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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.

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  • $\begingroup$ I'm not sure I understand the question: since the conjugate of a real number is itself, a real matrix is normal if and only if it is symmetric. (And all normal matrices are automatically diagonalizable, including symmetric real matrices.) $\endgroup$ – Greg Martin Mar 28 at 17:55
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    $\begingroup$ That’s not true: A real orthogonal matrix is also normal but is in general not symmetric (consider e.g. rotation by 90 degrees). A normal real matrix in general need not diagonalizable over the reals, as the example of orthogonal matrces shows (it is diagonalizable over the complex numbers, of course). $\endgroup$ – Sascha Baer Mar 28 at 17:57
  • $\begingroup$ Ack, you're completely right. I was confusing normal with Hermitian. $\endgroup$ – Greg Martin Mar 28 at 19:53
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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.

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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.)

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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 $$A^T=P^{-T}DP^T=PDP^{-1}=A.$$

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