Let $A \in M_n(\mathbb R)$.

Show that if $A(\,{}^t\!A)A$ is symmetric, then $A$ is also symmetric.

My attempt:

If $A \in Gl_n(\mathbb{R})$,

We have : ${}^t(A^{-1})=(\,{}^t\!A)^{-1}$ Then : $${}^t\!AA\,{}^t\!A= A{}^tAA A\,{}^t\!A=(\,{}^\!tA)^{-1}A\,{}^t\!AA$$

This does not allow me to continue ...

I started with $A$ invertible for use after density. However, it may be a bad idea, what do you think?

Thank you

Edit : ${}^tA$ is the transpose of A.

  • $\begingroup$ What does $^tA$ denote? $\endgroup$ – Git Gud Apr 3 '14 at 9:21
  • $\begingroup$ @GitGud The transpose of $A$ $\endgroup$ – ah11950 Apr 3 '14 at 9:22
  • 2
    $\begingroup$ What about $A^t$? $\endgroup$ – Git Gud Apr 3 '14 at 9:24
  • 2
    $\begingroup$ @GitGud In many countries they denote ${}^tA$. Like France $\endgroup$ – user119228 Apr 3 '14 at 16:36

Hint: Let me denote the transpose of $A$ by $A^t$. Since $AA^tA$ is symmetric,

$$AA^tA=A^tAA^t\Longrightarrow (AA^t)^3=(A^tA)^3 \Longrightarrow AA^t=A^tA.$$

That is to say, $A$ is normal. Then use the fact that $A$ unitarily similar to a diagonal matrix to show $A$ is symmetric.

Update: Let me extend the statement to the complex valued case and add some details to the proof. To begin with, let me denote the Hermitian conjugate of $A\in M_n(\Bbb C)$ by $A^*$.

Claim: Given $A\in M_n(\Bbb C)$, if $AA^*A$ is Hermitian, then $A$ is also Hermitian.

Proof: Denote $B:=AA^*A$. As shown in the hint, since $B=B^*$, $$(AA^*)^3=BB^*=B^*B=(A^*A)^3.$$ Since $AA^*$ and $A^*A$ are Hermitian, the equality above implies that $AA^*=A^*A$. This is due to a general fact: if $H$ is Hermitian and $f:\Bbb R\to\Bbb R$ is a function, then there is a natural way to uniquely define a Hermitian matrix $f(H)$. In our situation, $H=(AA^*)^3=(A^*A)^3$, $f(x)=x^{\frac{1}{3}}$.

We have seen that $A$ is a normal matrix, so $A=U\Lambda U^*$ for some unitary $U$ matrix and some diagonal matrix $\Lambda$. Then $$\Lambda\Lambda^*\Lambda=U^*BU=U^*B^*U=\Lambda^*\Lambda\Lambda^*.$$ Since $\Lambda$ is diagonal, the equality above implies that it is real valued, so $A$ is Hermitian. $\quad\square$

  • $\begingroup$ @Nico: In contrast to your opinion, I think only this step is trivial in the whole argument, because other steps are based on certain facts in linear algebra but this step only needs direct calculation: $(AA^t)^3=(AA^tA)(A^tAA^t)=(A^tAA^t)(AA^tA)=(A^tA)^3$. $\endgroup$ – Hu Zhengtang Apr 3 '14 at 20:27
  • $\begingroup$ Arf, Silly me! Thank you very much. Very smart. $\endgroup$ – user117932 Apr 3 '14 at 20:29
  • $\begingroup$ (I will accept your answer later because I think it deserve more that two upvotes) $\endgroup$ – user117932 Apr 3 '14 at 20:31
  • $\begingroup$ @Nico: You are welcome. Thank you for your consideration. $\endgroup$ – Hu Zhengtang Apr 3 '14 at 20:33
  • $\begingroup$ @HuZhengtang I don't get $(AA^t)^3=(A^tA)^3 \Longrightarrow AA^t=A^tA$. Can you please clarify? $\endgroup$ – Git Gud Apr 3 '14 at 21:43

By an orthogonal change of basis, we may assume that the singular value decomposition of $A$ is $US$, where $U$ is real orthogonal and $S$ is nonnegative and diagonal. Then the given condition implies that $US^3$ is symmetric. Hence $(US^3)(S^3U^T) = (S^3U^T)(US^3)$ (the credit of considering $(AA^TA)(A^TAA^T)=(A^TAA^T)(AA^TA)$ here goes to Hu Zhengtang), or $US^6U^T = S^6$. Since positive semidefinite matrices have unique positive semidefinite $n$-th roots, it follows that $USU^T = S$. Therefore both $U$ and $U^T$ commute with $S$, and in turn also with $S^+$, the Moore-Penrose pseudoinverse of $S$. Consequently, $$ A=US=US^3(S^+)^2=(US^3)^T(S^+)^2=S^3U^T(S^+)^2=S^3(S^+)^2U^T=SU^T=A^T. $$


Here I only answer the question for case $A\in M_2(\mathbb{R})$ $$A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}\Longrightarrow AA^tA=\begin{bmatrix} \sim & a^2b+b^3+acd+bd^2\\ a^2c+abd+c^3+c^2d & \sim \end{bmatrix}$$ $$a^2c+abd+c^3+c^2d=a^2b+b^3+acd+bd^2\Longrightarrow a^2(c-b)+ad(b-c)+(c^2+bc+b^2)(c-b)+d^2(b-c)=0$$ $$\Longrightarrow c=d\text{ or } a^2-ad+c^2+cb+b^2-d^2=0$$ But pay attention that we are in real numbers so squares can't be negative. Thus; $$a^2+b^2+c^2+d^2=ad-bc\leq (\max\{a,d\})^2\leq a^2+d^2\Longrightarrow b^2+c^2=0\Longrightarrow b=c=0$$

Therefore $A$ is asymmetric.


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