Determinant of block diagonal matrix

Question

Suppose $X\in U(2)$. We can write $X=A+iB$.

Let $M:=\begin{bmatrix} A & -B \\ B & A\end{bmatrix}$. I want to to show that $\det M=1$, but I'm having trouble.

We don't know if $A$ and $B$ commute, so we can't say $\det(M)=\det(A^2+B^2)$.

WolframAlpha didn't give me a helpful expression either.

Is there some algebraic trick to help me out?

Answer 1

$X$ is unitary. Therefore $(A+iB)(A^T-iB^T)=I$. Equating the real parts and imaginary parts on both sides, we see that $MM^T=I$ and $\det M=\pm 1$. Since the set of all $M$s is connected (because $U(2)$ is path-connected) and $\det M=1$ when $X=I$, we must have $\det M=1$ for all $M$s.

Alternatively, note that $$ \pmatrix{I&0\\ iI&I}\pmatrix{A&-B\\ B&A}\pmatrix{I&0\\ -iI&I}=\pmatrix{A+iB&-B\\ 0&A-iB}. $$ Therefore $\det M=\det(A+iB)\det(A-iB)=|\det(X)|^2=1$.