How to prove that $A^2$ is a symmetric matrix?

Question

Conjecture 1 :

Let $A$ be a real matrix such that $A^5=A A^T A A^T A$. Then $A^2$ is a symmetric matrix.

(here $A^T$ denotes the transpose of a matrix A).

I guess that the following is also true :

Conjecture 2 :

If $A^{2n+1}=AA^TAA^T\cdots AA^TA$ then $A^n $ is symmetric.

PS: This second conjecture has been shown to be false when $A$ is invertible, see Robert Israel's answer below. But I still think that it's true when $A$ is not invertible.

The post if matrix such $AA^T=A^2$ then $A$ is symmetric? solves the $n=1$ case.

Answer 1

The answer to the second question is no if $n > 2$. $A$ could be an orthogonal matrix (so $A A^T = A^T A = I$) with $A^{2n} = I$, e.g. a rotation by $\pi/n$ $$ \pmatrix{\cos(\pi/n) & \sin(\pi/n)\cr -\sin(\pi/n) & \cos(\pi/n)\cr} $$

Answer 2

The OP had silently modified the conclusion of conjecture 2 from "$A^2$ is symmetric" (which was refuted by Robert Israel's elegant counterexample) to "$A^n$ is symmetric". With this change, we will see that the answer to conjecture 2 is now affirmative, and in turn, conjecture 1 is also true. Using the ideas from loup blanc's answer, we give a proof as follows.

Since the condition $A^{2n+1}=AA^TAA^T\cdots AA^TA$ is invariant under a change of orthonormal basis, we may assume that $$ A=\pmatrix{X&Y\\ 0&0}, $$ where $X$ is a square submatrix and $\pmatrix{X&Y}$ has full row rank. Now, from $A^{2n+1}=AA^TAA^T\cdots AA^TA$, we obtain $A^{2n+1}A^T=(AA^T)^{n+1}$, i.e. $$ X^{2n}(XX^T+YY^T) = (XX^T+YY^T)^{n+1}.\tag{1} $$ As $\pmatrix{X&Y}$ has full row rank, $XX^T+YY^T$ is positive definite. Hence $(1)$ implies that $X$ is invertible and $\det(X^{2n})=\det(XX^T+YY^T)^n$. Therefore $Y=0$ and the problem boils down to proving that $X^n$ is symmetric.

Since $X$ is invertible and $X^{2n+1}=XX^TXX^T\cdots XX^TX$, we obtain from $X^{2n+1}X^{-1}=X^{-1}X^{2n+1}$ that $(XX^T)^{n+1}=(X^TX)^{n+1}$. So, by the uniqueness of positive definite $(n+1)$-th root, we must have $XX^T=X^TX$. But then $X^{2n+1}=XX^TXX^T\cdots XX^TX=X^{n+1}(X^T)^n$. Therefore $X^n=(X^T)^n$. Consequently, $A^n$ is symmetric.

Answer 3

Assume that $A\in M_3(\mathbb{R})$. Since the cases when $rank(A)=1,3$ are solved, assume that $rank(A)=2$.

Prop. Under the hypothesis above, $A^5=AA^TAA^TA$ implies that $A^2$ is symmetric.

Proof. Let $A=\begin{pmatrix}a&b&c\\d&e&f\\g&h&k\end{pmatrix}$. We may assume that $AA^T=diag(u,v,0)$ where $u,v>0$. Clearly $g=h=k=0$ and $A^2=\begin{pmatrix}a^2+bd&b(a+e)&ac+bf\\d(a+e)&bd+e^2&cd+ef\\0&0&0\end{pmatrix}$.

Since $A$ is defined up to a real factor, we may assume $u=1$ and finally we study the system $AA^T=diag(1,v,0)$, $A^5=\begin{pmatrix}a&b&c\\v^2d&v^2e&v^2f\\0&0&0\end{pmatrix}$. We use the Grobner basis software of Maple ; here the difficulty is that there are an infinity of solutions over $\mathbb{C}$ and over $\mathbb{R}$ and consequently, we must work also with hand !

We obtain $ad+be+cf=0$ and $d(v^2-1)=b(v^2-1)=cf(v^2-1)=0$.

Case 1. $v=1$. Then $a^2+b^2+c^2=d^2+e^2+f^2=1$ and $(c^2+f^2)(c^2+f^2-2)=0$.

Case 1.1. $c^2+f^2=2$. Then $c=\pm 1,f=\pm 1$, $a=b=d=e=0$ that is contradictory because $ad+be+cf\not=0$.

Case 1.2. $c^2+f^2=0$. Then $A^2$ is symmetric.

Case 2. $v\not=1$ and $d=b=cf=0$. We obtain $c^2(c^2-2)=0,a^2+c^2=1$ that implies $c=0,a^2=1$.

Case 2.1. $f=0$. then $A^2$ is symmetric.

Case 2.2. $f\not=0$. Then $2e^2+f^2=0$, that is contradictory.

Answer 4

Let $\lambda_k$ be the eigenvalues of $A$ sorted in descending modulus order, i.e. $\vert \lambda_1\vert \geq \vert \lambda_2\vert \geq \dots \geq \vert \lambda_m\vert$ and let $\sigma_k$ be the singular values of $A$ in descending (modulus) order as well.

(i.) zero eigenvalues
$r=\text{ rank}\Big(A^{2n+1}\Big)$ $=\text{rank}\Big(A^{2n+1}(A^{2n+1})^*\Big) =\text{rank}\Big((AA^*)^{2n+1}\Big) =\text{rank}\Big(AA^*\Big) =\text{rank}\Big(A\Big)$
tells us that that eigenvalue $0$ for $A$ is semi-simple and via rank-nullity we see its geometric multiplicity is equal to that of $\big(AA^*\big)$ which necessarily has semi-simple eigenvalues (Spectral Theorem). Conclude $A\text{, }A^{2n}$ and $\big(AA^*\big)$ have semi-simple eigenvalue $0$ with multiplicity $m-r$.

(ii.) non-zero eigenvalues
For $1\leq j \leq r$ we have an orthnormal set of eigenvectors $\big\{\mathbf v_j\big\}$ of $AA^*$ with eigenvalue $\sigma_j^2\neq 0$. Now $\mathbf v_j \in \text{image }A$ so $\mathbf v_j = A\mathbf x_j$ yielding
$\sigma_j^{2n}\mathbf v_j =\big(AA^*\big)^{n}\mathbf v_j=\big(AA^*\big)^{n}A\mathbf x_j=\big(A^{2n} \big)A\mathbf x_j= A^{2n} \mathbf v_j$

$\implies A^{2n}$ has eigenvalues $\lambda_k^{2n}=\sigma_k^{2n}\geq 0$ for $1\leq k \leq m$ [combine (i.) and (ii.)]
$\implies \vert \lambda_k\vert = \sigma_k$ for $1\leq k \leq m$
$\implies \sum_{k=1}^m \vert \lambda_k\vert^2 = \sum_{k=1}^m \sigma_k^2 = \big \Vert A \big \Vert_F^2$ i.e. Schur's Inequality is met with equality hence $A$ is normal, hence $A^n$ is normal with all eigenvalues in $\mathbb R$, i.e. $A^n$ is Hermitian (or real symmetric in the case of the OP).

$\big[$ Alternative argument for normality: $\sum_{k=1}^m \vert \lambda_k\vert = \sum_{k=1}^m \sigma_k$ so $A$ is normal per "deferred proof of the singular value inequality" in Kronecker Product of Normal Matrices $\big]$