Kernel and Image of Matrix AB

[bubbles]

[SOLVED] Kernel and Image

Homework Statement

Ker(A) = Im(B)
AB = ?

A is an m x p matrix. B is a p x n matrix.

Homework Equations

The Attempt at a Solution

Since Ker(A) is the subset of the domain of B and Im(B) is the subset of the codomain of B, AB = I. I = identity matrix.

Is this right? It doesn't seem to make sense. There must be a mathematical (symbolic) way to solve for AB, right?

 

[Dick]

I'm assuming this means B:R^n->R^p. A:R^p->R^m. That way at least Ker(A) and Im(B) live in the same space. Now what do you say AB:R^n->R^m is?

 

[bubbles]

Is AB the p x m identity matrix?

 

[Dick]

Bad guess. AB is nxm. Consider AB(v). B(v) is in Im(B)=Ker(A). Guess again. Uh, what do you mean by 'identity matrix' anyway. There is no pxm identity matrix. Do you mean the zero matrix?

 

[bubbles]

Since Im(B) = Ker(A), does that imply that A and B are invertible? Then if B(v) is in the Ker(A) and Ker(A) = {0}, then AB = 0. Is that right?

 

[Dick]

Only one thing you said is right. Why should anything be invertible and why should Ker(A)={0}? If B(v) is in Ker(A) what's A(B(v))? If you think AB=0 then try and put together an argument that will convince me. In your own words.

 

[bubbles]

Since B(v) is in Ker(A) and T(v) = A * ker(A) = 0, then AB must be zero.

 

[Dick]

What's T? I think you MIGHT know what you are trying to say, but that's just a guess. If that's what you are planning to turn in as a solution, I don't think it will work. Can't you put that more clearly?

 

[bubbles]

The book defines the kernel as a subspace x and T(x) = Ax = 0 and ker(A) = x. T(x) is the linear transformation and A is the matrix. Should I not put the book's equation T(x) = Ax in my solution?

So in the solution I wrote in my homework, I wrote: BX is in ker(A) and AX = 0, X is in ker(A), therefore AB = 0, is that a good explanation?

 

[Dick]

I don't have your book. What's the linear transformation associated with Bx? Is it also T? What's the linear transformation associated with AB(x). If it's T, I'll scream. You aren't expressing yourself clearly. That's all I'll say.

 

[Dick]

So in the solution I wrote in my homework, I wrote: BX is in ker(A) and AX = 0, X is in ker(A), therefore AB = 0, is that a good explanation?

It's getting there. Try saying this. For any vector x, Bx is in Im(B). Since Im(B)=Ker(A) then Bx is in Ker(A). If Bx is in Ker(A) then A(Bx)=0. So ABx=0 for all x. So AB=0. How does that sound?

 

[bubbles]

Thank you for your help! I understand what you mean about my vague explanation. I will try to clarify things a bit on my paper.

 

[Dick]

You're welcome. Thanks for letting me scream. Good luck!