# Calculate the pseudo inverse of the matrix

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.

• BTW since $A$ has full column rank, you can compute it's pseudoinverse as $(A^TA)^{-1}A^T$. Commented Jan 18, 2014 at 18:36
• @Algebraic Pavel : yes you are right, that is a better way to calculate the pseudo inverse. thank you very much! Commented Jan 19, 2014 at 1:58

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, $\dfrac{2}{\sqrt{6}} = \sqrt{\dfrac{2}{3}}$.