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Prove that if $A$ is an $m\times n$ matrix and $B$ is an $n\times p$ matrix, then $\operatorname{rank} AB$ is less than or equal to $\operatorname{rank} B$.

The hint is: prove that if the $k$th column of $B$ is not a pivot column, then the $k$th column of $AB$ is not a pivot column of $AB$.

I don't think I am allowed to use row space, column space or the Sylvester's inequality as we haven't learned them yet in the class.

I have been stuck here for ages, so hints or solutions are very appreciated.

Thank you for your time.

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You are not allowed to use the concepts of row space and column space? What is the definition of the rank of a matrix then? –  user1551 Mar 18 '13 at 16:22
    
Well I guess my teacher haven't named those spaces yet... will first appear 2 chapters later according to the book. Though I do get ur point there. –  Patrick Olsen Mar 18 '13 at 16:27
    
That's fine, but what exactly is your textbook definition of the rank of a matrix? Without knowing that, people cannot help. –  user1551 Mar 18 '13 at 16:30

1 Answer 1

up vote 3 down vote accepted

I am assuming that you are using the fact that the rank of the matrix is the number of linearly independent columns of the matrix.

The trick is to use block multiplication. If we write $B$ columnwise as $$B=\begin{pmatrix}\mathbf{b}_1 & \cdots & \mathbf{b}_p\end{pmatrix}$$ then we may express $AB$ in the same way as $$AB = \begin{pmatrix}A\mathbf{b}_1 & \cdots & A\mathbf{b}_p\end{pmatrix}$$ To this end, it follows that if column $i$ of $B$ is originally a linear combination of the other columns of $B$, then we also have column $i$ of $AB$ as a linear combination of the other columns of $AB$. This shows that the number of linearly dependent columns of $AB$ is at least the number of linearly dependent columns of $B$. In other words, the number of linearly independent columns of $AB$ is at most the number of linearly independent columns of $B$, i.e. $$\rm rank(B) \ge \rm rank(AB)$$ You'll probably need to air out the above argument a bit, but this is the general idea.

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This was just simple and great man! Will surely help me on my next problem as well. Thank you. –  Patrick Olsen Mar 18 '13 at 16:34
    
@PatrickOlsen You're welcome! –  EuYu Mar 18 '13 at 16:35

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