# 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.) Mar 28, 2021 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). Mar 28, 2021 at 17:57
• Ack, you're completely right. I was confusing normal with Hermitian. Mar 28, 2021 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 $$A^T=P^{-T}DP^T=PDP^{-1}=A.$$