# Kernel of the Transpose

1. Apr 28, 2008

### 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?

2. Apr 28, 2008

### Hurkyl

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

3. Apr 29, 2008

### johndoe3344

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?

4. Apr 29, 2008

### 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 $A^TA$?