if A*=2A, then A=0

Question

let A be a matrix of size n×n such that A* = 2A. prove that A=0

Suppose that $A^* =2A$ and r is an eigenvalue of $A$, then $Ax=rx$ for some eigenvector x

$$r(x,x) = (rx,x)=(Ax,x)=(x,A^*x)=(x,2Ax)=2(A^*x,x)=4(Ax,x)$$ So, $$(Ax,x)=4(Ax,x) \implies (A-4A)(x,x)=0 \implies -3A=0 $$ then $A=0$.

Is it right?

Answer 1

I see now you want a solution verification. Your solution is correct up until you say that $(Ax,x)=4(Ax,x)\implies (A-4A)(x,x)=0$. All that you can conclude is $((A-4A)x,x)=0$, but you cannot "pull out" the $(A-4A)$. You could do this if $(A-4A)$ was a scalar, but here it is a matrix. I do not know if your exact method can be salvaged.

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Hint

Using the relation $A^*=2A$, what can you deduce about $(A^{*})^{*}$? Use this, and the fact that $(A^*)^*=A$, to prove $A=0$. If you use only the equations I have stated, without using any vectors $x$ or inner products $(Ax,x)$, then you can find a very short proof.