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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.

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  • $\begingroup$ (Addendum to bounty information). Well, when $n>2$, we can append Robert Israel's counterexample with a zero matrix to form a counterexample. So, we only need to consider the case $n=2$ (and the conjecture boils down to conjecture 1) with a singular $A$ of size at least $3\times3$. $\endgroup$
    – user1551
    Commented Feb 3, 2015 at 11:40

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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 $\pi/n$ $$ \pmatrix{\cos(\pi/n) & \sin(\pi/n)\cr -\sin(\pi/n) & \cos(\pi/n)\cr} $$

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    $\begingroup$ If you put that $2 \times 2$ matrix in the upper left corner of a $3 \times 3$ matrix, and fill in with zeroes, you get a noninvertible example for $n > 2$. So the "PS" conjecture seems to be wrong as well. $\endgroup$ Commented Dec 10, 2014 at 13:25
  • $\begingroup$ For $2\times2$ matrices, however, that "PS" is correct if $A$ is noninvertible. Let $A=uv^T$ for some vectors $u$ and $v$. The condition $A^{2n+1}=(AA^T)^nA$ gives $(v^Tu)^{2n} uv^T = (v^Tv)^n (u^Tu)^n uv^T$. By Cauchy-Schwarz inequality, one can assert that $u,v$ are linearly dependent. Hence $A$ is symmetric and so is $A^2$. This result generalises to the case of a general rank-1 square matrix $A$ as well. $\endgroup$
    – user1551
    Commented Dec 10, 2014 at 16:41
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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.

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  • $\begingroup$ @ user1551 , thanks for the bounty. Yet, I do not understand the decomposition $A=\pmatrix{X&Y\\ 0&0}$ where $X$ is invertible ; that implies that $A$ is one to one over $im(A)$. How do you do when $A=\begin{pmatrix}1&1&4\\2&2&5\\0&0&0\end{pmatrix}$ ? $\endgroup$
    – user91684
    Commented Feb 4, 2015 at 18:52
  • $\begingroup$ @loupblanc Thanks for pointing out the error. I messed up the order of proof after a series of cut-and-pastes before posting the answer. That $X$ is invertible should be a consequence rather than an assumption. See my new edit. $\endgroup$
    – user1551
    Commented Feb 4, 2015 at 21:25
  • $\begingroup$ @ user1551 , well-done. $\endgroup$
    – user91684
    Commented Feb 5, 2015 at 10:51
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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.

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