
# 15: Diagonalizing Symmetric Matrices

Symmetric matrices have many applications. For example, if we consider the shortest distance between pairs of important cities, we might get a table like this:

$\begin{array}{c|ccc}& Davis & Seattle & San\; Francisco \\ \hline Davis & 0 & 2000 & 80 \\Seattle & 2000 & 0 & 2010 \\San \;Francisco & 80 & 2010 & 0\end{array}$

Encoded as a matrix, we obtain:

$M=\begin{pmatrix}0 & 2000 & 80 \\2000 & 0 & 2010 \\80 & 2010 & 0\end{pmatrix}=M^{T}.$

Definition

A matrix is $$\textit{symmetric}$$ if it obeys

$M=M^{T}.$

One nice property of symmetric matrices is that they always have real eigenvalues. Review exercise 1 guides you through the general proof, but here's an example for $$2\times 2$$ matrices:

Example 128

For a general symmetric $$2\times 2$$ matrix, we have:

\begin{eqnarray*}P_{\lambda} \begin{pmatrix} a & b \\ b& d \end{pmatrix}&=&\det\begin{pmatrix}\lambda-a&-b\\-b&\lambda-d \end{pmatrix}\\&=& (\lambda-a)(\lambda-d)-b^{2} \\&=& \lambda^{2}-(a+d)\lambda-b^{2}+ad\\\Rightarrow \lambda &=& \frac{a+d}{2}\pm \sqrt{b^{2}+\left(\frac{a-d}{2}\right)^{2}}.\end{eqnarray*}

Notice that the discriminant $$4b^{2}+(a-d)^{2}$$ is always positive, so that the eigenvalues must be real.

Now, suppose a symmetric matrix $$M$$ has two distinct eigenvalues $$\lambda \neq \mu$$ and eigenvectors $$x$$ and $$y$$:

$Mx=\lambda x, \qquad My=\mu y.$

Consider the dot product $$x\cdot y = x^{T}y = y^{T}x$$ and calculate:

\begin{eqnarray*}x^{T}M y &=& x^{T}\mu y = \mu x\cdot y, \textit{ and }\\x^{T}M y &=& (y^{T}Mx)^{T} \textit{ (by transposing a $$1\times 1$$ matrix)}\\&=& x^{T}M^{T}y \\&=& x^{T}My \\&=& x^{T}\lambda y \\&=& \lambda x\cdot y.\end{eqnarray*}

Subtracting these two results tells us that:

\begin{eqnarray*}0 &=& x^{T}My-x^{T}My=(\mu-\lambda)\,x\cdot y.\end{eqnarray*}

Since $$\mu$$ and $$\lambda$$ were assumed to be distinct eigenvalues, $$\lambda-\mu$$ is non-zero, and so $$x\cdot y=0$$. We have proved the following theorem.

Theorem

Eigenvectors of a symmetric matrix with distinct eigenvalues are orthogonal.

Example 129

The matrix $$M=\begin{pmatrix}2&1\\1&2\end{pmatrix}$$ has eigenvalues determined by

$\det(M-\lambda I)=(2-\lambda)^{2}-1=0.$

So the eigenvalues of $$M$$ are $$3$$ and $$1$$, and the associated eigenvectors turn out to be $$\begin{pmatrix}1\\1\end{pmatrix}$$ and $$\begin{pmatrix}1\\-1\end{pmatrix}$$. It is easily seen that these eigenvectors are orthogonal:

$\begin{pmatrix}1\\1\end{pmatrix} \cdot \begin{pmatrix}1\\-1\end{pmatrix}=0$

In chapter 14 we saw that the matrix $$P$$ built from any orthonormal basis $$(v_{1},\ldots, v_{n} )$$ for $$\mathbb{R}^{n}$$ as its columns,

$P=\begin{pmatrix}v_{1} & \cdots & v_{n}\end{pmatrix}\, ,$

was an orthogonal matrix:

$P^{-1}=P^{T}, \textit{ or } PP^{T}=I=P^{T}P.$

Moreover, given any (unit) vector $$x_{1}$$, one can always find vectors $$x_{2}, \ldots, x_{n}$$ such that $$(x_{1},\ldots, x_{n})$$ is an orthonormal basis. (Such a basis can be obtained using the Gram-Schmidt procedure.)

