Spectral Theorem (real version)

If \(A\) is a real symmetric matrix, then it is orthogonally diagonalizable. In particular, there exists an orthogonal matrix \(Q\) and a diagonal matrix \(D\) such that:

\[ A=QDQ^{T} \]

If the order of \(A\) is \(n\), we can express \(Q\) and \(D\) as:

\[ Q=\begin{bmatrix}\vert & & \vert\\ q_{1} & \cdots & q_{n}\\ \vert & & \vert \end{bmatrix},\,\,\,\,D=\begin{bmatrix}\lambda_{1}\\ & \ddots\\ & & \lambda_{n} \end{bmatrix} \]

The columns of \(Q\) are the eigenvectors of \(A\). The corresponding eigenvalues are to be found on the diagonals of \(D\). Therefore, \((\lambda_{i},q_{i})\) is an eigenpair of \(A\), that is, \(Aq_{i}=\lambda_{i}q_{i}\). Also, \(q_{1},\cdots,q_{n}\) form an orthonormal basis for \(\mathbb{R}^{n}\). Recall that:

\[ Q^{T}Q=QQ^{T}=I \]

We can also express \(A\) as the sum of \(n\) outer products:

\[ A=\sum\limits_{i=1}^{n}\lambda_{i}q_{i}q_{i}^{T} \]

One way of seeing this is to treat the product \(QDQ^{T}\) as \((QD)Q^{T}\). Since \(D\) is diagonal, \(QD\) would just result in scaling the columns of \(Q\) by the corresponding diagonal entries in \(D\):

\[ QD=\begin{bmatrix}\vert & & \vert\\ \lambda_{1}q_{1} & \cdots & \lambda_{n}q_{n}\\ \vert & & \vert \end{bmatrix} \]

The outer product now follows.

Here is an example of a symmetric matrix and its spectral decomposition:

\[ A=\begin{bmatrix}2 & 0 & -1\\ 0 & 2 & 0\\ -1 & 0 & 2 \end{bmatrix},\,\,\,\,Q=\frac{1}{\sqrt{2}}\begin{bmatrix}1 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & -1 \end{bmatrix},\,\,\,\,D=\begin{bmatrix}1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 3 \end{bmatrix} \]