Is the determinant of $\left( \begin{array}{rr} A & B\\ -B & A \end{array}\right) $ non-negative? I'd like to know if $$\left(
\begin{array}{rr} A &amp; B\\
-B &amp; A \end{array}\right)
$$
has non-negative determinant, where $A,B$ are $n \times n$ real matrices, not necessarily commuting.
I'm trying to show almost complex structure $J$ admits an orientation given by $\{e_1,\cdots ,e_n,Je_1,\cdots Je_n \}$. To show this orientation is well defined, I need to pick another such basis and show the transition matrix has positive determinant. And above is the transition matrix.
Thanks in advance for anyone who could help.
Note that in here the answer is only provided when $A$ and $B$ commute.
$\newcommand{\overbar}[1]{\mkern 1.5mu\overline{\mkern-1.5mu#1\mkern-1.5mu}\mkern 1.5mu}$
As noted in the comment, taking determinants on both sides of the identity
\begin{align*}
\begin{pmatrix} I_{(n)} &amp; iI_{(n)} \\ 0 &amp; I_{(n)} \end{pmatrix} 
\begin{pmatrix} A &amp; B \\ -B &amp; A \end{pmatrix}
\begin{pmatrix} I_{(n)} &amp; -iI_{(n)} \\ 0 &amp; I_{(n)} \end{pmatrix} =
\begin{pmatrix} A - iB &amp; 0 \\ -B &amp; A + iB \end{pmatrix} 
\end{align*}
yields
\begin{align*}
\det\begin{pmatrix} A &amp; B \\ -B &amp; A \end{pmatrix} &amp;= \det(A + iB)\det(A - iB) = \det(A + iB)\det(\overbar{A + iB}) \\
&amp;= \det(A + iB)\overbar{\det(A + iB)} = |\det(A + iB)|^2 \geq 0,
\end{align*}
where we used the condition that $A$ and $B$ are real matrices, and for any $C \in \mathbb{C}^{n \times n}$, $\det(\overbar{C}) = \overbar{\det(C)}$, which can be easily verified by the definition of determinant.
We know:
det (X+Y) ≥ det(X) + det (Y)
this means: det(A^2 + B^2) ≥ det(A^2) + det(B^2)
which is ==&gt; |A^2 + B^2| ≥ |A^2| + |B^2|; EQN 1
Using another property of matrices: |A|^n = |A^n|
Thus, RHS of EQN 1 = |A|^2 + |B|^2
Also, determinant of any nxn real matrix always exists and is a real number. So, square of a real number will always be non-negative!
Since RHS is a non-negative quantity and LHS is greater than equal to RHS, LHS WILL always be a non-negative quantity too!
Hope this addresses your concerns..