Pseudoinverse of $$$\left[\begin{array}{cccc}0 & 0 & 0 & 0\\0 & 2 & 1 & 2\\2 & 1 & 0 & 1\\2 & 0 & 1 & 4\end{array}\right]$$$
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Find the Moore-Penrose pseudoinverse of $$$\left[\begin{array}{cccc}0 & 0 & 0 & 0\\0 & 2 & 1 & 2\\2 & 1 & 0 & 1\\2 & 0 & 1 & 4\end{array}\right]$$$.
Solution
Find the transpose of the matrix: $$$\left[\begin{array}{cccc}0 & 0 & 0 & 0\\0 & 2 & 1 & 2\\2 & 1 & 0 & 1\\2 & 0 & 1 & 4\end{array}\right]^{T} = \left[\begin{array}{cccc}0 & 0 & 2 & 2\\0 & 2 & 1 & 0\\0 & 1 & 0 & 1\\0 & 2 & 1 & 4\end{array}\right]$$$ (for steps, see matrix transpose calculator).
Multiply the original matrix by its transpose:
$$$\left[\begin{array}{cccc}0 & 0 & 0 & 0\\0 & 2 & 1 & 2\\2 & 1 & 0 & 1\\2 & 0 & 1 & 4\end{array}\right]\cdot \left[\begin{array}{cccc}0 & 0 & 2 & 2\\0 & 2 & 1 & 0\\0 & 1 & 0 & 1\\0 & 2 & 1 & 4\end{array}\right] = \left[\begin{array}{cccc}0 & 0 & 0 & 0\\0 & 9 & 4 & 9\\0 & 4 & 6 & 8\\0 & 9 & 8 & 21\end{array}\right]$$$ (for steps, see matrix multiplication calculator).
Find the row space of the resulting matrix: $$$\left\{\left[\begin{array}{c}0\\1\\0\\0\end{array}\right], \left[\begin{array}{c}0\\0\\1\\0\end{array}\right], \left[\begin{array}{c}0\\0\\0\\1\end{array}\right]\right\}$$$ (for steps, see row space calculator).
The columns of the matrix $$$P$$$ are the found vectors: $$$P = \left[\begin{array}{ccc}0 & 0 & 0\\1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{array}\right]$$$.
Find the transpose of the matrix: $$$P^{T} = \left[\begin{array}{ccc}0 & 0 & 0\\1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{array}\right]^{T} = \left[\begin{array}{cccc}0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{array}\right]$$$ (for steps, see matrix transpose calculator).
Multiply the transpose by the original matrix:
$$$\left[\begin{array}{cccc}0 & 0 & 2 & 2\\0 & 2 & 1 & 0\\0 & 1 & 0 & 1\\0 & 2 & 1 & 4\end{array}\right]\cdot \left[\begin{array}{cccc}0 & 0 & 0 & 0\\0 & 2 & 1 & 2\\2 & 1 & 0 & 1\\2 & 0 & 1 & 4\end{array}\right] = \left[\begin{array}{cccc}8 & 2 & 2 & 10\\2 & 5 & 2 & 5\\2 & 2 & 2 & 6\\10 & 5 & 6 & 21\end{array}\right]$$$ (for steps, see matrix multiplication calculator).
Find the row space of the resulting matrix: $$$\left\{\left[\begin{array}{c}1\\0\\0\\\frac{2}{3}\end{array}\right], \left[\begin{array}{c}0\\1\\0\\- \frac{1}{3}\end{array}\right], \left[\begin{array}{c}0\\0\\1\\\frac{8}{3}\end{array}\right]\right\}$$$ (for steps, see row space calculator).
The columns of the matrix $$$Q$$$ are the found vectors: $$$Q = \left[\begin{array}{ccc}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\\\frac{2}{3} & - \frac{1}{3} & \frac{8}{3}\end{array}\right]$$$.
The pseudoinverse of a matrix $$$A$$$ is $$$A^{+} = Q \left(P^{T} A Q\right)^{-1} P^{T}$$$.
