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\begin{array} %\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} a_{11}b_{11} & a_{12}b_{12} \\ a_{21}b_{21} & a_{22}b_{22} \end{pmatrix} =(a_{ij})(b_{ij}) \end{array} Can anybody help me in separating $a_{ij}$ and $b_{ij}$? In other words, what will be the two matrices (in multiplication) of order 2 by 2 each but one contain only $a_{ij}$ and other contain only $b_{ij}$ ?

Your help will be appreciated.

PS: None of $a_{ij}$ and $b_{ij}$ are zero.

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    Setting all $a_{ij}$ values to $0$ is one solution. – 2017-02-09
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    From your wording I'm presuming that you want a decomposition that works in general. Why do you think it's possible to do so? – 2017-02-09
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    @epimorphic please educate me about it. I have no idea. – 2017-02-09
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    @epimorphic mathematically I can't justify but it is my expectation if it happens matrices looks simple. – 2017-02-09

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I am not totally sure if that is what you are asking since your notation is a little confusing; see that the result of the multiplication above is:

\begin{array} %\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} a_{11}b_{11} & a_{12}b_{12} \\ a_{21}b_{21} & a_{22}b_{22} \end{pmatrix} = \begin{pmatrix} a_{11}b_{11} + a_{21}b_{21} && a_{12}b_{12} + a_{22}b_{22} \\ a_{11}b_{11} + a_{21}b_{21} && a_{12}b_{12} + a_{22}b_{22} \end{pmatrix} \end{array}

If you want two matrices $2\times 2$, the first one with only $a_{ij}$ terms and the second with only $b_{ij}$ terms, then the first row and column respectively should be:

\begin{array} %\begin{pmatrix} a_{21} & a_{11} \\ ... & ... \end{pmatrix} \begin{pmatrix} b_{21} & ... \\ b_{11} & ... \end{pmatrix} \text{or} \begin{pmatrix} a_{11} & a_{21} \\ ... & ... \end{pmatrix} \begin{pmatrix} b_{11} & ... \\ b_{21} & ... \end{pmatrix} \end{array}

Take the second case for example. Then the second row of $A$ must be, again

\begin{array} %\begin{pmatrix} a_{11} & a_{21} \\ a_{11} & a_{21} \end{pmatrix} \begin{pmatrix} b_{11} & ... \\ b_{21} & ... \end{pmatrix} \end{array}

And that conflicts with the fact that you need the $a_{12}, a_{22}$ somehwere. Same would happen if you pick the first case.

In conclusion, they cannot be separated the way you suggest