find matrix $A$ that satisfies $A^2=A$and $A^T=-A$ How should I find matrix $A$ that satisfies $A^2=A$ and $A^T=-A$?
I tried from $Ax=\lambda x$ where $\lambda$ is the eigenvalue of $A$ and $x$ is eigenvector of $A$. But that wasn't seems fit for this problem. Can someone tell me how to show this?
 A: Look at the range/column space of $A$, $\text{range}(A)=\{Ax:x\in\Bbb F^n\}$. The matrix $A$ acts the same as the identity matrix on this subspace due to $A^2x=Ax$. Now for any vector $y\in\text{range}(A),$ $$y^Ty=(y^TA^T)y=y^T(-Ay)=y^T(-y)=-y^Ty.$$
If $\Bbb F=\Bbb Q,\Bbb R$, this is possible only when $y$ is the zero vector, meaning $\text{range}(A)$ contains only the zero vector.
Edit: The argument above does not exactly work for $\Bbb F=\Bbb C$, because $y^Ty=0$ does not imply $y$ is the zero vector (for example if $y=\begin{bmatrix}i\\1\end{bmatrix}$, $y^Ty=i^2+1^2=0$). But we can make a modification:
For any vector $y\in\text{range}(A)$ and any vector $x\in\Bbb F^n$, $$y^Tx=(y^TA^T)x=y^T(-Ax)=(y^TA^T)(-Ax)=y^T((-A)^2x)=y^T(Ax)=(y^T\cdot -A^T)x=-y^Tx.$$ Now this holds for any $x$, meaning $y^Tx=0$ for all $x$. In particular put x to be a standard basis vector $$x=e_i=\begin{bmatrix}0\\\vdots\\1\\\vdots\\0\end{bmatrix}$$ ($i$-th entry is $1$ and all else $0$) to see that the $i$-th entry of $y$ is $y^Te_i=0$.
A: $$A^T=-A \implies A^T A=-A^2 \implies A^T A=-A \text { since } A^2=A$$
But $A^TA$  is symmetric, so $A$ is symmetric as well. Therefore $A^T=A$. But since $A^T=-A$, we must have $$A=-A$$
