Pseudoinverse of $2\times 2$ matrix 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?
 A: 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.
A: 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,
$$\begin{bmatrix} 0 & a\\ 0 & b\end{bmatrix}^+ = \lim_{t \to 0} \begin{bmatrix} 0 & 0\\ \frac{a}{a^{2} + b^{2} + t} & \frac{b}{a^{2} + b^{2} + t}\end{bmatrix} = \begin{bmatrix} 0 & 0\\ \frac{a}{a^{2} + b^{2}} & \frac{b}{a^{2} + b^{2}}\end{bmatrix}$$
A: 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.
