Determinant of a $2 \times 2$ complex block matrix is nonnegative

Question

Let $n \geq 1$ and $A, B \in M_n(\mathbb C)$. Form the matrix

$$g= \begin{bmatrix} A & -B \\ \overline B & \overline A \end{bmatrix} \in M_{2n}(\mathbb C)$$

I would like to prove that the matrix $g$ has non negative determinant. Actually, I can prove this in the case $A$ and $B$ have real entries, this is a classic exercise. To do this, I would make some operations on columns and lines to reduce to an upper-triangular matrix by blocks, which would result in the identity $\det(g)=\det(A+iB)\det(A-iB)\geq0$. However, this method seems to fail in the case of complex entries.

Could someone give me a hand with this exercise?

EDIT: Using density of invertible matrices, we may assume that $A$ is invertible. Using the formula given by Schur complement, I can reduce this problem to the following. Given $X$ a square matrix with complex entries, we have $\det(I+X\overline{X})\geq 0$. I am currently trying to prove this, but I have not been able to conclude yet. Note that if I use the notation of the initial problem, then $X = A^{-1}B$.

Answer 1

Following the steps described in the exercise p.94 in Mneimné and Testard's book "Introduction à la théorie des groupes de Lie classiques" suggested by Loup Blanc in the comments, I was able to write down a proof for this problem. For the sake of completeness, I will describe the steps below.

First of all, as described in my Edit, it is enough to treat the case where $A$ is invertible. This follows from the density of invertible matrices and by continuity of the determinant. Using the Schur complement formula, we obtain $$\det(g)=\det(\overline A)\det(A+B\overline{A}^{-1}\overline B)=\underbrace{\overline{\det(A)}\det(A)}_{\geq \,0}\det(I+(A^{-1}B)\overline{A^{-1}B})$$ Thus, we are reduced to proving that $\det(I+X\overline X)\geq 0$ for every complex square matrix $X$. To do this, we follow multiple steps.

The first step is to justify that the characteristic polynomial of $X\overline X$ has real coefficients. It is enough to prove that $X\overline X$ and $\overline X X$ share the same characteristic polynomial. The way I proved this is by describing the coefficients of the polynomial of $X\overline X$ in terms of the sums of principal minors of $X\overline X$. Using Cauchy-Binet's formula to further decompose these minors, I end up with an expression which is indeed symmetric in $X$ and $\overline X$.

The second step is to justify that the set $E$ of matrices $X\in M_n(\mathbb C)$ such that $X\overline X$ has $n$ distinct eigenvalues is dense in $M_n(\mathbb C)$. For this, consider the application sending $X$ to the discriminant of the characteristic polynomial of $X\overline X$. This application can be seen as a polynomial in the $2n^2$ variables $\operatorname{Re}(x_{i,j})$ and $\operatorname{Im}(x_{i,j})$ where $X=(x_{i,j})$ (it is not directly a polynomial in the $x_{i,j}$ because of the complex conjugation). The set $E$ is the locus where this application does not vanish. If $E$ was not dense, there would exist some non empty open subset $U$ which does not meet $E$. On this open subset, our polynomial application would be $0$, hence this application would be $0$ everywhere, that is we would have $E=\emptyset$. This is absurd (for instance, $\operatorname{diag}(1,2,\ldots,n)\in E$).

Third and last step, we see now that it is enough to consider the case $X\in E$. The eigenvalues of $I+X\overline X$ are just $1 +$ the eigenvalues of $X\overline X$. Then $\det(I+X\overline X)$ is just the product of all of them (with multiplicities, but these are all $1$ since $X\in E$). Because the characteristic polynomial of $X\overline X$ has real coefficients, the non real eigenvalues come by pair $\mu$ and $\overline \mu$. The products $(1+\mu)(1+\overline{\mu})$ are all non negative, so we only need to look at real eigenvalues. If $\lambda$ is a real eigenvalue of $X\overline X$ and $v$ is an associated eigenvector, because the associated eigenspace has dimension $1$, there exists $r\in \mathbb C$ such that $X\overline v = r v$. From this, we easily deduce that $\lambda = |r|^2\geq 0$, which eventually allows us to conclude.

Answer 2

Consider $\mathbb{H}$, the $\mathbb{R}$ algebra of quaternions. Every quaternion can be written uniquely as $$q = a+ b j$$ where $a$, $b\in \mathbb{C}$. Note that $j c = \bar c j$ for all $c \in \mathbb{C}$.

Consider now $M_n(\mathbb{H})$ the $\mathbb{R}$ algebra of $n\times n$ matrices with quaternionic entries. Every quaternionic matrix can be written as $$Q = A + B j$$ where $A$, $B \in M_n(\mathbb{C})$. Moreover, we have $$Q_1 Q_2 = (A_1 + B_1 j)(A_2+ B_2j) = A_1 A_2 + B_1j A_2 + A_1 B_2j + B_1j B_2j = (A_1 A_1 - B_1 \bar B_2) + (B_1 \bar A_2 + A_1 B_2) j$$ We conclude that the map $$M_n(\mathbb{H}) \ni A+ Bj \mapsto \colon =\left ( \begin{matrix} A & - B \\ \bar B & \bar A \end{matrix} \right) $$ is a morphism of $\mathbb{R}$ algebras. Note also that $$j (A + B j) j^{-1} = \bar A + \bar B j$$ It follows that $A+ Bj \mapsto \phi(A+Bj) \colon = \det \left ( \begin{matrix} A & - B \\ \bar B & \bar A \end{matrix} \right)\in \mathbb{R}$.

We have a morphism of Lie groups $GL_n(\mathbb{H}) \to \mathbb{R}^{\times}$, given by the restriction of $\phi$. Since $GL_n(\mathbb{H})$ is connected ( use elementary transformations and $\mathbb{H}^{\times}$ connected), we conclude that $\phi(GL(n, \mathbb{H}) ) \subset (0, \infty)$. Now use that $GL_n(\mathbb{H})$ is dense in $M_n(\mathbb{H})$ and get $\phi(M_n(\mathbb{H})) \subset [0, \infty)$.

Note: the solution uses some topology. Let's compare this with the similar problem: $\det \left ( \begin{matrix} A & - B \\ \ B & A \end{matrix} \right)$ is positive of $A$, $B \in M_n(\mathbb{R})$. The solution in this case uses the fact that $$\det \left ( \begin{matrix} A & - B \\ B & A \end{matrix} \right)= \det (A+i B)\det (A-iB)= \det(A+i B) \cdot \overline{\det(A+iB)}$$. Now, in this case the solution is purely algebraic. In fact, we show that for any commutative ring $R$ and $A$, $B\in M_n(R)$ the determinant $\det \left ( \begin{matrix} A & - B \\ \bar B & \bar A \end{matrix} \right)$ is a sum of two squares of elements obtained from $A$, $B$. Can we do a similar thing in our problem, now using a sum of $4$ squares?