Equal determinants

Let $A, B \in \mathbb{R}^{n \times n}$ that are diagonizable in $\mathbb{R}$. If $\det (A^2+B^2)=0$ and $AB=BA$ , then prove that $\det A = \det B =0$.

Solution

We will use the fact that $A, B$ are simultaneously diagonalizable. (The proof of this can be found in the texts of advanced Linear Algebra but for the moment we are taking it for granted.)

Since the above happens there exists an invertible matrix $T$ and diagonizable $P=(p_k), Q=(q_k)$ such that:

$$A=TPT^{-1}, \;\;\;\; B=TQT^{-1}$$

Hence:

$$\begin{align*}
0 &=\det \left ( A^2+B^2 \right ) \\
 &= \det \left ( T P^2 T^{-1} + TQ^2 T^{-1} \right )\\
 &= \det \left ( T\left ( P^2+Q^2 \right )T^{-1} \right ) \\
 &= \det \left ( P^2+Q^2 \right ) \\
 &=\prod_{k=1}^{n}\left ( p_k^2+ q_k^2 \right )
\end{align*}$$

Hence for some $k$ holds $p_k^2 + q_k^2 =0 $ . Since $p_k, \; q_k$ are real this means that $p_k=q_k=0$. The result now follows, since:

$$\det A= \det P=0$$

and similarly $\det B= \det Q=0 $.