$AA^\top + A + A^\top = 0$, then $|\det A|\leq 2^n$

Question

Let $A\in \mathbb R^{n\times n}$ such that $AA^\top + A + A^\top = 0$. Prove that $|\det A|\leq 2^n$.

$AA^\top + A + A^\top = 0$ rewrites as $(A+I_n)(A^\top +I_n) = I_n$, hence $A+I_n$ is an orthogonal matrix and $\det(A+I_n)\in \{-1,1\}$.

If $\lambda$ is a real eigenvalue of $A$ (with an eigenvector in $\mathbb R^n$) then $\lambda\in \{0,-2\}$. However $A$ may have complex eigenvalues, or the eigenvectors may have complex entries.

I cannot make further progress. I'm not supposed to know that an orthogonal matrix can be diagonalized over $\mathbb C$ with eigenvalues having modulus $1$, I'm thus looking for a solution which does not leverage this fact.

All you need is that every eigenvalue $\lambda$ of $A$ satisfies $\vert \lambda\vert\leq 2$ (then we can use that the determinant is the product of the eigenvalues). — Severin Schraven, Mar 13 '24 at 07:57
You've almost done everything. If $A+I$ is orthogonal, we have $|\lambda|-1\leq|\lambda+1|=1$ for all (complex) eigenvalues of $A$. — Jochen, Mar 13 '24 at 07:59
Can someone provide a short proof that the complex eigenvalues of an orthogonal matrix in $\mathbb R^n$ have modulus $\leq 1$ ? — brised by Linear Algebra, Mar 13 '24 at 08:04

score 1 · Answer 1 · answered Mar 17 '24 at 16:32

I think one should be able to argue without eigenvalues, but rather just using the fact that an orthogonal matrix is an isometry, or equivalently that its columns form an orthonormal basis of $\mathbb R^n$: If $B=A+I$ then the condition on $A$ implies $BB^{\intercal}=I_n$, that is, $B$ is an orthogonal matrix.

Write $B = (b_1|b_2|\ldots |b_n)$ so that the $b_i$ are the columnn vectors of $B$. Then if $\{e_i: 1\leq i \leq n\}$ denote the standard basis of $\mathbb R^n$, it follows that $A$ has columns $b_i-e_i$. Since the determinant is the volume of higher-dimensional parallelogram given by the column vectors, and this is bounded by the product of the lengths of the column vectors, the result follows from the triangle inequality.

[Note the claims about the volume-like properties of $\det$ follow from the alternating property and induction (essentially using Gram-Schmidt), so they are elementary.]

score 0 · Accepted Answer · answered Mar 17 '24 at 15:01

The challenge for me was not to use auxiliary results on complex eigenvalues of orthogonal matrices (see Ben Grossmann's question).

A replacement for such results is the canonical form. Subtracting $I_2$ to each $2\times 2$ block, each block is easily seen to contribute by a factor $\leq 4$ to the determinant.

$AA^\top + A + A^\top = 0$, then $|\det A|\leq 2^n$

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