Theorem 7.3.1.
A symmetric matrix \(A\) is orthogonally diagonalizable.
Section 7.3 Orthogonally Diagonalization
A square matrix \(A\) is called symmetric if \(A^T=A\text{.}\) For example,
\begin{equation*}
\begin{pmatrix}
1 \amp 2 \amp 3 \\
2 \amp 4 \amp 5 \\
3 \amp 5 \amp 6
\end{pmatrix}
\end{equation*}
is symmetric.
A square matrix \(P\) is called orthogonal if \(P^TP=PP^T=I_n\text{.}\) That is, \(P^{-1}=P^T\text{.}\) For example,
\begin{equation*}
\begin{pmatrix}
\cos(\pi/4) \amp -\sin(\pi/4) \amp 0 \\
\sin(\pi/4) \amp \cos(\pi/4) \amp 0 \\
0 \amp 0 \amp 1
\end{pmatrix}
\end{equation*}
is orthogonal.
A square matrix \(A\) is called orthogonally diagonalizable if there exists an orthogonal matrix \(P\) such that
\begin{equation*}
P^{-1}AP=P^TA P=D,
\end{equation*}
where \(D\) is a diagonal matrix.
Note that if \(A\) is orthogonally diagonalizable, then \(A=PDP^T\text{.}\) Thus \(A\) is symmetric because \(A^T=(PDP^T)^T=PD^TP^T=PDP^T=A\text{.}\)
Objective.
Orthogonally diagonalize the symmetric matrix
\begin{equation*}
A = \begin{pmatrix}
5 \amp -8 \amp 4 \\
-8 \amp 5 \amp -4 \\
4 \amp -4 \amp -1
\end{pmatrix}
\end{equation*}
Exercise: Orthogonally diagonalize the symmetric matrix
\begin{equation*}
A=\left(\begin{array}{rrrr}
1 \amp -2 \amp 0 \amp 0 \\
-2 \amp 1 \amp 0 \amp 0 \\
0 \amp 0 \amp 1 \amp -2 \\
0 \amp 0 \amp -2 \amp 1
\end{array}\right)
\end{equation*}
