# How to Find Moore Penrose Inverse

I have a matrix: $$A= \begin{bmatrix} -1 & 0 & 1 & 2 \\ -1 & 1 & 0 & -1 \\ 0 & -1 & 1 & 3 \\ 0 & 1 & -1 & -3 \\ 1 & -1 & 0 & 1 \\ 1 & 0 & -1 & -2 \\ \end{bmatrix}$$ I want to find a Moore Penrose generalized inverse for it, but I only know how to do this using computer software. My attempt: I think the inverse is supposed to be of the form $$C'(CC')^{-1}(B'B)^{-1}B'$$ where $A$ has dimensions $m\times n$, $B$ has dimensions $m\times k$, $C$ has dimensions $k\times n$, and all three have rank $k$. I just don't understand how to actually find this inverse matrix.

• I suggest you review the wikipedia article on Moore-Penrose. There are a few different formulae there. I would not recommend a hand calculation of the result if you do not need to perform one, en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse – JMJ Jan 28 '18 at 3:30

After doing Singular-value decomposition of the matrix, it would be very easy to find pseudo inverse of the matrix.

Steps:

Step 1: Find the eigenvalues of $A^TA$ or $AA^T,$ whichever is of less size.

In this case, $A^TA$ would be of less size ($4\times 4$, rather than $6\times 6$).

Step 2: Extract the singular values and frame the $\Sigma$ matrix which is formed by framing a diagonal matrix with diagonal elements equal to the squareroot of each of these eigenvalues and extending it to the shape of $A$ by introducing additional zeroes.

The eigenvalues of $A^TA$ are $\sigma_1^2=34$, $\sigma_2^2=6$, $\sigma_3^2=0$, $\sigma_4^2=0$. Shape of $\Sigma$=Shape of A=$6\times4$.

$\therefore \Sigma=\left[\begin{matrix}\sigma_1 & 0 & 0 & 0\\0 & \sigma_2 & 0 & 0\\0 & 0 & \sigma_3 & 0\\0 & 0 & 0 & \sigma_4\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{matrix}\right]=\left[\begin{matrix}5.8311 & 0 & 0 & 0\\0 & 2.4495 & 0 & 0\\0 & 0 & 0.0000 & 0\\0 & 0 & 0 & 0.0000\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{matrix}\right]$.

Step 3: Frame V as the matrix with columns as linearly independent orthonormal vectors spacing the domain of $A^TA$.

Since the orthonormal eigenvectors span the domain of $A^TA$, let's find out the eigenvectors of $A^TA$.

The eigenvector corresponding to $\lambda_1=34$ is $v_1=\left[\begin{matrix}0.0648\\0.2593\\-0.3241\\-0.9075\end{matrix}\right]$.

The eigenvector corresponding to $\lambda_2=6$ is $v_2=\left[\begin{matrix}-0.8018\\0.5345\\0.2673\\0.0\end{matrix}\right]$.

The eigenvector corresponding to $\lambda_3=0$ is $v_3=\left[\begin{matrix}0.5773\\0.5773\\0.5773\\0.0\end{matrix}\right]$.

It can be noticed that an independent eigenvector for $\lambda_4=0$ does not exist. Therefore, the other independent vector that together helps in spanning the entire domain can be found by using Gram–Schmidt process. (It is basically taking a random vector and removing the components of other orthonormal vectors from it.)

Therefore, by taking a random vector $r_4=\left[\begin{matrix}1\\1\\1\\1\end{matrix}\right]$,

$v'_4=r_4-(r_4.v_1)v_1-(r_4.v_2)v_2-(r_4.v_3)v_3$ and $v_4=\frac{v'_4}{\left|v'_4\right|}$.

