Linear algebra rank and dimensions

[SpiffyEh]

Homework Statement

Prove Rank A + dim Nul A^T = m where A is in R^(mxn)

Homework Equations

The Attempt at a Solution

I honestly can't figure out where to go with this. I know that Rank A + dim Nul A = n, but I don't know if there is a relationship between the two.

 

[hgfalling]

Hint: If rank A = x, what does rank A^T equal?

 

[SpiffyEh]

Does rank A^T = x as well?

 

[hgfalling]

Yes.

 

[SpiffyEh]

I don't see how that relationship works out, i saw it in a theorem online but i wasn't able to find it in my book. Also I don't see where to go from there

 

[hgfalling]

Does your book talk about "row rank" and "column rank" and then say we just call it "rank" because they are always equal? Anyway, now just use your rank-nullity theorem on A^T. (ie, if A is mxn, then A^T is nxm, etc)

 

[SpiffyEh]

Yeah it has a section about rank but it never mentions the transpose. So, If i use the rank-nullity theorem on A^T could I just transpose the rank and the nullity and make it equal to m instead of n? Then since the transpose of the rank is equal to it not being transposed just switch it to that? Would that be enough to prove the concept?

 

[hgfalling]

Yes. Recall that the column rank of A is the dimension of the column space of A, and the row rank is the dimension of the row space of A. These are equal. The proof of this is longer than this sentence, so I refer you to other references for that.

Now the row space of A^T is the column space of A, right? Because the rows of A^T are the columns of A. So the rank of A is the rank of A^T.

 

[SpiffyEh]

ok ok thank you that makes sense. We were told to ignore the whole concept of row space so I didn't even think of using that