Is a real matrix that is both normal and diagonalizable symmetric? If so, is there a proof of this not using the spectral theorem? 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.
 A: Here is a proof avoiding the spectral Theorem.
Let $\lambda _1, \lambda _2, \ldots , \lambda _k$ be the distinct eigenvalues of $A$ and let $E_1, E_2,  \cdots ,  E_k$ be the projections onto the
corresponding eigenspaces.
By seeing things from the point of view of a (not necessarily orthonormal) basis of eigenvectors,  it is very very easy  to
prove that (don't let the long expression scare you)
$$
   E_i = \frac{(A-\lambda _1)\ldots \widehat{(A-\lambda _i)}\ldots (A-\lambda _k)}{(\lambda _i-\lambda _1)\ldots \widehat{(\lambda _i-\lambda _i)}\ldots (\lambda _i-\lambda _k)},
  $$
where the hat means omission.
From this it is clear that each $E_i$ is also normal.

Lemma.  Any normal, idempotent, real matrix is symmetric.
Proof.  Let $E$  be such a matrix.   We first claim that $\text{Ker}(E)=\text{Ker}(E^TE)$.  To see this, observe that
the inclusion $\text{Ker}(E)\subseteq \text{Ker}(E^TE)$ is evident.  On the other hand, if $x\in \text{Ker}(E^TE)$, then
$$
  \|Ex\|^2 =   \langle Ex, Ex\rangle  =  \langle E^TEx, x\rangle  =0,
  $$
so $x\in  \text{Ker}(E)$.
We then have that
$$
  \text{Ker}(E)=\text{Ker}(E^TE) =\text{Ker}(EE^T) =\text{Ker}(E^T).
  $$
Recalling that the range $R(A^T)$,  of the transpose of  a matrix $A$, coincides with
$\text{Ker}(A)^\perp$, we then have that
$$
  R(E^T) =
  \text{Ker}(E)^\perp =
  \text{Ker}(E^T)^\perp =
  R(E).
  $$
We then see that $E$ and $E^T$ are projections sharing range and kernel, so necessarily $E=E^T$.  QED

Back to the question, we then have that
$$
  A=\sum_{i=1}^k \lambda _kP_k,
  $$
so we conclude that $A$ is symmetric.
A: You can also do this with simultaneous diagonalizability and two classical inequalities: Cauchy-Schwarz and rearrangement inequality.
$A$ is real diagonalizable and so is $A^T$. Now $A$ and $A^T$ commute, hence they are simultaneously diagonalizable such that $A=SD_1S^{-1}$ and $A^T = SD_2S^{-1}$.  Since the trace is invariant to conjugation, this implies
(i.)
$\text{trace}\big(A^2\big)=\text{trace}\big(D_1^2\big)\geq \text{trace}\big(D_2D_1\big)=\text{trace}\big(A^TA\big)  =\big\Vert A\Big \Vert_F^2$
by rearrangement inequality
(ii.)
$\text{trace}\big(A^2\big)\leq \text{trace}\big(A^TA\big)  =\big\Vert A\Big \Vert_F^2$
by Cauchy-Schwarz
Thus $\text{trace}\big(A^2\big)= \big\Vert A\Big \Vert_F^2$ so Cauchy-Schwarz is met with equality$\implies A = \eta\cdot A^T$, where $\eta \in\big\{-1,1\big\}$ because $A$ and its transpose have the same Frobenius norms.  Thus $A$ is either symmetric or skew symmetric.  (And supposing $A\neq \mathbf 0$ we can eliminate skew symmetry because $\mathbf x^T A\mathbf x=0$ for all $\mathbf x\in \mathbb R^n$ by skew symmetry, so $A$ could not have any non-zero real eigenvalues, which we know isn't true.)
A: 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
$$A^T=P^{-T}DP^T=PDP^{-1}=A.$$
