1
$\begingroup$

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

$y = BA^{-1}x$

where:

$y$ = $n \times 1$ 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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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?

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.