# A square matrix A with ker(A^2)= ker (A^3)

1. Sep 20, 2011

### yeland404

1. The problem statement, all variables and given/known data
a square matrix A with ker(A^2)= ker (A^3), is ker(A^3)= ker (A^4),verify...

2. Relevant equations

3. The attempt at a solution

2. Sep 20, 2011

### micromass

Staff Emeritus

3. Sep 20, 2011

### yeland404

it means that A^2*vector x= 0 and A*x=0 has same result,then I really confused how to do the next step

4. Sep 20, 2011

### micromass

Staff Emeritus
No, it means that

$$A^2x=0~\Leftrightarrow~A^3x=0$$

You need to prove that

$$A^3x=0~\Leftrightarrow~A^4x=0$$

5. Sep 20, 2011

### yeland404

emm...., to the difinition it says that ker(A) is T(x)=A(x)=0

6. Sep 20, 2011

### micromass

Staff Emeritus
Yes, but you're not working with ker(A) here, but with ker(A2).

7. Sep 20, 2011

### yeland404

should define a matrix A and write A^2 as dot product of A & A,and also to A^3...it seems to be so complex, or use the block to devide the matrix into some small matrix?

8. Sep 20, 2011

### micromass

Staff Emeritus
Do you understand my Post 4??

9. Sep 20, 2011

### yeland404

so times A on both side of the equation?

10. Sep 20, 2011

### micromass

Staff Emeritus
No, you need to prove two things:

If $A^3x=0$, then $A^4x=0$. This can indeed be accomplished by multiplying sides with A.

But you also need to prove that if $A^4x=0$, then $A^3x=0$.

11. Sep 26, 2011

### yeland404

so Ker (A^2)=0 can lead to Ker(A^4)=0,then?

12. Sep 26, 2011

### HallsofIvy

Staff Emeritus
NO ONE has said that Ker(A^2)= 0 so I do not understand why you are asking this question. You seem, frankly, to have no idea what the question is saying. ker(A^2)= ker(A^3) means, as micromass said, "A^2x= 0 if and only if A^3x= 0". You want to use that to prove "A^3x= 0 if and only if A^4x= 0". As micromass said, the first part is easy: if A^3x= 0 then, applying A to both sides, A^4x= A0= 0 which proves that ker(A^3) is a subset of ker(A^4). You still need to prove the other way: if A^4x= 0, then A^3= 0. You cannot just multiply by A^{-1} because you have no reason to think that A is invertible. But notice that ker(A^2)= ker(A^3) has not yet been used so it might help to note that A^4x= A^2(A^2x).