Kernel of the Transpose

  • #1

Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
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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
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?
 
  • #4
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Let me use Latex for your convenience (you can click on them to learn how to write in case you don't know)...

[tex]B = A^T[/tex]

Then,

[tex]\ker{(A)} \Rightarrow Ax=0[/tex] and [tex]\ker{(BA)} \Rightarrow BAx = 0[/tex]

Now plug B in

[tex]\ker{(A^TA)} \Rightarrow A^TAx = 0[/tex]

And what do you know about [itex]A^TA[/itex]?
 

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