Findind the Pseudoinverse of a $2\times 2$-matrix (real-valued) As is said in the title I would like to find the pseudoinverse of the real-valued matrix $$A = \left(\begin{array}{cc}1 & 1\\0 & 0\end{array}\right)$$
I've never done this before but I looked up some different methods but none worked..

*

*First I tried the SVD but ended up stuck with an impossible system to find $U$. I used:

$$AA^{\bot} = \left(\begin{array}{cc}2 & 0\\0 & 0\end{array}\right)$$
Which gave me $\Sigma = \left(\begin{array}{cc}\sqrt{2} & 0\\0 & 0\end{array}\right)$ and $V = \left(\begin{array}{cc}1 & 0\\0 & 1\end{array}\right) = I_2$
By construction we should have
$$A = U\Sigma V^* \implies A = U\Sigma$$ (since $V = I_2$), but such $U$ can't exist, and so I'm stuck..
But I would really like to understand why, is it because $A, A^{\bot}$ or their product didn't fulfill some initial conditions I missed?

*

*Anyway since it didn't get me where I wanted I tried using the formula: $A^+ = (A^{\bot}A)^{-1}A^{\bot}$, which I believe only requires $A^{\bot}A$ to be inversible. But once more I ended up with a contradiction, since $(A^{\bot}A)^{-1} = \left(\begin{array}{cc}0 & 1\\1 & - 1\end{array}\right)$ and $A^{\bot} = \left(\begin{array}{cc}1 & 1\\0 & 0\end{array}\right)$, giving $A^+$ should be $\left(\begin{array}{cc}1 & 0\\0 & 0\end{array}\right)$, which doesn't satisfy the property (for example) $A = A^+AA^+$.

So if anyone has some time to look it up and just tell me if I missed an important condition to apply those methods, or could give me another one to finish this question myself, that would be great!

You don't need the SVD if you are able to find a full-rank decomposition, in this case
$$
A=\begin{bmatrix} 1 \\ 0 \end{bmatrix}\begin{bmatrix} 1 & 1 \end{bmatrix}
$$
When you have such a full-rank decomposition $A=BC$, where $B$ is right-invertible and $C$ is left-invertible, then $A^+=C^+B^+$ and
$$
B^+=(B^TB)^{-1}B^T\qquad C^+=C^T(CC^T)^{-1}
$$
For your case $B^TB=[1]$ and $CC^T=[2]$, so
$$
A^+=\frac{1}{2}\begin{bmatrix} 1 \\ 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \end{bmatrix}=\begin{bmatrix} 1/2 & 0 \\ 1/2 & 0 \end{bmatrix}
$$
