To give some context, I'm trying to solve the following problem:

$y = BA^{-1}x$


$y$ = $n \times 1$ vector -- is known

$x$ = $3 \times 1$ vector -- is unknown

$B$ = $3 \times n$ matrix -- is known

$A$ = $n \times n$ singular matrix -- So $A$ is known and cannot be inverted to solve the problem.

If I could compute a generalized inverse for $B$ (let's call it $B^\dagger$) than I could avoid dealing with $A^{-1}$ altogether. The solution would simply be:

$x = AB^\dagger y$

However for this to work $B^\dagger$ would have to be a left inverse, i.e. $B^\dagger B = I$.

Any ideas on a generalized inverse method that would produce a left inverse for a short (less rows than columns) matrix? Is this even possible?

  • $\begingroup$ What do you mean by $A^{-1}$ is $A$ is not invertible? And the size of your vectors and matrices are inconsistent with the product you write (maybe the sizes of $x$ and $y$ are inverted?), unless $n=3$.... $\endgroup$ – Arnaud D. Oct 22 '16 at 10:17

Forgive my confusion. The text states $\mathbf{A}$ is singular, the formula uses $\mathbf{A}^{-1}$. The answers are for different assumptions.

If the matrix product $\mathbf{C}=\mathbf{B}\,\mathbf{A}$, is know, build the psuedoinverse from the singular value decomposition: $$ \mathbf{C} = \mathbf{U}\, \Sigma\, \mathbf{V}^{*} \qquad \Rightarrow \qquad \mathbf{C}^{+} = \mathbf{V}\, \Sigma^{+} \,\mathbf{U}^{*} $$

Both matrices $\mathbf{A}$ and $\mathbf{B}$ will also have a pseudoinverse.

This post shows when the pseudoinverse is a right inverse, a left inverse, and a classic inverse: generalized inverse of a matrix and convergence for singular matrix

This post connects classic inverse and pseudoinverse: When pseudo inverse and general inverse of a invertible square matrix will be equal or not equal?

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