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

Question

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

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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.

Answer 1

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: $A=B=0.$

Answer 2

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 $A=0$.

Answer 3

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. $A=A^T=-A^T \implies A=\mathbf 0=-B = B$