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I am trying to solve some matrix multiplications, but I would like to know If I am allowed to take a common factor from matrices like this

C - ABC = (1 - AB)*C

where A is m*n and B is n*n. And if yes, what matrix will be the 1 matrix? It cannot be an identity as the AB is m*n.

Thanks a lot

K. Stasko
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3 Answers3

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You can write $C-ABC=(I-AB)C$ with $I$ the identity matrix (a matrix with zeroes off-diagonal and ones on the main diagonal) all day long as long as $C-ABC$ makes sense.

To see this, suppose that $C$ is a matrix with $m$ rows and $n$ columns. Then $A$ must have $m$ rows and $B$ must have $m$ columns because otherwise $C-ABC$ would not make sense. So $AB$ is necessarily square and you can compute the difference $I-AB$.

JPi
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If $AB$ is not square (e.g. if it is $m\times n$ with $m \not = n$) then the subtraction in your original expression will not work as $C$ and $ABC$ will have different dimensions.

Henry
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Why not factor first? $C = IC$ with $I$ square and same dimension as number of rows for $C$. From there regardless of all other dimensions, as long as your original equation makes sense and has conformable dimensions, then YES there is the common factor that can be factored: $$C - ABC$$ $$IC - ABC$$ $$(I - AB)C$$

adam W
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  • If the factored equation does not make sense, then your original must be wrong. $C$ is $m \times p$ because it has the same rows as $A$ in $C - ABC$ else the subtraction cannot be done. $C$ is $n \times p$ since $B$ is $n \times n$ and $BC$ is part of the equation. Thus you have $n = m$ and the square identity makes sense and is also necessary for your equation. – adam W Feb 10 '14 at 21:43