Let $Q$ and $P$ be $n\times n$ matrix, then $PQ$ and $QP$ have the same eigenvalues

Question

Is the following statement true ?

Let $Q$ and $P$ be $n\times n$ matrix, then $PQ$ and $QP$ have the same eigenvalues.

What if $Q$ is invertible ?

Edit:

We do not assume that $P$ and $Q$ have the same eigenvalues.

Answer 1

If $Q$ is an invertible matrix, then $PQ$ and $QP$ are in fact similar matrices. This is easy to see : $Q^{-1}(QP)Q = PQ$. Therefore, it obviously is the case that $QP$ and $PQ$ have the same eigenvalues, in fact up to multiplicity. (I leave you to see this yourself. Use arguments illustrated in the comments above).

Now, let $P$ and $Q$ not be invertible. We'll assume they are over the real numbers. Thanks to $GL_{n}(\mathbb R)$ being dense in $M_{n \times n}(\mathbb R)$, we have that there exist invertible matrices $Q_n \to Q$ (entrywise convergence). Since right and left matrix multiplication are both continuous operations, we see that $PQ_n \to PQ$ and $Q_nP \to QP$. Now, $Q_nP$ and $PQ_n$ have the same eigenvalues up to multiplicity, and therefore the same characteristic polynomial.

The characteristic polynomial of any matrix $A$ has coefficients which are polynomials in the entries of $A$. Therefore, if $A_n \to A$, then it also follows by continuity of polynomials that the coefficients of the characteristic polynomials of $A_n$ converges to the coefficients of the characteristic polynomial of $A$.

The above argument shows that the characteristic polynomial of $PQ$ is the limit of the characteristic polynomials of $PQ_n$, and similarly for $QP$ and $Q_nP$. But then $Q_nP$ and $PQ_n$ have the same characteristic polynomial, hence $PQ$ and $QP$ have the same characteristic polynomial.

In particular, for $P,Q \in M_{n \times n}$, $PQ$ and $QP$ have the same eigenvalues, up to multiplicity as well.

However, if $P \in M_{m \times n}$ and $Q \in M_{n \times m}$, then too $PQ$ and $QP$ are square matrices, then too you can ask about their eigenvalues. Of course, a similar argument to what we did above doesn't hold(since the concept of inverses will be defenestrated), but the result is still stunning : $\chi_{PQ}(t) = \chi_{QP}(t) \times t^{m-n}$, where $m>n$. Therefore, the non-zero eigenvalues of $PQ$ still equal those of $QP$ up to multiplicity, but what can at most occur beyond that is one having the zero eigenvalue and the other not having it. The proof of this can be given neatly using block matrices, and I can append it if you request.