Now suppose $$M$$ is a symmetric $$n\times n$$ matrix and $$\lambda_{1}$$ is an eigenvalue with eigenvector $$x_{1}$$ (this is always the case because every matrix has at least one eigenvalue--see review problem 3). Let the square matrix of column vectors $$P$$ be the following:

$P=\begin{pmatrix}x_{1} & x_{2} & \cdots & x_{n}\end{pmatrix},$

where $$x_{1}$$ through $$x_{n}$$ are orthonormal, and $$x_{1}$$ is an eigenvector for $$M$$, but the others are not necessarily eigenvectors for $$M$$. Then

$MP=\begin{pmatrix}\lambda_{1} x_{1} & Mx_{2} & \cdots & Mx_{n}\end{pmatrix}.$

But $$P$$ is an orthogonal matrix, so $$P^{-1}=P^{T}$$. Then:

\begin{eqnarray*}P^{-1}=P^{T} &=& \begin{pmatrix}x_{1}^{T}\\ \vdots \\ x_{n}^{T}\end{pmatrix} \\\Rightarrow P^{T}MP &=& \begin{pmatrix}x_{1}^{T}\lambda_{1}x_{1} & * & \cdots & *\\x_{2}^{T}\lambda_{1}x_{1} & * & \cdots & *\\\vdots & & & \vdots\\x_{n}^{T}\lambda_{1}x_{1} & * & \cdots & *\\\end{pmatrix}\\&=& \begin{pmatrix}\lambda_{1} & * & \cdots & *\\0 & * & \cdots & *\\\vdots & * & & \vdots\\0 & * & \cdots & *\\\end{pmatrix}\\&=& \begin{pmatrix}\lambda_{1} & 0 & \cdots & 0\\0 & & & \\\vdots & & \hat{M} & \\0 & & & \\\end{pmatrix}\, .\\\end{eqnarray*}

The last equality follows since $$P^{T}MP$$ is symmetric. The asterisks in the matrix are where “stuff'' happens; this extra information is denoted by $$\hat{M}$$ in the final expression. We know nothing about $$\hat{M}$$ except that it is an $$(n-1)\times (n-1)$$ matrix and that it is symmetric. But then, by finding an (unit) eigenvector for $$\hat{M}$$, we could repeat this procedure successively. The end result would be a diagonal matrix with eigenvalues of $$M$$ on the diagonal. Again, we have proved a theorem:

Theorem

Every symmetric matrix is similar to a diagonal matrix of its eigenvalues. In other words,

$M=M^{T} \Leftrightarrow M=PDP^{T}$

where $$P$$ is an orthogonal matrix and $$D$$ is a diagonal matrix whose entries are the eigenvalues of $$M$$.

To diagonalize a real symmetric matrix, begin by building an orthogonal matrix from an orthonormal basis of eigenvectors:

Example 130

The symmetric matrix

$$M=\begin{pmatrix}2&1\\1&2\end{pmatrix}\, ,$$ has eigenvalues $$3$$ and $$1$$ with eigenvectors $$\begin{pmatrix}1\\1\end{pmatrix}$$ and $$\begin{pmatrix}1\\-1\end{pmatrix} respectively. After normalizing these eigenvectors, we build the orthogonal matrix: $P = \begin{pmatrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\\frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}}\end{pmatrix}\, .$ Notice that \(P^{T}P=I$$. Then:

$MP = \begin{pmatrix}\frac{3}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\\frac{3}{\sqrt{2}} & \frac{-1}{\sqrt{2}}\end{pmatrix} = \begin{pmatrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\\frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}}\end{pmatrix} \begin{pmatrix}3 & 0 \\0 & 1\end{pmatrix}.$

In short, $$MP=DP$$, so $$D=P^{T}MP$$. Then $$D$$ is the diagonalized form of $$M$$ and $$P$$ the associated change-of-basis matrix from the standard basis to the basis of eigenvectors.