Proof: Matrix Rank 1 | 3x3 Matrix A = BC

[ptolema]

Homework Statement

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

Homework Equations

rank (BC)=rank (A)=1
rank (BC) $$\leq$$ rank (B) and rank (BC) $$\leq$$ rank (C)

The Attempt at a Solution

I prove that if B is a 3x1 matrix and C is a 1x3 matrix, then the 3x3 matrix BC has rank at most 1 (rank BC $$\leq$$ 1) in a different part of the problem. I'm not sure if that would be useful in this proof, though; this one is more like proving the converse. This is how far I got with this proof in particular:

define A=BC,where B is 3×n matrix and C is n×3 matrix

rank BC=1 ≤ rank B ≤ n,1 ≤ rank B ≤ 3
1 ≤ rank C ≤ n,1 ≤ rank C ≤ 3

from here, I need to show that n=1, but I don't know how to get to that point. A little help here?

 

[Dick]

If A has rank 1, then every column of A is a multiple of the same vector. Call the vector v and the first column av, the second column bv and the third column cv. Can you figure out an explicit way to factor that matrix?

 

[ptolema]

the vector v van be factored out, so that after some elementary row ops, bv and cv are rows of zeros. I think I might get what you're getting at, but how can I use that in terms of B and C?

 

[Dick]

Don't do row ops. If you are writing vectors in column form, then put B=v. What's C?

 

[ptolema]

the first entry of C is a, the second is b, and the third is c. When you put it like that, it seems so obvious. Thanks, I think I got it from here!