$\Longrightarrow v_4=\left[\begin{matrix}0.14\\0.5602\\-0.7001\\0.42\end{matrix}\right]$

$\therefore V=\left[\begin{matrix}v_1&v_2&v_3&v_4\end{matrix}\right]=\left[\begin{matrix}0.0648 & -0.8018 & 0.5774 & 0.14\\0.2593 & 0.5345 & 0.5774 & 0.5602\\-0.3241 & 0.2673 & 0.5774 & -0.7001\\-0.9075 & 0.0 & 0.0 & 0.42\end{matrix}\right]$.

Step 4: Frame U as the matrix with columns equal to the orthonormal eigenvectors of $AA^T$, and satisfy the equation A=$U\Sigma V^T$.

Let $U=\left[\begin{matrix}u_1&u_2&u_3&u_4&u_5&u_6\end{matrix}\right]$, where $u_1,u_2,u_3,u_4,u_5,u_6$ are the columns of $U$.

Now, $A=U\Sigma V^T \Longrightarrow AV=U\Sigma$

$\Longrightarrow \left[\begin{matrix}-2.2039 & 1.0691 & 0.0 & 0.0\\1.102 & 1.3363 & 0.0 & 0.0\\-3.3059 & -0.2672 & 0.0 & 0.0\\3.3059 & 0.2672 & 0.0 & 0.0\\-1.102 & -1.3363 & 0.0 & 0.0\\2.2039 & -1.0691 & 0.0 & 0.0\end{matrix}\right]=\left[\begin{matrix}\sigma_1 u_1& \sigma_2 u_2&0&0\end{matrix}\right]$.

$\Longrightarrow u_1=\frac{1}{\sqrt{34}}\left[\begin{matrix}-2.2039\\1.102\\-3.3059\\3.3059\\-1.102\\2.2039\end{matrix}\right]=\left[\begin{matrix}-0.378\\0.189\\-0.567\\0.567\\-0.189\\0.378\end{matrix}\right]$ and $u_2=\frac{1}{\sqrt{6}}\left[\begin{matrix}1.0691\\1.3363\\-0.2672\\0.2672\\-1.3363\\-1.0691\end{matrix}\right]=\left[\begin{matrix}0.4365\\0.5455\\-0.1091\\0.1091\\-0.5455\\-0.4365\end{matrix}\right]$.

To find $u_3,u_4,u_5$ and $u_6$, we again use Gram–Schmidt process.

By taking random vector as $rv=\left[\begin{matrix}1\\2\\3\\4\\5\\6\end{matrix}\right]$,

$u'_3=rv-(rv.u_1)u_1-(rv.u_2)u_2$ and $u_3=\frac{u'_3}{\left|u'_3\right|} \Longrightarrow u_3=\left[\begin{matrix}0.3884\\0.4272\\0.4272\\0.3884\\0.3884\\0.4272\end{matrix}\right]$.

Similarly by taking random vectors as $rv_4=\left[\begin{matrix}1\\0\\0\\0\\0\\0\end{matrix}\right]$,$rv_5=\left[\begin{matrix}1\\1\\0\\0\\0\\0\end{matrix}\right]$ and $rv_6=\left[\begin{matrix}1\\1\\1\\0\\0\\0\end{matrix}\right]$

$u'_4=rv_4-(rv_4.u_1)u_1-(rv_4.u_2)u_2-(rv_4.u_3)u_3$ and $u_4=\frac{u'_4}{\left|u'_4\right|} \Longrightarrow u_4=\left[\begin{matrix}0.7182\\-0.4631\\-0.4631\\0.0221\\0.0221\\0.2331\end{matrix}\right]$,

$u'_5=rv_5-(rv_5.u_1)u_1-(rv_5.u_2)u_2-(rv_5.u_3)u_3-(rv_5.u_4)u_4$ and $u_5=\frac{u'_5}{\left|u'_5\right|} \Longrightarrow u_5=\left[\begin{matrix}0.0\\0.5193\\-0.4435\\-0.6207\\0.342\\0.1774\end{matrix}\right]$

