1
$\begingroup$

$A=BC$, where $B$ is a singular and every matrix is a $n\times n$ matrix. Can we find an explicit form of $C$?

If $B$ is non-singular, $C=B^{-1}A$ is obvious. But I am not sure how to find out the form of $C$ when $B$ is singular. It can be $B^-A=B^-BC$, where $B^-$ is a generalized inverse of $B$ However, it doesn't result in an explicit form of $C$.

Thank you for all of your comments in advance.

  • 2
    $0=0C$ for all $C$, so you can't hope for something like this – 2017-02-14
  • 0
    If $B$ is singular, this equation doesn't have an unique solution for $C$. It may have no solution (example: $B$ is the zero matrix, $A$ is nonzero), or it may have infinitely many solutions (example: both $A$ and $B$ are the zero matrix). – 2017-02-14
  • 0
    There are many ways that the image of $C$ can be contained in the kernel (nullspace) of $B$, since $B$ is singular. This gives lots of solutions to the equation $0 = BC$. – 2017-02-14
  • 0
    I think this question is salvageable, despite the examples in comments above. Singular is equivalent to having a characteristic polynomial with a root at zero; perhaps we can determine some information about $C$ if we have a bound on the multiplicity of this root? – 2017-02-14
  • 0
    Thank you for all of quick and helpful answers! – 2017-02-14

1 Answers 1

0

There does not always (for an arbitrary $A$) exist such a matrix $C$ and when it does, it's never unique. To see this simply, you have to realize that $$BC=A$$ is $n$ separate equations: one for each column of $A$. If we denote the columns of $A$ by $a_i$ and the unknown columns of $C$ by $c_i$, then the $n$ equations are $$Bc_1=a_1$$ $$Bc_2=a_2$$ $$\cdots$$ $$Bc_n=a_n.$$ Each one of these equations only has a solution if $a_n$ is in the column space of $B$, which presumably is the case if $A$ was obtained as the product $BC$ However, then each equation has a solution of the form $$c_i=p_i+N_B,$$ where $p_i$ is a particular solution and $N_B$ is the null space of the matrix $B$.