# 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.

• 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.) – Greg Martin Mar 28 at 17:55
• 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). – Sascha Baer Mar 28 at 17:57
• Ack, you're completely right. I was confusing normal with Hermitian. – Greg Martin Mar 28 at 19:53

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.

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.)

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