# 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?

• How many votes does this question have? Aug 6, 2019 at 3:31
• Do you want zero matrix?
– edm
Aug 6, 2019 at 3:34
• @edm I want to see why this would be zero matrix Aug 6, 2019 at 3:35
• What does $A^2=A$ tell you about the eigenvalues? What does $A^t=-A$ tell you about the eigenvalues? What do these two results together tell you about the eigenvalues? Aug 6, 2019 at 3:38

$$A^T=-A \implies A^T A=-A^2 \implies A^T A=-A \text { since } A^2=A$$ But $$A^TA$$ is symmetric, so $$A$$ is symmetric as well. Therefore $$A^T=A$$. But since $$A^T=-A$$, we must have $$A=-A$$
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 $$y^Te_i=0$$.