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Let $\mathscr{M}$ be the set of all $n\times n$ matrices having entries $0$ and $1$ in such a way that there is one $1$ in each row and column.

(a) If $M\in\mathscr{M}$, describe $AM$ in terms of the rows and columns of A.

(b) If $M\in\mathscr{M}$, describe $MA$ in terms of the rows and columns of A.

I don't understand this question properly. So I want some help on what this question is exactly asking. I forgot to add this point before . Apologies for that.

creative
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    Have you tried to do this for, say, a small $3\times 3$ matrix $A$ and a permutation matrix $M$? –  Aug 30 '14 at 06:23
  • Hint: write the product on an example, with the usual setting ($A$ on the left of product $AB$, $B$ on top of it), to help visualize. A $4\times4$ example should be enough to understand. – Jean-Claude Arbaut Aug 30 '14 at 06:58
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    The simplest test case is $$M=\pmatrix{0&1\cr1&0\cr}.$$ Surely you have checked what happens when you multiply a generic $2\times2$ matrix by that from left or right? It amounts to an elementary row (resp. column) operation of a certain type :-) – Jyrki Lahtonen Aug 30 '14 at 07:01
  • Are you sure that should be an "I" (the letter $I$) and not just a "1" (the number $1$)? – Ben Grossmann Aug 30 '14 at 12:43
  • @Omnomnomnom, Thanx, you are right, I just missed that. I edited the question !! – creative Sep 01 '14 at 06:47

1 Answers1

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Let $A_1,\dots,A_n$ be the columns of $A$, and $M_1,\dots,M_n$ the columns of $M$. Note that each $A_i$ is one of the standard basis vectors, $\{e_1,\dots,e_n\}$, where $e_i = (\overbrace{0,\dots,0}^{i-1},1,0,\dots,0)^T$. Note the following two things:

  • For any matrix $M$ and $A$ as above, we can state that $$ MA = M \pmatrix{A_1 & \cdots & A_n} = \pmatrix{MA_1 & \cdots & MA_n} $$

  • For any basis vector $e_i$, we have $$ M e_i = \pmatrix{M_1 & \cdots & M_n} \pmatrix{0\\\vdots\\1\\ \vdots \\0} = M_i $$

That should be enough to figure out what $MA$ looks like. For the other case, note that $(AM)^T = M^TA^T$.

Ben Grossmann
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