# Which pseudo-inverse to compute when Inverse is not possible? (No linear solve)

Let us assume that we have a function, $$f(A)=\text{vec}(A^{-1})^\intercal B$$, dependent on $$A^{-1}$$. However, due to some machine-precision limitations, the programming language I'm using cannot invert $$A$$ even if it has $$Det(A)>0$$. So, I'm thinking of using a pseudo-inverse, instead.

However, I don't know if there's a 'distance' relationship between the true inverse and the pseudo-inverse. I just know that when there's an inverse, both are equal. But it would be nice, that when the original matrix A and the 'perturbed/truncated' version of A are close, that the pseudo-inverse be close to the inverse of the original...

I'm also looking at pseudo-inverses for which there's a relatively easy and efficient way to compute.

Any suggestions?

Edit: $$\dim(A)=DT\times DT$$ and $$\dim(B)=D^2T^2\times D^2$$

Following Federico Poloni's comment, I think I could actually recompute the function above as $$\left[\cdots Tr\left(\text{LinearSolve}\left(A^\intercal, C_i \right) \right) \cdots \right]_{1\times i=1,...,D^2}$$ where $$C_i=\text{Reshape}_{DT\times DT}\left(\text{Column}(B,i)\right)$$

• what's the action of $\rm{vec}(\cdot)$? is it to reshape the inverse into a one-dimensional (column?) vector? – GoHokies Apr 27 at 8:14
• @GoHokies Yes. It's an example of a function, among others, which I use with reshapping operations. ;) – An old man in the sea. Apr 27 at 8:23
• ok, so if $A$ is $N \times N$, then $B$ is an $N^2 \times N^2$ matrix - or a $N^2 \times 1$ vector? – GoHokies Apr 27 at 8:27
• Is $\det(A)>0$ coming from an exact formula, or is it just a numerical observation? Determinants give no easily readable information on the distance of a matrix from singularity, so in numerical practice they are seldom used. – Federico Poloni Apr 27 at 8:37
• Also: you write "no linear solve", but what you have here can be rewritten as linear solves: vectorize (each column of) $B$ to obtain a $N\times N$ matrix $C$, then compute $A^{-1}C$, and the quantity you are looking for is its trace. – Federico Poloni Apr 27 at 18:37

Pseudoinverses typically will be computed via some truncation procedure to determine the rank, so they are not close to the original inverse. Example: $$A = Q \begin{bmatrix} 1\\ & 10^{-12} \\ & & 10^{-18} \end{bmatrix}Q^{*}$$ has exact (pseudo)inverse $$A^+ = A^{-1} = Q \begin{bmatrix} 1\\ & 10^{12} \\ & & 10^{18} \end{bmatrix}Q^{*},$$ but any numerical procedure will have to take a decision on the rank and will likely truncate one or two of those diagonal entries to zero, returning for instance $$B = Q \begin{bmatrix} 1\\ & 10^{12} \\ & & 0 \end{bmatrix}Q^{*},$$ or $$B = Q \begin{bmatrix} 1\\ & 0 \\ & & 0 \end{bmatrix}Q^{*},$$ (depending on truncation thresholds), which are not close at all to $$A^+$$. That is an inherent limitation and I don't think you can overcome it. It's by design: typically, when one writes pinv(A) one wants $$B$$, not $$A^{-1}$$. If you really want $$A^{-1}$$, then just use inv(A) and forget about the warnings.