$u'_6=rv_6-(rv_6.u_1)u_1-(rv_6.u_2)u_2-(rv_6.u_3)u_3-(rv_6.u_4)u_4-(rv_6.u_5)u_5$ and $u_6=\frac{u'_6}{\left|u'_6\right|} \Longrightarrow u_6=\left[\begin{matrix}0.0\\0.0001\\0.2698\\-0.361\\-0.6309\\0.6316\end{matrix}\right]$

$\therefore U=\left[\begin{matrix}u_1&u_2&u_3&u_4&u_5&u_6\end{matrix}\right]$

$\Longrightarrow U=\left[\begin{matrix}-0.378 & 0.4365 & 0.3884 & 0.7182 & 0.0 & 0.0\\0.189 & 0.5455 & 0.4272 & -0.4631 & 0.5193 & 0.0001\\-0.567 & -0.1091 & 0.4272 & -0.4631 & -0.4435 & 0.2698\\0.567 & 0.1091 & 0.3884 & 0.0221 & -0.6207 & -0.361\\-0.189 & -0.5455 & 0.3884 & 0.0221 & 0.342 & -0.6309\\0.378 & -0.4365 & 0.4272 & 0.2331 & 0.1774 & 0.6316\end{matrix}\right]$

(In this case, alternative matrices for $U$ are possible. And any one amongst those can be considered to be $U$.)

Step 5: Frame the matrix $\Sigma^+$ as the matrix $\Sigma$ with its non-zero singular elements replaced by their corresponding reciprocals and then transposed.

$\therefore \Sigma^+=\left[\begin{matrix}\frac{1}{5.8311} & 0 & 0 & 0\\0 & \frac{1}{2.4495} & 0 & 0\\0 & 0 & 0.00 & 0\\0 & 0 & 0 & 0.00\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{matrix}\right]^T=\left[\begin{matrix}0.1715 & 0 & 0 & 0 & 0 & 0\\0 & 0.4082 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0\end{matrix}\right]$

Step 6: Finally calculate the Moore-Penrose inverse by the relation $A^+=V \Sigma^+ U^T$.

$\Longrightarrow A^+=\left[\begin{matrix}-0.1471 & -0.1765 & 0.0294 & -0.0294 & 0.1765 & 0.1471\\0.0784 & 0.1274 & -0.049 & 0.049 & -0.1274 & -0.0784\\0.0686 & 0.049 & 0.0196 & -0.0196 & -0.049 & -0.0686\\0.0588 & -0.0294 & 0.0882 & -0.0882 & 0.0294 & -0.0588\end{matrix}\right]$

• Could you explain why that matrix is so different than the one generated when I used ginv in R? $$\begin{bmatrix} -.14705882& -.17647059& .02941176& -.02941176& .17647059& .14705882\\ .07843137& .12745098& -.04901961& .04901961& -.12745098& -.07843137\\ .06862745& .04901961& .01960784& -.01960784& -.04901961& -.06862745 \\ .05882353& -.02941176& .08823529& -.08823529& .02941176& -.05882353\\ \end{bmatrix}$$ – Johnny Jan 28 '18 at 14:56
• And how did you get those eigenvalues? I had: > eigen(t(A)%*%A)$values [1] 3.400000e+01 6.000000e+00 1.515128e-15 0.000000e+00 – Johnny Jan 28 '18 at 15:15 • Sorry, I mistakenly considered$A=\left[\begin{matrix}-1 & 0 & 1 & 2\\-1 & 1 & 0 & -1\\0 & -1 & 1 & 3\\0 & 1 & -1 & -3\\0 & -1 & 0 & 1\\1 & 0 & -1 & -2\end{matrix}\right]$, and the numerical values ended up to be different. I have updated the answer now. Thanks for pointing it out. – Suneesh Jacob Jan 28 '18 at 22:15 1. Compute SVD$A=U\Sigma V^T$2.$A^+=V\Sigma^{+}U^{T}$where$\Sigma^{+}$can be computed by taking the inverse of the singular values, the rest of$\Sigma\$ should be zero and then transpose.