Why Every Real Symmetric Matrix Is Diagonalizable

Let $A$ be an $n\times n$ real matrix satisfying

$$A^T=A.$$

The claim is stronger than ordinary diagonalizability: $A$ has an orthonormal basis of real eigenvectors. Consequently, there is an orthogonal matrix $Q$ and a real diagonal matrix $D$ such that

$$Q^TAQ=D,$$

or equivalently,

$$A=QDQ^T.$$

This is the real spectral theorem.

The key idea: maximize a quadratic form

Consider the function

$$f(x)=x^TAx$$

on the unit sphere

$$S^{n-1}=\{x\in\mathbb R^n:x^Tx=1\}.$$

The sphere is compact and $f$ is continuous, so $f$ attains a maximum at some unit vector $v$. Write the maximum as

$$\lambda=f(v)=v^TAv.$$

We will show that $v$ is an eigenvector of $A$.

At the constrained maximum, the method of Lagrange multipliers gives

$$\nabla f(v)=\mu\nabla(x^Tx)\big|_{x=v}$$

for some scalar $\mu$. Since

$$\nabla(x^TAx)=(A+A^T)x=2Ax$$

when $A^T=A$, and $\nabla(x^Tx)=2x$, we obtain

$$2Av=2\mu v.$$

Thus

$$Av=\mu v.$$

Multiplying on the left by $v^T$ and using $v^Tv=1$ shows that

$$\mu=v^TAv=\lambda.$$

Therefore $v$ is a real unit eigenvector of $A$.

The orthogonal complement is invariant

Let

$$V=v^\perp=\{x\in\mathbb R^n:v^Tx=0\}.$$

We claim that $A$ maps $V$ into itself. Take any $x\in V$. Since $A$ is symmetric and $Av=\lambda v$,

$$\begin{aligned} v^T(Ax)&=(A^Tv)^Tx\\ &=(Av)^Tx\\ &=(\lambda v)^Tx\\ &=\lambda v^Tx\\ &=0. \end{aligned}$$

Hence $Ax\in V$. In other words, $V$ is invariant under $A$.

The restriction of $A$ to $V$ is still symmetric. Indeed, for $x,y\in V$,

$$\langle Ax,y\rangle=x^TA^Ty=x^TAy=\langle x,Ay\rangle.$$

The space $V$ has dimension $n-1$. Applying the same argument to the restriction of $A$ on $V$, we obtain a unit eigenvector in $V$. Repeating this process gives orthonormal eigenvectors

$$v_1,v_2,\ldots,v_n$$

that form a basis of $\mathbb R^n$.

Constructing the diagonalization

Let $\lambda_i$ be the eigenvalue corresponding to $v_i$, so that

$$Av_i=\lambda_i v_i.$$

Form the matrix whose columns are these orthonormal eigenvectors:

$$Q=\begin{bmatrix}v_1&v_2&\cdots&v_n\end{bmatrix}.$$

Because the columns are orthonormal,

$$Q^TQ=I,$$

so $Q$ is orthogonal. Also,

$$AQ=\begin{bmatrix}Av_1&Av_2&\cdots&Av_n\end{bmatrix} =\begin{bmatrix}\lambda_1v_1&\lambda_2v_2&\cdots&\lambda_nv_n\end{bmatrix}.$$

Therefore

$$AQ=QD,$$

where

$$D=\operatorname{diag}(\lambda_1,\lambda_2,\ldots,\lambda_n).$$

Multiplying by $Q^T$ on the left gives

$$Q^TAQ=D.$$

Thus $A$ is diagonalizable, and the diagonalizing matrix can be chosen orthogonal.

A concrete example

Consider

$$A=\begin{bmatrix}2&1\\1&2\end{bmatrix}.$$

This matrix is symmetric. Its characteristic polynomial is

$$\det(A-\lambda I) =\det\begin{bmatrix}2-\lambda&1\\1&2-\lambda\end{bmatrix} =(2-\lambda)^2-1 =(\lambda-1)(\lambda-3).$$

The eigenvalues are $1$ and $3$. For $\lambda=3$, an eigenvector is $(1,1)^T$, and for $\lambda=1$, an eigenvector is $(1,-1)^T$. Normalize them:

$$v_1=\frac{1}{\sqrt2}\begin{bmatrix}1\\1\end{bmatrix}, \qquad v_2=\frac{1}{\sqrt2}\begin{bmatrix}1\\-1\end{bmatrix}.$$

They are orthonormal, so

$$Q=\frac{1}{\sqrt2}\begin{bmatrix}1&1\\1&-1\end{bmatrix}$$

is orthogonal. With

$$D=\begin{bmatrix}3&0\\0&1\end{bmatrix},$$

we have

$$Q^TAQ=D.$$

Why symmetry matters

For arbitrary real matrices, eigenvectors need not exist in sufficient number, and even when all eigenvalues exist, the matrix need not be diagonalizable. Symmetry forces the orthogonal-complement argument above to work at every stage.

It also implies that eigenvectors belonging to distinct eigenvalues are automatically orthogonal. If $Au=\lambda u$ and $Av=\mu v$, then

$$\lambda u^Tv=(Au)^Tv=u^TA^Tv=u^TAv=\mu u^Tv.$$

Thus $(\lambda-\mu)u^Tv=0$. When $\lambda\ne\mu$, this gives $u^Tv=0$.

Common convention issue: real versus complex matrices

The statement is correct for a real symmetric matrix, meaning $A^T=A$ with real entries. For complex matrices, the corresponding condition is Hermitian:

$$A^*=A,$$

where $A^*$ is the conjugate transpose. Every complex Hermitian matrix is unitarily diagonalizable.

A complex matrix can satisfy $A^T=A$ without being diagonalizable. For example,

$$B=\begin{bmatrix}1&i\\i&-1\end{bmatrix}$$

is complex symmetric, but

$$B^2=0$$

while $B\ne0$. Hence its only eigenvalue is $0$, and it cannot be diagonalizable: a diagonalizable matrix with only eigenvalue $0$ would have to be the zero matrix.