# Kernel of the Transpose

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?

Staff Emeritus
Gold Member
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

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 $A^TA$?