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Let $\boldsymbol{A}$ is a real symmetric matrix, $\boldsymbol{B}$ is a real antisymmetric matrix, $\boldsymbol{A}^2 = \boldsymbol{B}^2$, prove $\boldsymbol{A} = \boldsymbol{B} = \boldsymbol{0}$.

I tried the second-order matrix. Let $\boldsymbol{A} = \begin{bmatrix} a_{11} & a_{12} \\ a_{12} & a_{22} \end{bmatrix}$, $\boldsymbol{B} = \begin{bmatrix} b_{11} & b_{12} \\ {-b_{12}} & b_{22} \end{bmatrix}$, $a_{ij}, b_{ij} \in \mathbb{R}$.

\begin{align} \boldsymbol{A}^2 &= \begin{bmatrix} {a_{11}^2 + a_{12}^2} & {a_{11}a_{12} + a_{12}a_{22}} \\ {a_{11}a_{12} + a_{12}a_{22}} & {a_{12}^2 + a_{22}^2} \end{bmatrix} \\ \boldsymbol{B}^2 &= \begin{bmatrix} {b_{11}^2 - b_{12}^2} & {b_{11}b_{12} + b_{12}b_{22}} \\ {-b_{11}b_{12} - b_{12}b_{22}} & {-b_{12}^2 + b_{22}^2} \end{bmatrix} \end{align}

Because $\boldsymbol{A}^2 = \boldsymbol{B}^2$, I got \begin{align} a_{11}^2 + a_{12}^2 &= b_{11}^2 - b_{12}^2 \\ a_{11}a_{12} + a_{12}a_{22} &= b_{11}b_{12} + b_{12}b_{22} \\ a_{11}a_{12} + a_{12}a_{22} &= -b_{11}b_{12} - b_{12}b_{22} \\ a_{12}^2 + b_{22}^2 &= -b_{12}^2 + b_{22}^2 \end{align}

Hence, \begin{align} a_{11}a_{12}+a_{12}a_{22}=b_{11}b_{12}+b_{12}b_{22}=0 \end{align}

I lost my momentum here.

KaiserKatze
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    What progress have you made on the question? On MSE we don't like to do your homework or to just solve problems for you, we're actually trying to help you figure things out, so it's good to show some effort. – Patrick Da Silva Feb 23 '21 at 15:16
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3 Answers3

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Let us denote the usual inner product on $ \mathbb R^n$ by $(\cdot|\cdot).$ Then

$$||Ax||^2=(Ax|Ax)=(A^TAx|x)=(A^2x|x)=(B^2x|x)=(Bx|B^Tx)=(Bx|-Bx)=-(Bx|Bx)=-||Bx||^2$$

for all $x \in \mathbb R^n.$

Hence $-||Bx||^2 \ge 0$ for all $x \in \mathbb R^n.$ This gives

$||Bx||^2 = 0$ for all $x \in \mathbb R^n$ and thus $||Ax||^2 = 0$ for all $x \in \mathbb R^n$ .

Cosequence: $A=B=0.$

Fred
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Let $\mu$ be a complex eigenvalue of $B$. Because $B$ is antisymmetric, then $\mu \in i\mathbb{R}$, so $\mu^2$ is a nonpositive real number. But $\mu^2$ is eigenvalue of $B^2$, so of $A^2$, but the eigenvalues of $A^2$ are all nonnegative since $A$ is symmetric. So $\mu=0$. You get $B=0$, so $A^2=0$, so $A$ is real symmetric and nilpotent, so $A=0$.

TheSilverDoe
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note the orthogonality
$-1\cdot \text{trace}\big(BA\big)= \text{trace}\big(B^TA\big)= \text{trace}\big(B^TA\big)^T= \text{trace}\big(A^TB\big)= \text{trace}\big(AB\big)= \text{trace}\big(BA\big)$
$\implies \text{trace}\big(BA\big)= 0$

using this we can compute the squared Frobenius norm of A+B
$\Big \Vert A+B\Big \Vert_F^2= \text{trace}\big(A^TA\big) +\text{trace}\big(B^TA\big)+\text{trace}\big(A^TB\big)+\text{trace}\big(B^TB\big) $
$= \text{trace}\big(A^2\big) +0+0 -\text{trace}\big(B^2\big) $
$=0 $
$\implies A+B = \mathbf 0\implies A = -B $
so $A$ is both symmetric and skew symmetric, i.e. $A=A^T=-A^T \implies A=\mathbf 0=-B = B$

user8675309
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