TITLE: Is a real matrix that is both normal and diagonalizable symmetric? If so, is there a proof of this not using the spectral theorem?
QUESTION [7 upvotes]: Given a quadratic real matrix $A$ for which we know it is diagonalizable ($D = P^{-1}AP$ for a diagonal matrix $D$) and that it is normal ($AA^T = A^TA$), is it true that $A$ is symmetric? By the spectral theorem over $\mathbb{C}$, $A$ is orthogonally diagonalizable (say via some unitary matrix $Q$). It then seems to me that from this we should get that the eigenspaces are orthogonal and hence there must also exist a real orthonormal basis of eigenvectors. Is this last correct? From this it would then follow by the real version of the spectral theorem that $A$ is symmetric.
If it is, is there a way to prove this statement directly, without appealing to the spectral theorem? This would give a nice justification/explanation for the difference in the two spectral theorems over $\mathbb{C}$ and $\mathbb{R}$ relating it directly to whether the matrix is diagonalizable in the first place.

REPLY [3 votes]: One can obtain that $A$ is symmetric by direct computation. The equality $A^TA=AA^T$ is
$$\tag1
PDP^{-1}P^{-T}DP^T=P^{-T}DP^TPDP^{-1}.
$$
Multiplying by $P^T$ on the left and by $P$ on the right,
$$\tag2
P^TPD(P^TP)^{-1}DP^TP=DP^TPD.
$$
Now by $(P^TP)^{-1}$ on the right,
$$\tag3
\big[P^TPD(P^TP)^{-1}\big]\,D=D\,\big[P^TPD(P^TP)^{-1}\big].
$$
We may assume without loss of generality that $A$ (and hence $D$) is invertible, by adding a suitable scalar multiple of the identity. By writing $D$ as a block-diagonal matrix with distinct eigenvalues it is easy to see that the matrices that commute with $D$ are block diagonal. So, with $$\tag4X=P^TPD(P^TP)^{-1}$$ we have $XD=DX$ and thus $X$ is block-diagonal. Knowing that $X$ is block-diagonal we rewrite $(4)$ as
$$
XP^TP=P^TPD.
$$
The (block) diagonal entries of this equality are
$$
X_{kk}(P^TP)_{kk}=\lambda_k\,(P^TP)_{kk}.
$$
Because $P^TP$ is positive definite, its block-diagonal entries are positive definite; in particular, invertible. We conclude that
$
X_{kk}=\lambda_{kk}\,I
$, that is $$X=D.$$
We can unravel this as
$$
D=P^TPD(P^TP)^{-1}=P^TPDP^{-1}P^{-T}.$$ Multiplying by $P^{-T}$ on the left and by $P^T$ on the right we obtain
$$A^T=P^{-T}DP^T=PDP^{-1}=A.$$