The subject is to calculate the pseudo inverse if matrix $\begin{equation*}
\mathbf{A} = \left(
\begin{array}{ccc}
1 & 0 \\
2 & 1 \\
0 & 1 \\
\end{array}
\right)
\end{equation*}$
My answer is as follows: (SVD decomposition)
First, $\begin{equation*}
\mathbf{A^TA} = \left(
\begin{array}{ccc}
5 & 2 \\
2 & 2 \\
\end{array}
\right)
\end{equation*}$, with eigenvalues $\lambda_1 = 6, \lambda_2 = 1$, and eigenvectors $\begin{equation*}
\mathbf{x_1} = \frac{1}{\sqrt{5}}\left(
\begin{array}{ccc}
2 \\
1 \\
\end{array}
\right)
\end{equation*}$,
$\begin{equation*}
\mathbf{x_2} = \frac{1}{\sqrt{5}}\left(
\begin{array}{ccc}
-1 \\
2 \\
\end{array}
\right)
\end{equation*}$, so the matrix $\begin{equation*}
\mathbf{V} = \frac{1}{\sqrt{5}}\left(
\begin{array}{ccc}
2 & -1 \\
1 & 2 \\
\end{array}
\right)
\end{equation*}$.
Second, $\begin{equation*}
\mathbf{AA^T} = \left(
\begin{array}{ccc}
1 & 2 & 0 \\
2 & 5 & 1 \\
0 & 1 & 1 \\
\end{array}
\right)
\end{equation*}$, with eigenvalues $\lambda_1 = 6, \lambda_2 = 1,\lambda_3 = 0$, and eigenvectors $\begin{equation*}
\mathbf{x_1} = \frac{1}{\sqrt{30}}\left(
\begin{array}{ccc}
2 \\
5 \\
1 \\
\end{array}
\right)
\end{equation*}$,
$\begin{equation*}
\mathbf{x_2} = \frac{1}{\sqrt{5}}\left(
\begin{array}{ccc}
-1 \\
0 \\
2 \\
\end{array}
\right)
\end{equation*}$,
$\begin{equation*}
\mathbf{x_3} = \frac{1}{\sqrt{6}}\left(
\begin{array}{ccc}
2 \\
-1 \\
1 \\
\end{array}
\right)
\end{equation*}$
, so the matrix $\begin{equation*}
\mathbf{U} = \left(
\begin{array}{ccc}
\frac{2}{\sqrt{30}} & – \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{6}} \\
\frac{5}{\sqrt{30}} & 0 & – \frac{1}{\sqrt{6}} \\
\frac{1}{\sqrt{30}} & \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{6}} \\
\end{array}
\right)
\end{equation*}$, and
$\begin{equation*}
\mathbf{\Sigma} = \left(
\begin{array}{ccc}
6 & 0 \\
0 & 1 \\
0 & 0 \\
\end{array}
\right)
\end{equation*}$.
Then, the pseudo inverse becomes: $A^+ = V \Sigma^+U^T$.
The problem comes to: when I was checking the SVD decomposition, I found $A\ne U\Sigma V^T$. However, I find nothing odds in the calculation. Please help me to point out the error.
Best Answer
Recall, for $\mathbf{\Sigma}$ we take the square roots of the non-zero eigenvalues and populate the diagonal with them, putting the largest in $\mathbf{\Sigma}_{11}$, the next largest in $\mathbf{\Sigma}_{22}$ and so on until the smallest value ends up in $\mathbf{\Sigma}_{mm}$.
$$\begin{equation*} \mathbf{\Sigma} = \left( \begin{array}{ccc} \sqrt{6} & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{array} \right) \end{equation*}$$
Everything else is correct (great job), although you can simplify some of the items in $\mathbf{U}$. For example, $\dfrac{2}{\sqrt{6}} = \sqrt{\dfrac{2}{3}}$.