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let matrix $A_{n\times n}$ is real matrix,such $AA^T=A^2$, The transpose of matrix $A$ is written $A^T$,

show that :

the matrix $A$ is Symmetric matrices

maybe this problem have more methos,because it is know that if matrix $A$ is symmetric,then we have $AA^T=A^2$,But for my problem,I can't prove it.Thank you

math110
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5 Answers5

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Although not an exact duplicate, the method for solving An equivalent condition for a real matrix to be skew-symmetric applies. Briefly speaking, for real square matrices, $\langle X,Y\rangle = \operatorname{trace}(Y^TX)$ defines an inner product and symmetric matrices are orthogonal to skew-symmetric matrices. Now, write $A=H+K$, where $H=\frac12(A+A^T)$ is the symmetric part and $K=\frac12(A-A^T)$ is the skew-symmetric part. Then $AA^T=A^2$ implies that $(H+K)K=0$ and in turn $HK=-K^2=K^TK$. Taking trace on both sides, we get $\langle K,H\rangle=\langle K,K\rangle$. Since $\langle K,H\rangle=0$, it follows that $\langle K,K\rangle=0$ and $K=0$, i.e. $A$ is symmetric.

Community
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user1551
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  • I think it's better to do this in 1 line with the (squared) Frobenius norm $\big \Vert A-A^T \big\Vert_F^2 = 2\cdot \text{trace}\big(AA^T\big) - \text{trace}\big(A^2\big) - \text{trace}\big((A^2)^T\big)=2 \cdot \text{trace}\big(A A^T \big) - 2\cdot \text{trace}\big(A^2\big)=0$ – user8675309 Nov 01 '24 at 00:51
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The solution suggested by user1551 is very nice, but, just for fun, here is a 2nd approach.

Assume that $AA^T = AA$. First we prove the following:

Claim: $AA = \underbrace{AA^T}_P = A^T A^T = \underbrace{A^TA}_Q$

Proof: Transposing $AA^T = AA$ gives $AA^T = A^T A^T$. Now we just need to worry about $Q$. With some fiddling around, we notice $$Q^2 = A^TAA^TA = A^TA^TA^TA = AAA^TA =AAAA= AA^T AA^T = P^2.$$ Since $P,Q$ are positive matrices, $P^2 = Q^2 \Rightarrow P=Q$ (uniqueness of positive square roots) and the claim is proved.

Since $A$ has real entries, $A^T$ equals its conjugate transpose $A^*$ so we have proved that $A$ is normal. At this point the problem is quite easy since $A = U D U^*$ for some diagonal $D$ and some unitary $U$. From $AA^* = A^2$ follows $DD^* = D^2$ which implies $D$ has real entries i.e. $D=D^*$. From here we calculate $$ A^T = A^* = (UDU^*)^* = UD^*U^* = UDU^* = A$$ so $A$ is symmetric.

Yimeng TANG
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Mike F
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Mike's answer is very nice, but for fun, here is a third approach. It is developed along the same line as Mike's, but using more basic techniques.

Observe that if $W$ is a symmetric or skew-symmetric matrix and $W^2=0$, then $W=0$. This is because $\|Wx\|_2^2=x^TW^TWx=\pm x^TW^2x=0$ for all $x$ and hence $Wx\equiv 0$.

Now, from the condition $AA^T=A^2$, we obtain $AA^T=AA=A^TA^T$ and in turn \begin{cases} AA^TAA^T = AAAA = AA^TA^TA,\\ A^TAAA^T = A^TA^TA^TA^T = AAAA = AAA^TA = A^TA^TA^TA = A^TAA^TA. \end{cases} Hence $(AA^T-A^TA)^2 = 0$. Since $W_1=AA^T-A^TA$ is symmetric and $W_1^2=0$, we infer that $W_1=0$, i.e. $AA^T=A^TA$.

But then we get $(A-A^T)^2 = AA-AA^T-A^TA+A^TA^T=0$. Since $W_2=A-A^T$ is skew-symmetric and $W_2^2=0$, we conclude that $W_2=A-A^T=0$, i.e. $A$ is symmetric.

Edit: As some users have pointed out that the statement is trivial when $A$ is invertible, we will also develop a proof along this line. By the given conditions, $\ker A=\ker A^T$:
  1. $\ker A \subseteq \ker A^T$ because $Ax=0 \Rightarrow \|A^Tx\|^2=x^TAA^Tx=x^TA(Ax)=0.$
  2. $\ker A^T \subseteq \ker A$ because $$ A^Tx=0 \Rightarrow AAx=AA^Tx=0 \Rightarrow Ax\in\ker A\subseteq\ker A^T \Rightarrow \|Ax\|^2=x^T(A^T(Ax))=0. $$

Therefore, by an orthonormal change of basis, we may assume that $A=\pmatrix{B&0\\ 0&0_{k\times k}}$, where $B$ is invertible, $BB^T=B^2$ and $k$ is the nullity of $A$. Hence $B^T=B^{-1}(BB^T)=B^{-1}(B^2)=B$ and $A$ is symmetric.

user1551
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  • Well done, very basic indeed! Already, what I posted I'd toyed with a bit to make it more elementary. It just goes to show, once you have a statement proved, finnessing the proof is a totally different ball game. You're psychologically free to try all sorts of funny things. Case in point, this proof is very nice, but someone who never saw another proof would probably read this and think "now how the hell did they come up that?!". – Mike F Nov 18 '13 at 10:49
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Just to add a different approach:

  1. We wish to prove the more general fact for complex matrices obtained replacing transpose with Hermitian transpose.

  2. use a Schur decomposition to reduce to the case of $A$ upper triangular.

  3. For $A$ upper triangular, computing the (1,1) terms of $AA^T$ and $A^2$ yields $A_{11}^2 = \sum_{j=1}^n |A_{1j}|^2$. By comparing absolute values, it is easier to see that $A_{11}$ must be real and $A_{1j}=0$ for all $j\neq 1$.

  4. go on like this by induction, each time proving that all terms in the $i$-th row are zero apart from the one on the diagonal

  5. You have proved that your upper triangular $A$ is real diagonal, hence Hermitian.

Federico Poloni
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Given $AA^t=A^2$. Premultiply by $A^{-1}$. You obtain $A^t=A$ which means that the matrix is symmetric. $AA^{-1}=A^{-1}A=I_d\,\,$identity matrix)

{The above result is true only if $A$ is invertible because only then premultiplying by $A^{-1}$ is defined..}

learner
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Apurv
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