# Prove:$A B$ and $B A$ has the same characteristic polynomial. [duplicate]

Suppose $A B$ are all $n\times n$ matrix. Prove:$A B$ and $B A$ has the same characteristic polynomial.

• If A is invertible

\begin{align}P(A B)=|\lambda E-A B|=\left|\lambda A\cdot A^{-1}-A B\right|=\left|A\left(\lambda A^{-1}-B\right)\right|=\left|A\left\|\text{$\lambda$A}^{-1}-B\right.\right|\end{align}

\begin{align}P(B A)=|\lambda E-B A|=\left|\lambda A^{-1}\cdot A-B A\right|=\left|\left(\lambda A^{-1}-B\right)A\right|=\left|\left.\text{$\lambda$A}^{-1}-B\right\|A\right|\end{align}

Is it true?

• If A is not invertible

How to prove?

## Edit:

I found a proof by Block Matrix Multiplication, it's more straightforward, is it right?

\begin{align}\left( \begin{array}{cc} E & 0 \\ -A & E \\ \end{array} \right).\left( \begin{array}{cc} \lambda E & B \\ \lambda A & \lambda E \\ \end{array} \right)=\left( \begin{array}{cc} \lambda E & B \\ 0 & \lambda E-A B \\ \end{array} \right)\end{align}

\begin{align}\left( \begin{array}{cc} \lambda E & B \\ \lambda A & \lambda E \\ \end{array} \right).\left( \begin{array}{cc} E & 0 \\ -A & E \\ \end{array} \right)=\left( \begin{array}{cc} \lambda E-B A & B \\ 0 & \lambda E \\ \end{array} \right)\end{align}

Another proof is by Sylvester determinant theorem which uses LU block decomposition.

Fill two matrices $A$ and $B$ with $2n^2$ distinct indeterminates. Observe $A$ and $B$ are invertible in the field $F=K(a_{ij},b_{ij})$ formed by adjoining these formal variables. Hence $\chi_{AB}(T)=\chi_{BA}(T)$ holds in the field $F[T]$ and so in the subring $K[a_{ij},b_{ij}][T]$, after which we can simply apply the evaluation map so that $\chi_{AB}(T)=\chi_{BA}(T)$ holds for any two matrices $A$ and $B$ with entries taken in $K$.

The beauty of this "universal" argument is that it bypasses the analytic concept of density, instead working purely algebraically and applying to any desired field $K$.

• However, a dense set in the Zariski topology is implicitly used to show that $A$ and $B$ are invertible matrices over $F$ (their determinants being nonzero). – Marc van Leeuwen Oct 6 '13 at 6:26
• This is like using an atomic bomb to kill a fly. – Raskolnikov Oct 6 '13 at 6:27
• @Raskolnikov The tools used would be considered flies in the broader context of modern algebra. – anon Oct 6 '13 at 6:28
• @MarcvanLeeuwen Would you say the fact that $x$ is nonzero in $K(x)$, where $x$ is an indeterminate, uses Zariski topology implicitly? – anon Oct 6 '13 at 6:29
• @Raskolnikov This reminds me of the legal proverb "Die Polizei soll nicht mit Kanonen auf Spatzen schießen" (Fritz Fleiner, Institutionen Des Deutschen Verwaltungsrecht, 8. Aufl., Tübingen, 1928). – user1551 Oct 6 '13 at 9:43

If $A$ is not invertible, $A_t=A+tI_n$ is invertible except for finitely many values of $t$. In particular, there is $\varepsilon > 0$ such that $A_t$ is invertible for every $t\in ]0,\varepsilon[$.

We therefore have $P(A_tB)=P(BA_t)$. Making $t\to 0$, we deduce $P(AB)=P(BA)$.

One way to do this is to note that $P(AB)-P(BA)$ is a polynomial in the entries of $A$ and $B$. You have shown that this vanishes for all invertible $A$, and want to show that it is precisely the zero polynomial. It suffices to note that the set of invertible $n\times n$ matrices over $\mathbb R$ or $\mathbb C$ is dense, and since polynomials are continuous it follows that $P(AB)-P(BA)$ is zero everywhere and thus the zero polynomial.