Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
Generally, there are lots of matrices that will map a given vector $b$ to a given vector $c$ (the sole exception is if $b=\mathbf{0}$ and $c\neq\mathbf{0})$. –  Arturo Magidin Nov 28 '11 at 3:36
    
Added that A is m by m, and is invertable... –  drozzy Nov 28 '11 at 3:37
1  
Even if $A$ is invertible, so long as $b\neq \mathbf{0}$ and $c\neq \mathbf{0}$, there are infinitely many matrices that will satisfy $A\mathbf{b}=\mathbf{c}$. –  Arturo Magidin Nov 28 '11 at 3:40
2  
It's an understatement to say that knowing that $A$ is diagonal helps a lot. –  Arturo Magidin Nov 28 '11 at 4:19

4 Answers 4

up vote 5 down vote accepted

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 $b_i=c_i=0$.

share|improve this answer
    
Thanks, turns out I didn't actually need this... I think. But it's good to know. –  drozzy Dec 5 '11 at 4:46

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 $A b = c$.

share|improve this answer
    
would the requirement that inverse(A) has to diagonal help? –  drozzy Nov 28 '11 at 4:12
1  
@drozzy: If $A^{-1}$ is diagonal, then so is $A$. If $A$ is diagonal with diagonal entries $(a_1,a_2,\ldots,a_m)$, then $Ab=[a_1b_1,a_2b_2,\ldots,a_mb_m]^\textrm{T}$. Then if all of the entries of $b$ are nonzero, you can determine each $a_k$ with the equation $a_k=c_k/b_k$. So, yes, if $A$ is diagonal and $b$ has all nonzero entries, then $A$ is determined by setting the diagonal entries to the entrywise quotients of $c$ by $b$. –  Jonas Meyer Nov 28 '11 at 4:17

[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.

share|improve this answer

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, $$ A=(cM)(bM)^{-1} $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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