$\det(I+A\bar{A}) \ge 0$ for all complex square matrices $A$? $\det(I+A\bar{A}) \ge 0$
Is it possible to prove the inequality is true for all complex square matrices $A$ where $I$ is the identity matrix and $\bar{A}$ is the complex conjugated matrix.
 A: Really it is a very difficult question dated 1980-81 (Am. Math. Monthly). Personally I did not find any solution. The simplest method is as follows. 
EDIT.


*

*Show the result when $A\bar{A}$ has no negative eigenvalues. 

*Show that the multiplicity of a negative eigenvalue of $A\bar{A}$ cannot be $1$.

*Show that the set of matrices $A$ that satisfy 1. is dense in $M_n(\mathbb{C})$.

*Conclude.
A: Eigenvalues of $A\bar{A}$ have been thoroughly studied in

D.C. Youla, A normal form for a matrix under the unitary congruence group, Canad. J. Math. 13(1961), 694-704.

In the paper, it is proved (in Lemma 5) that for any complex square matrix $A$, those eigenvalues of $A\bar{A}$ that are not real nonnegative must either be


*

*non-real and occur in conjugate pairs, or

*real, negative and have even multiplicities (so they also occur in "conjugate pairs").


It follows immediately that $\det(I+A\bar{A})$ is real nonnegative.
A: I'm going to give a complete proof of this result, which has a surprisingly conceptual basis, using the quaternions. I believe it can possibly be simplified further.
Let $\mathbb{H}$ be the division ring of the quaternions. It contains $\mathbb{C}$ as a subring. View $\mathbb{H}$ as a left-$\mathbb{H}$-vector space over itself. An arbitrary endomorphism of $\mathbb{H}$ consists of right multiplication by some quaternion $w$, which can be written uniquely as $w = a + bj$ for some complex numbers $a$ and $b$. If we view $\mathbb{H}$ as a left-$\mathbb{C}$ vector space with basis $(1, j)$, then the matrix of this mapping is easily seen to be $$\begin{pmatrix}a & -\overline{b} \\ b & \overline{a} \end{pmatrix}.$$ 
Now let $V = \mathbb{H}^n$, and let $f \colon V \to V$ be a left-$\mathbb{H}$-linear endomorphism of $V$. Let $v_1, v_2, \dots v_n$ be a basis of $V$ (which can be the standard basis). Then a basis of $V$ over $\mathbb{C}$ is given by $(v_1, jv_1, v_2, jv_2, \dots, v_n, jv_n)$. In view of the preceding remarks, the matrix of $f$ in this basis must be made up of $2 \times 2$ blocks of the form above. Now if we reorder the basis to be $(v_1, v_2, \dots, v_n, jv_1, jv_2, \dots, jv_n$), then the matrix becomes one of the form
$$T = \begin{pmatrix}A & -\overline{B} \\ B & \overline{A} \end{pmatrix},$$
where $A$ and $B$ are arbitrary $n \times n$ complex matrices. Furthermore, the foregoing arguments show that the $\mathbb{H}$-linearity of $f$ is equivalent to its matrix having this form. 
Henceforth, the argument will be based on two facts:


*

*The determinant of any matrix of the given form is a nonnegative real number.

*We note that $$\det (I + A\overline{A}) = \det \begin{pmatrix} I & -\overline{A} \\ A & I \end{pmatrix}$$
is the determinant of a matrix of the required form.  
Point 2 follows from the factorization
  $$\begin{pmatrix} I & O \\ -A & I \end{pmatrix}\begin{pmatrix} I & -\overline{A} \\ A & I \end{pmatrix} = \begin{pmatrix} I & -\overline{A} \\ O & I+A\overline{A} \end{pmatrix}.$$
Point 1 is where I wonder if some improvements are possible, but here is a proof. Let $f \colon V \to V$ again be left-$\mathbb{H}$-linear, and let $T$ be its matrix as above.
First we prove that $\det T$ is a real number. The conjugate matrix
$$\overline{T} = \begin{pmatrix} \overline{A} & -B \\ \overline{B} & A \end{pmatrix}$$
can be obtained from $T$ in the following steps: exchange the first $n$ rows with the last $n$ rows; exchange the first $n$ columns with the last $n$ columns; multiply the bottom $n$ rows by $-1$; multiply the last $n$ columns by $-1$. It follows that $\det T = \det \overline{T} = \overline{\det T}$, hence $\det T$ is real.
Now, for any real number $\lambda$, what was said about $T$ is equally applicable to any matrix $T - \lambda I$, which is still of the required form. It follows therefore that the characteristic polynomial $\chi(\lambda)$ takes only real values for real $\lambda$, hence has real coefficients.
This shows that the non-real eigenvalues of $T$ occur in conjugate pairs (including multiplicities), and since $\det T$ is the product of its eigenvalues we need only check that negative real eigenvalues have even multiplicity. However, this is true for any real eigenvalue $\lambda$ (whether positive or negative), for the following reason. 
The multiplicity of $\lambda$ is $\dim \ker (f - \lambda \operatorname{id})^{2n}$. Since $f$ is $\mathbb{H}$-linear, so too is $(f - \lambda \operatorname{id})^{2n}$. (That $\lambda$ is real is crucial for this, because left-multiplication by a complex number is not left-$\mathbb{H}$-linear, as $\mathbb{C}$ is not in the centre of $\mathbb{H}$.) Therefore the kernel of this operator is an $\mathbb{H}$-subspace of $V$, and it must therefore have even dimension over $\mathbb{C}$.
