Find all square matrices $A$ satisfying ${}^t\!A = -A$.

Question

I am given an $n \times n$ matrix $A$.

If $A = A^2$ and ${}^t\!A = -A$, I need to find matrix of $A$.

$$A^2=A⟹A^2⋅{}^t\!A=A⋅{}^t\!A⟹A^2⋅(-A)=A⋅(-A)$$

How do I show the matrix of A from here ?

Answer 1

$A^t=-A$, hence $A=A^2=(-A^t)^2=(A^t)^2=(A^2)^t=A^t$.

Can you proceed ?

Answer 2

The transpose of $A^{2}$ is the square of the transpose of $A$. From the given hypothesis you get $A=-A$ so $A=0$.

Answer 3

Another way to solve this is to look at the (monic) minimal polynomial $\mu(x)$ of $A$. Note that $A^2=A$ implies that $\mu(x)$ divides $x^2-x$. Thus, any eigenvalue of $A$ can only be $0$ or $1$. Now, if $\lambda$ is an eigenvalue of $A$, then $A^t=-A$ implies that $-\lambda$ is also an eigenvalue of $A$ (unless your matrix $A$ is defined over a field of characteristic $2$, where $-\lambda=\lambda$). This proves that $1$ cannot be an eigenvalue of $A$. That is, $0$ is the only eigenvalue of $A$, whence $\mu(x)=x$, and so $A=0$.

In characteristic $2$, $A$ can be any diagonalizable symmetric matrix whose spectrum is a subset of $\{0,1\}$. There are many such matrices $A$ in this case.