# Homework Help: Finding basis for null(A) and null(AP)

1. Jan 26, 2010

### jumbogala

1. The problem statement, all variables and given/known data
A is a mxn matrix, and P is an invertible nxn matrix.

So I want to prove that the bases of null(A) and null(AP) have the same number of elements.

2. Relevant equations

3. The attempt at a solution
I was going to start off by assuming that {X1, X2, ... Xm} is a basis of null(A). This is my first issue. Can I really just assume that?

So AXi = 0, following the definition of null space. Then V-1Xi is in null(AV), since AV(V-1Xi) = 0. Somehow I need to get from that to the fact that {V-1X1, V-1X2, ... V-1Xm} is independent. But how?

I can prove that {V-1X1, V-1X2, ... V-1Xm} spans null(A), I just need to know that it's independent for it to be a basis. And if it is, then it has m elements and so does null(A)'s basis.

2. Jan 26, 2010

### HallsofIvy

Yes, null(A) is a vector space and every vector space has a basis.

Where did "V" come from? There was no mention of V before. Did you mean P?

I can prove that {V-1X1, V-1X2, ... V-1Xm} spans null(A), I just need to know that it's independent for it to be a basis. And if it is, then it has m elements and so does null(A)'s basis.[/QUOTE]
Proof by contradiction: Suppose $\{P^{-1}X1, P^{-1}X2, ..., P^{-1}Xm}$ is NOT independent. Then there exist numbers $a_1, a_2, ..., a_m$, not all 0, such that $a_1P^{-1}X1+ a_2P^{-1}X2+ ...+ a_mP^{-1}Xm= 0$. Since P, and so $P^{-1}$, is a linear tranformation, that is the same as $P^{-1}a_1X1+ a_2P^{-1}X2+ ...a_mP^{-1}Xm= P^{-1}(a_1X1+ a_2X2+ ...+ a_mXm)= 0$. Now take P of both sides.

3. Jan 26, 2010