$$

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

$m\ne n$, it is impossible for

$AB$ and

$BA$ to be similar. However, the trick is to show that

$$\left[\begin{array}{cc}AB& O_{m,n}\\ B& O_{n,n}\end{array}\right]\text{ and }\left[\begin{array}{cc}{O}_{m,m}& O_{m,n}\\ B& BA\end{array}\right]$$

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

$$\left[\begin{array}{cc}{I}_m& -A\\ O_{n,m}& I_n\end{array}\right]\left[\begin{array}{cc}AB& O_{m,n}\\ B& O_{n,n}\end{array}\right]=\left[\begin{array}{cc}{O}_{m,m}& O_{m,n}\\ B& O_{n,n}\end{array}\right].$$

It is straightforward to find

$${\left[\begin{array}{cc}{I}_m& -A\\ O_{n,m}& I_n\end{array}\right]}^{-1}=\left[\begin{array}{cc}{I}_m& A\\ O_{n,m}& I_n\end{array}\right].$$

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

$$\left[\begin{array}{cc}{O}_{m,m}& O_{m,n}\\ B& O_{n,n}\end{array}\right]\left[\begin{array}{cc}{I}_m& A\\ O_{n,m}& I_n\end{array}\right]=\left[\begin{array}{cc}{O}_{m,m}& O_{m,n}\\ B& BA\end{array}\right].$$

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

Sample answer

By direct computation, we have

$$\left[\begin{array}{cc}{I}_m& -A\\ O_{n,m}& I_n\end{array}\right]\left[\begin{array}{cc}AB& O_{m,n}\\ B& O_{n,n}\end{array}\right]\left[\begin{array}{cc}{I}_m& A\\ O_{n,m}& I_n\end{array}\right]=\left[\begin{array}{cc}{O}_{m,m}& O_{m,n}\\ B& BA\end{array}\right],$$

so

$$\left[\begin{array}{cc}AB& O_{m,n}\\ B& O_{n,n}\end{array}\right]\text{ and }\left[\begin{array}{cc}{O}_{m,m}& O_{m,n}\\ B& BA\end{array}\right]$$

are similar, and

$AB$ and

$BA$ have the same nonzero eigenvalues.

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