How to prove that Real normal matrix with real Eigenvalues is symmetric

Question

I wanted to show above. I know that if instead of real matrix of we have complex matrix then I can do using spectral theorem. But I don't able to do this. Any help will be appreciated.

Answer 1

1. Since $A^TA$ is real symmetric, we may split $\mathbb R^n$ into a direct sum of orthogonal subspaces $V_1\oplus\cdots\oplus V_k$, where each $V_i$ is the eigenspace for a distinct eigenvalue $s_i$ of $A^TA$.
2. Now, the normality condition implies that $(A^TA)A = (AA^T)A = A(A^TA)$, i.e. $A$ commutes with $A^TA$. Hence each $V_i$ is an invariant subspace of $A$. The similar holds for $A^T$.
3. However, on each $V_i$ we have $A^TA=s_iI$. When $s_i=0$, this means $A$ is the zero map on $V_i$, which is self-adjoint. When $s_i\ne0$, we have $(\frac1{\sqrt{s_i}}A)^T(\frac1{\sqrt{s_i}}A)=I$ on $V_i$ and hence the restriction of $A$ on $V_i$ is $\sqrt{s_i}Q_i$ for some real orthogonal map $Q_i$. Since $A$ has a real spectrum, $Q_i$ must have a real spectrum too.
4. Therefore, the problem boils down to showing that every real orthogonal matrix $Q$ with a real spectrum is symmetric. This can be easily proved by mathematical induction on the size of $Q$.

Answer 2

By spectral theorem $A=U^*DU$, U unitary, D diagonal and real by your hypothesis. Then $A^*=(U^*DU)^*=U^*D^*U=U^*DU=A$ since $D^*=D$ being D real and diagonal. But $A^*=A^t$ since A is also real. Hence $A^t=A$