MathAlgebraLinear algebraMatrix decomposition

Pseudoinverse

Getting the pseudoinverse

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The SVD of the matrix: $A= \begin{pmatrix} 0 & 2 \\ 1 & 1 \\ -2 & 0 \\ \end{pmatrix}$ is given by: $U = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & \sqrt{\frac{2}{3}} \\ -\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \end{pmatrix} \qquad \Sigma= \begin{pmatrix} \sqrt{6} & 0 \\ 0 & 2 \\ 0 & 0 \end{pmatrix} \qquad V=\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1 \\ 1 & 1 \\ \end{pmatrix}$ Calculate the pseudoinverse of $A$ .

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A^{\dag}= \frac{1}{6} \begin{pmatrix} -\frac{1}{2} & 1 & -\frac{5}{2} \\ \frac{5}{2} & 1 & \frac{1}{2} \\ \end{pmatrix}

A^{\dag}= \frac{1}{3} \begin{pmatrix} -2 & 1 & -\frac{5}{2} \\ 3& \frac{3}{2} & \frac{-5}{2} \\ \end{pmatrix}

A^{\dag}= \frac{1}{12} \begin{pmatrix} -\frac{1}{12} & -\frac{5}{12} & -\frac{5}{12} \\ \frac{5}{12} & 1 & \frac{1}{12} \\ \end{pmatrix}

A^{\dag}= \begin{pmatrix} -\frac{3}{2} & 1 \\ \frac{5}{2} & \frac{3}{2} \\ \end{pmatrix}

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