Multiply the matrices:
$$$\left[\begin{array}{cccc}0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{array}\right]\cdot \left[\begin{array}{cccc}0 & 0 & 0 & 0\\0 & 2 & 1 & 2\\2 & 1 & 0 & 1\\2 & 0 & 1 & 4\end{array}\right] = \left[\begin{array}{cccc}0 & 2 & 1 & 2\\2 & 1 & 0 & 1\\2 & 0 & 1 & 4\end{array}\right]$$$ (for steps, see matrix multiplication calculator).
$$$\left[\begin{array}{cccc}0 & 2 & 1 & 2\\2 & 1 & 0 & 1\\2 & 0 & 1 & 4\end{array}\right]\cdot \left[\begin{array}{ccc}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\\\frac{2}{3} & - \frac{1}{3} & \frac{8}{3}\end{array}\right] = \left[\begin{array}{ccc}\frac{4}{3} & \frac{4}{3} & \frac{19}{3}\\\frac{8}{3} & \frac{2}{3} & \frac{8}{3}\\\frac{14}{3} & - \frac{4}{3} & \frac{35}{3}\end{array}\right]$$$ (for steps, see matrix multiplication calculator).
Find the inverse matrix: $$$\left[\begin{array}{ccc}\frac{4}{3} & \frac{4}{3} & \frac{19}{3}\\\frac{8}{3} & \frac{2}{3} & \frac{8}{3}\\\frac{14}{3} & - \frac{4}{3} & \frac{35}{3}\end{array}\right]^{-1} = \left[\begin{array}{ccc}- \frac{17}{78} & \frac{6}{13} & \frac{1}{78}\\\frac{14}{39} & \frac{7}{26} & - \frac{10}{39}\\\frac{5}{39} & - \frac{2}{13} & \frac{2}{39}\end{array}\right]$$$ (for steps, see matrix inverse calculator).
Finally, multiply the matrices:
$$$\left[\begin{array}{ccc}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\\\frac{2}{3} & - \frac{1}{3} & \frac{8}{3}\end{array}\right]\cdot \left[\begin{array}{ccc}- \frac{17}{78} & \frac{6}{13} & \frac{1}{78}\\\frac{14}{39} & \frac{7}{26} & - \frac{10}{39}\\\frac{5}{39} & - \frac{2}{13} & \frac{2}{39}\end{array}\right] = \left[\begin{array}{ccc}- \frac{17}{78} & \frac{6}{13} & \frac{1}{78}\\\frac{14}{39} & \frac{7}{26} & - \frac{10}{39}\\\frac{5}{39} & - \frac{2}{13} & \frac{2}{39}\\\frac{1}{13} & - \frac{5}{26} & \frac{3}{13}\end{array}\right]$$$ (for steps, see matrix multiplication calculator).
$$$\left[\begin{array}{ccc}- \frac{17}{78} & \frac{6}{13} & \frac{1}{78}\\\frac{14}{39} & \frac{7}{26} & - \frac{10}{39}\\\frac{5}{39} & - \frac{2}{13} & \frac{2}{39}\\\frac{1}{13} & - \frac{5}{26} & \frac{3}{13}\end{array}\right]\cdot \left[\begin{array}{cccc}0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{array}\right] = \left[\begin{array}{cccc}0 & - \frac{17}{78} & \frac{6}{13} & \frac{1}{78}\\0 & \frac{14}{39} & \frac{7}{26} & - \frac{10}{39}\\0 & \frac{5}{39} & - \frac{2}{13} & \frac{2}{39}\\0 & \frac{1}{13} & - \frac{5}{26} & \frac{3}{13}\end{array}\right]$$$ (for steps, see matrix multiplication calculator).
Answer
$$$\left[\begin{array}{cccc}0 & 0 & 0 & 0\\0 & 2 & 1 & 2\\2 & 1 & 0 & 1\\2 & 0 & 1 & 4\end{array}\right]^{+} = \left[\begin{array}{cccc}0 & - \frac{17}{78} & \frac{6}{13} & \frac{1}{78}\\0 & \frac{14}{39} & \frac{7}{26} & - \frac{10}{39}\\0 & \frac{5}{39} & - \frac{2}{13} & \frac{2}{39}\\0 & \frac{1}{13} & - \frac{5}{26} & \frac{3}{13}\end{array}\right]\approx \left[\begin{array}{cccc}0 & -0.217948717948718 & 0.461538461538462 & 0.012820512820513\\0 & 0.358974358974359 & 0.269230769230769 & -0.256410256410256\\0 & 0.128205128205128 & -0.153846153846154 & 0.051282051282051\\0 & 0.076923076923077 & -0.192307692307692 & 0.230769230769231\end{array}\right]$$$A