A = B * C, how do I solve for B

Question

$$A = B C$$

where $A$'s dimension = $n$ x $1$, $B$ 's dimension = $n$ x $n$, C 's dimension = $A$ = $n$ x $1$

I know A and C.

How do I solve for B?

Attempt: I was thinking about multiply by the inverse of C. But C is non invertiable because its dimension nx1.

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Context: statweb.stanford.edu/~tibs/sta305files/Rudyregularization.pdf (page 26, I am looking to solve for the Smoother Matrix S).

Answer 1

You can't solve for a unique $B$. For example let $A=kC$, therefore$$BC=kC$$then the answer $B$ to the above equation includes all matrices that have an eigenvalue $k$ and a corresponding eigenvector $C$.

Note that if $B$ is not symmetric, then its trace is not unique. For example if$$A=C=\begin{bmatrix}1\\1\end{bmatrix}$$and$$B=\begin{bmatrix}a&1-a\\b&1-b\end{bmatrix}$$then the equation $A=BC$ is satisfied for any $a,b\in \Bbb R$, but if $B$ is symmetric, we from $A=BC$ we can write $$C^TA=C^TBC$$therefore$$tr(C^TA)=tr(C^TBC)=tr(BC^TC)=C^TC\cdot tr(B)$$hence$$tr(B)={A^TC\over C^TC}$$where I exploited the properties of trace.