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

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 $$X = A^{-1}B$$.

• You can prove it when $A$ and $B$ are real? Try $A = 1$, $B = -1$. Dec 16, 2019 at 3:53
• @RobertIsrael I have just noticed I made a typo in the expression of the matrix. The entries of the second row are supposed to be reversed. I have edited it just now. Dec 16, 2019 at 3:57
• Since you read french language, there is another proof in the book (in french) 'Introduction à la théorie des groupes de Lie classiques", p. 94.
– user91684
Dec 16, 2019 at 20:54
• Yes it is well-known that it suffices to show that det(I+AA¯)≥0; The original proof of these results is an exercise in the American Mathematical Monthly (dated 1980-81). I think that there is a proof in the present website. Look for it! – l
– user91684
Dec 16, 2019 at 20:55
• Very nice ! Merci beaucoup @loupblanc for your reference. I was able to write down a solution to the problem following the steps described in the book of Mneimné and Testard. For the sake of completeness, I will write down the steps of the proof below. Dec 17, 2019 at 3:25

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.

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?