Divide by a vector? When doing matrix multiplication can I carry a vector to the other side?
For example if I have:
$Ab = c$
where A is m by m invertable matrix, and b is m by 1 col vector, c m by 1. Can I do something like this:
$A = c/b$
And what does that mean...
I just need to find matrix A, as I have b and c vectors.
P.S. Also, I know that inverse(A) is diagonal. If it helps.
 A: Yes, having the inverse of $A$ be diagonal helps immensely!
If the inverse of $A$ is diagonal,
$$A^{-1} = \left(\begin{array}{ccc}
\lambda_1 & \cdots & 0\\
\vdots & \ddots & \vdots\\
0 & \cdots & \lambda_n,\end{array}\right)$$
then $A$ is diagonal with
$$A = \left(\begin{array}{ccc}
\frac{1}{\lambda_1} & \cdots & 0\\
\vdots & \ddots & \vdots\\
0 & \cdots & \frac{1}{\lambda_n}
\end{array}\right).$$
That means that if
$$\mathbf{b}= \left(\begin{array}{c}b_1\\b_2\\\vdots \\b_m\end{array}\right),\qquad \mathbf{c} = \left(\begin{array}{c}c_1\\c_2\\\vdots \\b_m\end{array}\right),$$
then you need 
$$\frac{1}{\lambda_i}b_i = c_i.$$
For this to be possible, you need $b_i=0$ if and only if $c_i=0$; but if this is the case, then $A$ is determined uniquely in all rows corresponding to nonzero entries of $\mathbf{b}$ and $\mathbf{c}$, and can be any nonzero entry in the rows where $b_i=c_i=0$. 
A: Unfortunately, it is not easy to interpret $c/b$ where $c$ and $b$ are vectors.
Consider the following example: $b = c = \begin{pmatrix}3\\4 \end{pmatrix}$.
It is easy to see that $A = \begin{pmatrix}1 & 0\\0 & 1 \end{pmatrix}$ gives us $Ab = c$.
However note that $\tilde{A} = \begin{pmatrix}0 & \frac34\\\frac43 & 0 \end{pmatrix}$ also gives us $\tilde{A} b = c$.
Hence, $c/b$ is not well-defined if we want to interpret $c/b$ as $A$ such that $A b = c$.
A: [Edit: The question was later modified to say that $A$ is diagonal, which changes things quite a bit.  This answer is for the general case. See Arturo's answer for the diagonal case.]
Typically, no, and Sivaram has already given an illustrative example.  In your situation, the only exception is when $m=1$ and $b\neq 0$. The reason is that if $b$ is an $m$-by-$1$ vector and $m>1$, then $Ab$ never determines $A$ completely.  One way to see this is to note that there exists an $m$-by-$m$ matrix $B$ such that $B$ is not the zero matrix, but $Bb=0$.  Then $A+B\neq A$, but $(A+B)b=Ab$.  Thus, whatever "$c/b$" might mean, it would have to be equally valid that it is equal to $A$ and to $A+B$, which is impossible.
In order for the equation $Ab=c$ to uniquely determine $A$, $b$ must be right invertible.  If we generalized a bit to allow $b$ to have other sizes than $m$-by-$1$, say $m$-by-$k$, this will be possible when $b$ has at least as many columns as rows ($k\geq m$) and has maximal rank ($m$).  This would correspond to $b$ representing a surjective linear transformation, and then $Ab$ determines $A$ because if you know $Ab$, then you know what $A$ does to everything in the range of $b$, namely everything.  Algebraically speaking, if $d$ is a matrix such that $bd=I_m$, the $m$-by-$m$ identity matrix, then $A=AI_m=Abd=cd$.  Thus, $cd$ plays the role of "$c/b$".  Note however that unless $m=k$, $d$ is not uniquely determined by $b$.  If $m=k$, then $d=b^{-1}$ is the inverse matrix of $b$, and $A=cb^{-1}$ looks more like the division you'd hope for.  And again, if $k<m$ or for any other reason $b$ has rank less than $m$, $d$ does not exist.
A: It might help if you would think about a matrix M that you could right-multiply both sides of your equation:
$$
(Ab)M = cM
$$
$$
A(bM) = cM
$$
then, if bM is invertible,
$$
A=(cM)(bM)^{-1}
$$
