# Normal Matrix with Real Eigenvalues is Hermitian [duplicate]

Let $A$ be a normal matrix. Then I want to show that, if $A$ has real eigenvalues, $A$ is Hermitian. (Notation: * denotes the complex conjugate, T denotes the transpose, and $\dagger$ denotes the conjugate transpose.)

Suppose that $A$ has only real eigenvalues and let $\lambda,v$ be a nonzero eigenvalue-eigenvector pair of $A$. Then we have

$\lambda(v,v)=(v,\lambda v)=(v,Av)=(A^\dagger v,v)$.

$\lambda(v,v)=\lambda^*(v,v)=(\lambda v,v)=(Av,v)$.

Thus, $(Av,v)=(A^\dagger v,v)$ and hence $\sum_i (Av)_i^*v_i=\sum_i (A^\dagger v)_i^* v_i$.

Write this as a matrix product: $(A^*v^*)^Tv=(A^{\dagger*}v^*)^Tv\implies (A^*v^*)^Tv-(A^{\dagger*}v^*)^Tv=0\implies ((A^*v^*)^T-(A^{\dagger*}v^*)^T)v=0$.

Since $v$ is nonzero, we have $(A^*v^*)^T-(A^{\dagger*}v^*)^T=0\implies A^*v^*-A^{\dagger*}v^*=0\implies (A^*-A^{\dagger*})v^*=0$

Since $v$ was nonzero, we have $v^*$ is also nonzero, so $(A^*-A^{\dagger*})=0\implies A^*=A^{\dagger*}$. Thus, we have $A=A^\dagger$ so $A$ is hermitian.

I've seen (here, for example) a much simpler proof. However, I can't find where I use that $A$ is normal, but I know that normality is required. Could anyone could point out where I made this assumption or where I've made a mistake?

You use the "fact" that if $(Bx,y)=0$ and $y\neq 0$, then $Bx =0$. You wrote it as $$((A - A^\dagger) \nu, \nu ) = 0 \implies (A- A^\dagger) \nu = 0$$