# $n \times m$ matrix conversion?

Is it possible to convert an $n\times m$ matrix $A$ such that

$$A=CB$$

where $B$ is a $1\times m$ matrix which contains all elements of $A$, and $C$ is a $n\times 1$ matrix. I'm assuming no since this might give a special case of matrices.. but i am not so sure. If this is not possible, is it possible to extend matrix $A$ to ($n$ by $m$) by ($n$ by $m$) so the same conditions are met, yet the matrix is replicated and the result needs to be unique. Just to give a reason for this, i figured out a way to make $A$ into a $1$ by ($n$ by $m$) vector $B$, but to find an inverse of this, i need to solve $A=CB$, which is what's giving me problems.

• What do you mean by "elements?" There are $nm$ entries in an $n\times m$ matrix, and only $m$ entries in a $1\times m$ matrix, so the entries of the latter can't include all of the entries of the former, in gneral. – Thomas Andrews Apr 15 '15 at 18:15

## 3 Answers

It is not possible in general case.

Let $A=CB$, $C\in M_{n\times1}(\Bbb R)$ and $B\in M_{1\times m}(\Bbb R)$. Take a hyperplan in $\Bbb R^m$ orthogonal to $B$: $$H=\{x\in\Bbb R^m:\,Bx=0\}.$$ After that it easy to see that $\ker A=H$ and $im A=span(C)$; among other things, $rank(A)=1$.

So, whenever you have a matrix such that $\dim\ker A\ne m-1$ (or, equivalently, $rank(A)\ne 1$) you can not factorise $A=CB$.

Regarding your intial question. No, it isn't be possible:

Consider an nxm matrix of random values - it has n.m independent points of data. Where as a 1xm and a nx1 matrix together have only n+m independent points of data.

There's less data being stored so the two cannot be equivalent.

I don't think I can answer the rest of your question. Sorry.

assume ellements of $C$ and $B$ in variable form like $C_i$ and $B_j$ where $i$ represents row and $j$ represents column. now for any $A_{ij}=C_i\cdot B_j$ thus lets say for first row $A_{1j}=C_1\cdot B_j$ thus ratio of the elements of $B$ is the same as the ratio of first row elements of $A$ but wait $A_{2j}=C_1\cdot B_j$ thus its a contradiction , unless all row have same ratios of their elements it is impossible to form $A=CB$ as for the second part i can always take $C=kA$ and $B=\frac{I}{k}$ where $k$ is a constant and $I$ is the identity matrix.