Findind the Pseudoinverse of a $2\times 2$-matrix (real-valued)

Question

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!

Answer 1

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 $$ A^+=\frac{1}{2}\begin{bmatrix} 1 \\ 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \end{bmatrix}=\begin{bmatrix} 1/2 & 0 \\ 1/2 & 0 \end{bmatrix} $$

Answer 2

It should be $U$, not $V$, that is equal to $I_2$ since the eigenvectors of $AA^T$ are the columns of $U$. Instead the eigenvectors of $A^TA$ are the columns of $V$: $$A^TA=\begin{bmatrix}1&1\\1&1\end{bmatrix}\qquad V=\frac1{\sqrt2}\begin{bmatrix}1&-1\\1&1\end{bmatrix}$$ Thus $$A^+=V\Sigma^+U^T=\begin{bmatrix}1/2&0\\1/2&0\end{bmatrix}$$

Answer 3

Given a rank-one matrix of any dimension $$\eqalign{ &A = xy^H \quad\;{\rm where}\; &x\in{\mathbb C}^{m\times 1},\;y\in{\mathbb C}^{n\times 1} }$$ there is a closed-form expression for its Moore-Penrose inverse $$A^+ = \frac{A^H}{(x^Hx)(y^Hy)}$$ In the current problem, $A$ is obviously a rank-one matrix with
$$\eqalign{ &x = \left(\begin{array}{rr}1\\0\end{array}\right), \quad &x^Hx = 1 \qquad &y= \left(\begin{array}{rr}1\\1\end{array}\right), \quad &y^Hy = 2 \\ }$$ Therefore $$\eqalign{ A^+ &= \frac{A^H}{(1)(2)} = \frac 12\pmatrix{1&0\\1&0} \\ }$$