nonzero eigenvalues.

## Thought

Note that $AB$ and $BA$ are $m\times m$ and $n\times n$ matrices, respectively. When $m\neq n$, it is impossible for $AB$ and $BA$ to be similar. However, the trick is to show that

$$
\begin{bmatrix}
AB & O_{m,n} \\
B & O_{n,n}
\end{bmatrix}
\text{ and }
\begin{bmatrix}
O_{m,m} & O_{m,n} \\
B & BA
\end{bmatrix}
$$

are similar. Thus, the eigenvalues of $AB$ and $BA$ are the same, except for the number of zeros.

To see this, we use the block row operation

$$
\begin{bmatrix}
I_m & -A \\
O_{n,m} & I_n
\end{bmatrix}
\begin{bmatrix}
AB & O_{m,n} \\
B & O_{n,n}
\end{bmatrix} =
\begin{bmatrix}
O_{m,m} & O_{m,n} \\
B & O_{n,n}
\end{bmatrix}.
$$

It is straightforward to find

$$
\begin{bmatrix}
I_m & -A \\
O_{n,m} & I_n
\end{bmatrix}^{-1} =
\begin{bmatrix}
I_m & A \\
O_{n,m} & I_n
\end{bmatrix}.
$$

When applying this matrix one the right, it stands for a block column operation. That is,

$$
\begin{bmatrix}
O_{m,m} & O_{m,n} \\
B & O_{n,n}
\end{bmatrix}
\begin{bmatrix}
I_m & A \\
O_{n,m} & I_n
\end{bmatrix} =
\begin{bmatrix}
O_{m,m} & O_{m,n} \\
B & BA
\end{bmatrix}.
$$

With these intuition behind the scene, this problem can be solved by a one-line proof.

## Sample answer

By direct computation, we have

$$
\begin{bmatrix}
I_m & -A \\
O_{n,m} & I_n
\end{bmatrix}
\begin{bmatrix}
AB & O_{m,n} \\
B & O_{n,n}
\end{bmatrix}
\begin{bmatrix}
I_m & A \\
O_{n,m} & I_n
\end{bmatrix} =
\begin{bmatrix}
O_{m,m} & O_{m,n} \\
B & BA
\end{bmatrix},
$$

so

$$
\begin{bmatrix}
AB & O_{m,n} \\
B & O_{n,n}
\end{bmatrix}
\text{ and }
\begin{bmatrix}
O_{m,m} & O_{m,n} \\
B & BA
\end{bmatrix}
$$

are similar, and $AB$ and $BA$ have the same nonzero eigenvalues.

*This note can be found at Course website > Learning resources.*