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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.

Widawensen
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3 Answers3

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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. $\quad\square$

Hu Zhengtang
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  • @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: $(AA^t)^3=(AA^tA)(A^tAA^t)=(A^tAA^t)(AA^tA)=(A^tA)^3$. – Hu Zhengtang Apr 03 '14 at 20:27
  • Arf, Silly me! Thank you very much. Very smart. –  Apr 03 '14 at 20:29
  • (I will accept your answer later because I think it deserve more that two upvotes) –  Apr 03 '14 at 20:31
  • @Nico: You are welcome. Thank you for your consideration. – Hu Zhengtang Apr 03 '14 at 20:33
  • @HuZhengtang I don't get $(AA^t)^3=(A^tA)^3 \Longrightarrow AA^t=A^tA$. Can you please clarify? – Git Gud Apr 03 '14 at 21:43
  • @GitGud: I am not clear about your background knowledge. Can you see if $S\in M_n(\Bbb R)$ is symmetric , then there exists a unique symmetric $T\in M_n(\Bbb R)$, such that $T^3=S$? – Hu Zhengtang Apr 04 '14 at 02:54
  • That's not exactly true due to permutation of the eigenvalues. Modulo this, yes, I can see it. – Git Gud Apr 04 '14 at 06:44
  • @GitGud: It is exactly true. Please verify this fact by yourself. Once you accept this fact, can you answer your own question? – Hu Zhengtang Apr 04 '14 at 09:16
  • @HuZhengtang You're right, I had missed something, thanks. – Git Gud Apr 04 '14 at 09:23
  • @GitGud: You are welcome. – Hu Zhengtang Apr 04 '14 at 09:38
  • @HuZhengtang Real normal matrices are unitarily (but not necessarily orthogonally) similar to complex (but not necessarily real) diagonal matrices. I think there is still a large gap between the last step in your hint and a complete solution. – user1551 Apr 04 '14 at 17:50
  • @user1551: I don't think the gap is so large(as I claimed my answer is only a hint rather than a full solution). Let me denote the Hermitian conjugate of $X\in M_n(\Bbb C)$ by $X^$. Then $A=U\Lambda U^$ for some unitary $U$ and some diagonal $\Lambda$. Then $AA^tA=AA^A=U\Lambda \Lambda^\Lambda U^$ is real and symmetric implies that $\Lambda \Lambda^\Lambda$ is Hermitian, and hence real. I think you can complete the argument from here. – Hu Zhengtang Apr 04 '14 at 18:26
  • @HuZhengtang Can you please elaborate on the uniqueness of the cubic root of a symmetric matrix? I'm stuck proving it. – Gabriel Romon Apr 04 '14 at 20:59
  • @GabrielR.: I will only provide a sketch. If $S$ and $T$ are real symmetric and $S=T^3$, then since $ST=TS$, for any eigenvalue $\lambda$ of $S$, the associated eigenspace $V_\lambda:={v: Sv=\lambda v}$ it $T$-invariant, i.e. $T(V_\lambda)\subset V_\lambda$. Then with out loss of generality, we may assume $S=\lambda I$, where $I$ is the identity matrix. Now can your prove that if $T$ is real symmetric and $T^3=\lambda I$, then $T=\lambda^{\frac{1}{3}}I$? – Hu Zhengtang Apr 05 '14 at 04:46
  • Thanks for your answer. By diagonalizing T I can prove your last claim. However I can't figure out why you can suppose without loss of generality that $S=\lambda I$. Can you enlighten me about this? – Gabriel Romon Apr 05 '14 at 05:08
  • @GabrielR.: I think I have explained in my last comment that restricted to any eigenspace $V_\lambda$ of $S$, $S$ acts like $\lambda I$ and $T(V_\lambda)\subset V_\lambda$. Maybe I should also have mentioned that the whose space is a direct sum of eigenspaces of $S$. – Hu Zhengtang Apr 05 '14 at 05:13
  • Hi, @GabrielR., sorry for bothering. I found you also commented under this answer of mine. May I know if you downvoted my answer? If so, since I have corrected the mistake in the answer pointed by you(and since that is an inessential mistake), would you mind canceling your downvote? I am sorry for being rude(I care so much about the downvote because it is the only downvote I received); if either you didn't downvote or you don't want to cancel it, please just ignore my message. I apologize again for my rudeness. – Hu Zhengtang Apr 06 '14 at 13:30
  • @HuZhengtang Wow this goes way back... I did downvote it, but cancelled it now since it is correct. – Gabriel Romon Apr 06 '14 at 13:59
  • @GabrielR.: Thank you so much for cancelling the downvote and for the upvote. – Hu Zhengtang Apr 06 '14 at 14:07
2

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, $$ A=US=US^3(S^+)^2=(US^3)^T(S^+)^2=S^3U^T(S^+)^2=S^3(S^+)^2U^T=SU^T=A^T. $$

user1551
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0

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.

mle
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AmirHosein Sadeghimanesh
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