Rank and inverse of matrices

I have some more linear algebra problems...

First: Prove that if B is a 3x1 matrix and C is a 1x3 matrix, then the 3x3 matrix BC has rank at most 1. Conversely, show that if A is any 3x3 matrix having rank 1, then there exist a 3x1 matrix B and a 1x3 matrix C such that A=BC

The first part is easy (it follows from a theorem). Im not sure how to do the "Conversely" part, and im also curious about whether it generalizes to mxn matrices and what the linear transformation analogy to this would be.

Second: Let A be an mxn matrix with rank m. Prove that there exists an nxm matrix B s.t AB = I
Also, let B be an nxm matrix with rank m. Prove that there exists an mxn matrix A such that AB = I

Im not sure about these. I know that since the rank is m, there are m linearly independent rows...

Any help would be useful. Thanks.

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For the first one, just to reiterate what you've already said, since the rank is one, there is one linearly independent row. So the other two rows are linear combinations of this row. What does that mean?

Oh ok, i got it. the matrix B could be any of the columns of A. And the matrix C could contain in each corresponding column, the scalar multiple needed to create the corresponding column of A. Thanks. I can see how it would work with nxn, nx1 and 1xn matrices, but with mxn matrices i guess theres something different and more complicated.

Still can't get the second part...

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If there are m linearly independent vectors in Rm, these must span the space, and so any vector can be written as a linear combination of them. Does this help?

Ok i think i have it. Each column in AB must be a linear combination of the columns of A with the coefficients being the appropriate column in B. By choosing appropriate values for the entries in B, any column vector can be generated for AB (element of R^m). Choose the B entries so that the jth column of AB has zeroes everywhere except at the jth spot a 1. And its similar for the other part of this question. Right?

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