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If $A$ is a real $n\times r$-matrix such that its column vectors are linearly independent, and $B$ be a real $r\times n$-matrix such that its row vectors are linearly independent. Is any of the two following statements true?

a) $\mathrm{ran}(AB) = \ker(B)$.

b) $\mathrm{ran}(AB) = \mathrm{ran}(A)$.

I was thinking that we know that $\mathrm{ran}(A)=0$ for the first one, but I do not see how one can proceed, I feel that I do not know enough about $A$ or $B$ to justify or falsify the statements. I guess for the second one the first step is to prove that $\mathrm{ran}(B)=0$, but I do not see why this is true either.

This is homework, so any tips or suggestions would be good. Or some references to where I can read more about this.

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At the sizes you have given, the product $AB$ is undefined. Do you mean $B$ is an $r\times n$ matrix? – Daryl Oct 10 '12 at 10:20
Yes, I fixed it now! – N3buchadnezzar Oct 10 '12 at 10:27
up vote 1 down vote accepted

By range you mean image, right?

You know that the column vectors of a matrix form the image of a base? If you know that they are linearly independent, what can you conclude? On the other hand, the row vectors of a matrix are the column vectors of its transpose. Whan can you conclude for its tranpose? What can you conclude for the matrix itself?

If haven't had dual maps, do you know that rank = column rank = row rank? If there are linearly independent rows/columns, a $k \times l$-matrix is of full rank, which is $\min(k,l)$ = number of rows/columns.

Look for an easy counterexample for a) in dimension $1$. Try to argue that b) is correct by using a surjectivity argument.

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