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

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?

## marked as duplicate by anomaly, Cameron Williams, Adriano, The Phenotype, Guy FsoneJan 28 '18 at 6:06

• Let $A, B$ be 1x1 matrices. Then the left hand side is $A^2 + B^2$, while the right hand side is $(A+iB)^2$, which is not equal to left hand side. – xyzzyz Jan 27 '18 at 20:57
• @xyzzyz: It's $|\det|^2$, not $\det^2$ itself. – anomaly Jan 27 '18 at 20:58
• it makes no difference to my counterexample. it's probably rather what @Guy Fsone says. – xyzzyz Jan 27 '18 at 21:02
• I think there is no need to have de module on the right hand side – Guy Fsone Jan 27 '18 at 21:05
• @GuyFsone: The right side isn't even real without the absolute value. – anomaly Jan 27 '18 at 21:05

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.

• That requires computing the determinant of the block matrix on the far right (which behaves as described, by the OP might not have proved that yet.) – anomaly Jan 27 '18 at 23:56

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 , $$\det( A^2+B^2)=\det\left((A+iB)\cdot(A-iB)\right)=\det( A+iB)\cdot\det( A-iB) \\=\det( A+iB)\cdot\overline{\det( A+iB)}= \bigg|\det(A+iB)\bigg|^2$$

• Why does $\det$ commute with the map you describe? – anomaly Jan 27 '18 at 21:02
• I don't understand your point of view for this proof... If can explain a little more i can be very grateful. – Bvss12 Jan 27 '18 at 21:05
• The exercise is fine don't have error. – Bvss12 Jan 27 '18 at 21:13
• Why is it necessary for $A$ and $B$ to commute? Do you have a counterexample? It follows in general just from the abstract nonsense of the map $M_n(\mathbb{C}) \to M_{2n}(\mathbb{R})$ induces on the action of $\mathbb{C}$ on itself by multiplication; the question is basically just showing that the determinant behaves as expected under a change of the ground field. – anomaly Jan 27 '18 at 21:37
• The OP's work is wrong. The problem as originally written, presumably from a linear algebra textbook, is correct. – anomaly Jan 27 '18 at 21:45