Symmetric Matrices: Real Eigenvalues & Orthogonal Eigenvectors

An $n\times n$ matrix $A$ is orthogonally diagonalizable (Definition 9.1) if there is an orthogonal matrix $S$ (so $S^{-1}=S^T$) with $S^{-1}AS=S^TAS=D$ diagonal; equivalently, $A$ has an orthonormal eigenbasis. This is the gold standard of diagonalization — the change-of-basis matrix is a rigid motion that preserves lengths and angles. Which matrices earn it? Remark 9.1 supplies half the answer at once. If $A=SDS^T$ with $S$ orthogonal and $D$ diagonal, then $A^T=(SDS^T)^T=(S^T)^T D^T S^T=SDS^T=A,$ using $D^T=D$. So every orthogonally diagonalizable matrix is symmetric. The Spectral Theorem (Theorem 9.1) will prove the striking converse — that symmetry is also sufficient — but the rest of this lesson gathers the eigenvalue and eigenvector facts that make that converse possible.

A general real matrix can have complex eigenvalues: the rotation $\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ has eigenvalues $\pm i$. Proposition 9.2 says this never happens for a symmetric matrix — a symmetric $n\times n$ matrix $A$ has $n$ real eigenvalues, counted with algebraic multiplicity. The proof takes a possibly complex eigenvalue $\lambda\in\mathbb{C}$ with eigenvector $v$ and shows $\lambda=\bar\lambda$. Since $A$ is real, conjugating $Av=\lambda v$ gives $A\bar v=\bar\lambda\bar v$, so $\bar v$ is an eigenvector for $\bar\lambda$. Writing $v=u+iw$ with $u,w\in\mathbb{R}^n$, note $\bar v\cdot v=\|u\|^2+\|w\|^2=\|v\|^2>0$. Now evaluate $\bar v\cdot(Av)$ two ways: directly $\bar v\cdot(Av)=\lambda\,(\bar v\cdot v)=\lambda\|v\|^2$, while using $A=A^T$ gives $\bar v\cdot(Av)=(A\bar v)\cdot v=\bar\lambda\,(\bar v\cdot v)=\bar\lambda\|v\|^2$. Because $\|v\|\neq0$, cancelling yields $\lambda=\bar\lambda$, i.e. $\lambda\in\mathbb{R}$. The degree-$n$ characteristic polynomial $c_A(\lambda)=\det(A-\lambda I)$ therefore has all real roots.

Here is the fact that makes symmetric matrices geometrically beautiful. Lemma 9.1: if $A$ is symmetric and $v_1,v_2$ are eigenvectors for distinct eigenvalues $\lambda_1\neq\lambda_2$, then $v_1\perp v_2$. The proof computes $v_1\cdot(Av_2)$ in two ways. On one hand $Av_2=\lambda_2 v_2$, so $v_1\cdot(Av_2)=\lambda_2\,(v_1\cdot v_2).$ On the other hand, using symmetry $A^T=A$, $v_1\cdot(Av_2)=(Av_2)^T v_1=v_2^T A^T v_1=v_2^T(Av_1)=\lambda_1\,(v_1\cdot v_2).$ Subtracting, $(\lambda_1-\lambda_2)(v_1\cdot v_2)=0$. Since $\lambda_1\neq\lambda_2$ the factor $\lambda_1-\lambda_2\neq0$, forcing $v_1\cdot v_2=0$. In words: different eigenspaces of a symmetric matrix are perpendicular. No Gram–Schmidt is needed between eigenspaces — their orthogonality comes for free, which is exactly why a symmetric matrix can be diagonalized by an orthogonal change of basis.

Lemma 9.1 handles distinct eigenvalues; the general argument needs one more geometric tool. Lemma 9.2: if $v$ is an eigenvector of a symmetric $A$ for a real eigenvalue $\lambda$, then the orthogonal complement $\operatorname{span}(v)^\perp=\{w\in\mathbb{R}^n\mid v\cdot w=0\}$ is $A$-invariant and has dimension $n-1$. Invariance is a one-line computation: for $w\perp v$, $v\cdot(Aw)=(Aw)^T v=w^T A^T v=w^T(Av)=\lambda\,(w^T v)=\lambda\,(v\cdot w)=0,$ so $Aw$ is again orthogonal to $v$. This lets the Spectral Theorem proceed by induction: peel off one unit eigenvector $u_1$, restrict $A$ to the $(n-1)$-dimensional invariant subspace $\operatorname{span}(u_1)^\perp$ (where it is again symmetric), and repeat. Remark 9.2 records the easy case: when the $n$ real eigenvalues are all distinct, Lemma 9.1 already makes the eigenbasis orthogonal, and normalizing produces an orthonormal eigenbasis — so $A$ is orthogonally diagonalizable. (Proposition 9.1 extends this to any diagonalizable symmetric matrix by running Gram–Schmidt inside each eigenspace.) The full theorem, and the explicit diagonalizing matrix, come next lesson.