简单计算可知
\begin{equation*}
A^HA = \begin{pmatrix}
80 & 100 & 40\\
100 & 170 & 140\\
40 & 140 & 200
\end{pmatrix}.
\end{equation*}
\(A^HA\)的三个特征值分别是360、90和0。取
\begin{equation*}
v_1 = \left(\frac{1}{3},\frac{2}{3},\frac{2}{3}\right)^T,
\end{equation*}
\begin{equation*}
v_2 = \left(\frac{2}{3},\frac{1}{3},-\frac{2}{3}\right)^T,
\end{equation*}
\begin{equation*}
v_3 =\left(\frac{2}{3},-\frac{2}{3},\frac{1}{3}\right)^T
\end{equation*}
分别是360、90和0的单位特征向量,于是
\begin{equation*}
Q = (v_1,v_2,v_3) = \begin{pmatrix}
\frac{1}{3} & \frac{2}{3} & \frac{2}{3}\\
\frac{2}{3} & \frac{1}{3} & -\frac{2}{3}\\
\frac{2}{3} & -\frac{2}{3} & \frac{1}{3}
\end{pmatrix}.
\end{equation*}
取
\begin{equation*}
u_1 = \frac{Av_1}{\sqrt{360}} = \left(\frac{3\sqrt{10}}{10},\frac{\sqrt{10}}{10} \right)^T,
\end{equation*}
\begin{equation*}
u_2 = \frac{Av_2}{\sqrt{90}} = \left(-\frac{\sqrt{10}}{10},\frac{3\sqrt{10}}{10} \right)^T,
\end{equation*}
于是取
\begin{equation*}
P = (u_1,u_2) = \begin{pmatrix}
\frac{3\sqrt{10}}{10} &
-\frac{\sqrt{10}}{10}\\
\frac{\sqrt{10}}{10} &
\frac{3\sqrt{10}}{10}
\end{pmatrix}.
\end{equation*}
矩阵\(A\)的SVD分解为
\begin{equation*}
A= \begin{pmatrix}
\frac{3\sqrt{10}}{10} &
-\frac{\sqrt{10}}{10}\\
\frac{\sqrt{10}}{10} &
\frac{3\sqrt{10}}{10}
\end{pmatrix}
\begin{pmatrix}
6\sqrt{10} & 0 & 0\\
0 & 3\sqrt{10} & 0
\end{pmatrix}
\begin{pmatrix}
\frac{1}{3} & \frac{2}{3} & \frac{2}{3}\\
\frac{2}{3} & \frac{1}{3} & -\frac{2}{3}\\
\frac{2}{3} & -\frac{2}{3} & \frac{1}{3}
\end{pmatrix}.
\end{equation*}