Let $A,B\in M_n(\mathbb{R})$, Prove $\det\begin{pmatrix} A & -B\\ B & A \end{pmatrix}=|\det(A+iB)|^2$

Question

Let $A,B\in \mathcal M_n(\mathbb{R})$, Prove $\det\begin{pmatrix} A & -B\\ B & A \end{pmatrix}=\bigg|\det(A+iB)\bigg|^2$

My work:

We know:

$$|\det(A+iB)|^2=\sqrt{(\det(A+iB))^2}^2=\det(A+iB)^2=\det(A+iB)\det(A+iB)=\det((A+iB)(A+iB))=\det(A^2+AiB+iBA+i^2B^2)=\det(A^2+AiB+iBA-B^2)$$

Here I'm stuck. Can someone help me?

Answer 1

Answer: For A and B commuting then we have $$\color{red}{\det \left(\begin{matrix} A& -B\\B&A \end{matrix}\right)= \bigg|\det(A+iB)\bigg|^2 }$$

Similarly like here For every $a \in \mathbb{R}$ evaluate $ \lim_{n \to \infty}\left(\begin{smallmatrix} 1&\frac{a}{n}\\\frac{-a}{n}&1\end{smallmatrix}\right)^{n}.$ by putting ourselves in the complex plan where we identify

$$ 1 \equiv \left(\begin{matrix} 1& 0\\0&1 \end{matrix}\right)~~~\text{and}~~~~ i \equiv \left(\begin{matrix} 0& -1\\1&0 \end{matrix}\right)\implies M=\left(\begin{matrix} A& -B\\B&A \end{matrix}\right)= A+iB$$ Now assume that $A$ is invertible (The case where A is not invertible will follows by density argument ): then from this

If A and B commute we obtain

$$\det \left(\begin{matrix} A& -B\\B&A \end{matrix}\right)=\det(A+BA^{-1}B)\det(A)= \det(A^2+BA^{-1}BA)=\det(A^2+B^2)$$

But we have, $$\color{red}{|A+iB|^2 =(A+iB)(A-iB)= A^2+B^2 }$$

And hence since $|z|^2=z\bar z$ we obtain , $$\det( A^2+B^2)=\det\left((A+iB)\cdot(A-iB)\right)=\det( A+iB)\cdot\det( A-iB) \\=\det( A+iB)\cdot\overline{\det( A+iB)}= \bigg|\det(A+iB)\bigg|^2$$

Answer 2

This follows directly from taking determinants of both sides of the following identity $$ \begin{pmatrix} 1 & 0 \\ i & 1\end{pmatrix}\begin{pmatrix} A & -B \\ B & A\end{pmatrix}\begin{pmatrix}1 & 0 \\ -i & 1\end{pmatrix}=\begin{pmatrix} A + iB & -B \\ 0 & A - iB\end{pmatrix} $$ and noting $\det (A + iB) $ and $\det(A-iB)$ are conjugates.