The kernel and range of a product of matrices.

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.

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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

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.