What is the relationship between ker(A) and ker(A^TA)?

[johndoe3344]

B= A transpose

What is the relation between ker(BA) and ker(A)? I was told that they are equal to each other, but I can't figure out why.

ker(A) => Ax = 0
ker(BA) => BAx = 0 so that BA is a subset of A. This shows that ker(BA) =0 whenever ker(A) = 0, but how does this also show that they are equal?

 

[Hurkyl]

It might be the case that your shorthand is obscuring things -- try writing things more precisely, and maybe the answer will become more clear.

 

[johndoe3344]

Hi, thanks for your response.

What do you mean my shorthand? I only said that B = A transpose because I didn't know how to write the superscript T on the forums (is that what you meant?)

Does showing that ker(A^T*A) is a subset of ker(A) show that they are equal?

 

[trambolin]

Let me use Latex for your convenience (you can click on them to learn how to write in case you don't know)...

B = A^T

Then,

\ker{(A)} \Rightarrow Ax=0 and \ker{(BA)} \Rightarrow BAx = 0

Now plug B in

\ker{(A^TA)} \Rightarrow A^TAx = 0

And what do you know about A^TA?