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How can I find the Moore-Penrose pseudoinverse of the $2 \times 2$ complex matrix

$$A=\begin{pmatrix}0&a\\0&b\end{pmatrix}$$

for $a \neq 0$ and $b \neq 0$?

Here I want to use the limit formula

$$A^+=\lim_{\epsilon \to 0} (\epsilon I+A^*A)^{-1}A^*$$

since $\mbox{rank}(A)=1$, which is not full rank. Any help, please?

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  • $\begingroup$ Can you define 'pseudoinverse'? I'm aware of a few different definitions. $\endgroup$ – The Count Feb 12 '17 at 19:22
  • $\begingroup$ @The Count I edited it. $\endgroup$ – Sam Feb 12 '17 at 19:24
  • $\begingroup$ Which part are you having trouble with? The adjoint? The inverse? The limit? $\endgroup$ – Roland Feb 12 '17 at 19:42
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Using SymPy:

>>> from sympy import *
>>> a, b = symbols('a b')
>>> M = Matrix([[0,a],[0,b]])
>>> t = Symbol('t')
>>> (t * eye(2) + (M.T * M))**-1 * M.T
Matrix([
[                  0,                   0],
[a/(a**2 + b**2 + t), b/(a**2 + b**2 + t)]])

Hence,

$$\begin{bmatrix} 0 & a\\ 0 & b\end{bmatrix}^+ = \lim_{t \to 0} \begin{bmatrix} 0 & 0\\ \frac{a}{a^{2} + b^{2} + t} & \frac{b}{a^{2} + b^{2} + t}\end{bmatrix} = \begin{bmatrix} 0 & 0\\ \frac{a}{a^{2} + b^{2}} & \frac{b}{a^{2} + b^{2}}\end{bmatrix}$$

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Computing eigendecompositions using SymPy:

>>> from sympy import *
>>> a, b = symbols('a b')
>>> M = Matrix([[0,a],[0,b]])
>>> (M.T * M).eigenvects()
[(0, 1, [Matrix([
[1],
[0]])]), (a**2 + b**2, 1, [Matrix([
[0],
[1]])])]
>>> (M * M.T).eigenvects()
[(0, 1, [Matrix([
[-b/a],
[   1]])]), (a**2 + b**2, 1, [Matrix([
[a/b],
[  1]])])]

We now build the matrices in the SVD:

>>> U = (1/sqrt(a**2 + b**2)) * Matrix([[a,-b],[b,a]])
>>> S = diag(sqrt(a**2 + b**2),0)
>>> V = Matrix([[0,1],[1,0]])
>>> U * S * V.T
Matrix([
[0, a],
[0, b]])

The SVD is

$$\begin{bmatrix} 0 & a\\ 0 & b\end{bmatrix} = \begin{bmatrix} \frac{a}{\sqrt{a^{2} + b^{2}}} & - \frac{b}{\sqrt{a^{2} + b^{2}}}\\ \frac{b}{\sqrt{a^{2} + b^{2}}} & \frac{a}{\sqrt{a^{2} + b^{2}}} \end{bmatrix} \begin{bmatrix} \sqrt{a^{2} + b^{2}} & 0\\ 0 & 0\end{bmatrix} \begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix}^{\top}$$

Hence, the pseudoinverse is

$$\begin{bmatrix} 0 & a\\ 0 & b\end{bmatrix}^{+} = \begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{a^{2} + b^{2}}} & 0\\ 0 & 0\end{bmatrix} \begin{bmatrix} \frac{a}{\sqrt{a^{2} + b^{2}}} & - \frac{b}{\sqrt{a^{2} + b^{2}}}\\ \frac{b}{\sqrt{a^{2} + b^{2}}} & \frac{a}{\sqrt{a^{2} + b^{2}}} \end{bmatrix}^{\top} = \color{blue}{\begin{bmatrix} 0 & 0\\ \frac{a}{a^{2} + b^{2}} & \frac{b}{a^{2} + b^{2}}\end{bmatrix}}$$

Verifying,

$$\begin{bmatrix} 0 & a\\ 0 & b\end{bmatrix} \begin{bmatrix} 0 & 0\\ \frac{a}{a^{2} + b^{2}} & \frac{b}{a^{2} + b^{2}}\end{bmatrix} \begin{bmatrix} 0 & a\\ 0 & b\end{bmatrix} = \begin{bmatrix} 0 & a\\ 0 & b\end{bmatrix}$$

$$\begin{bmatrix} 0 & 0\\ \frac{a}{a^{2} + b^{2}} & \frac{b}{a^{2} + b^{2}}\end{bmatrix} \begin{bmatrix} 0 & a\\ 0 & b\end{bmatrix} \begin{bmatrix} 0 & 0\\ \frac{a}{a^{2} + b^{2}} & \frac{b}{a^{2} + b^{2}}\end{bmatrix} = \begin{bmatrix} 0 & 0\\ \frac{a}{a^{2} + b^{2}} & \frac{b}{a^{2} + b^{2}}\end{bmatrix}$$

Also, both products of the given matrix and its pseudoinverse are symmetric, as required.

This is the real case. The complex case should be easy to tackle.

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  • 1
    $\begingroup$ Ohh I got my mistake. I missed some calculation. It should be $A^+=\lim_{\varepsilon \to 0} \begin{pmatrix}0&0 \\a/(\epsilon +a^2+b^2)&b/(\epsilon +a^2+b^2)\end{pmatrix}= \begin{pmatrix}0&0 \\a/(a^2+b^2)&b/(a^2+b^2)\end{pmatrix}$. Thanks! $\endgroup$ – Sam Feb 12 '17 at 20:29
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A neat fact about rank-one matrices, such as this one $$\eqalign{ &a,b\in{\mathbb C} \\ &x=\pmatrix{a\\b},\quad y=\pmatrix{0\\1} \\ &A=xy^H,\quad{\rm rank}(A)={\tt1} }$$ is that there's a closed-form expression for its Moore-Penrose inverse $$A^+ = \frac{yx^H}{(x^Hx)\,(y^Hy)} = \frac{A^H}{|a|^2+|b|^2}$$ This result is not limited to $2\times 2$ matrices.

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