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Linear Algebra- Kernel and images of a matrix

  1. Oct 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider a square matrix A:

    a. What is the relationship between ker(A) and ker(A^2)? Are they necessarily equal? Is one of them necessarily contained in the other? More generally, What can you say about ker(A), ker(A^2), ker(A^3), ker(A^4),...?

    b. What can you say about im(A), im(A^2), im(A^3), im(A^4),...?


    2. The attempt at a solution

    So i believe if A is invertible nxn matrix, than ker(A)={<0,0,0>} and so will ker(A^2) and so on. And the image of A if A is invertible is im(A)=R^n and so will the im(A^2) and so on, but im not sure what it would be for other conditions of A at least thats what I think this question wants.
     
  2. jcsd
  3. Oct 20, 2009 #2
    a. If [tex]x \in \ker(A)[/tex], what can you say about [tex]A^2 x[/tex]? What does that tell you about [tex]\ker(A^2)[/tex]?

    b. For any [tex]x[/tex], [tex]A^2x[/tex] is in [tex]\operatorname{im}(A^2)[/tex]. If you rewrite [tex]A^2x[/tex] as [tex]A(Ax)=Ay[/tex], what can you say about [tex]\operatorname{im}(A)[/tex]?
     
  4. Oct 20, 2009 #3
    for A: A^2 * x would equal the zero vector which means the ker(A^2) contains x but is not necessarily equal because it may contain another vector

    B. Is that saying the im(A) contains the im(A^2) but is not necessarily equal to the im(A^2) because it will span more vectors?
     
  5. Oct 20, 2009 #4
    In other words, we have [tex]\ker(A)\subset \ker(A^2)[/tex].

    Yes. You could also write : [tex]\operatorname{im}(A^2)\subset \operatorname{im}(A)[/tex]

    Now, how can you apply this to [tex]\ker(A^3)[/tex], [tex]\ker(A^4)[/tex]... and [tex]\operatorname{im}(A^3)[/tex], [tex]\operatorname{im}(A^4)[/tex]...?
     
  6. Oct 20, 2009 #5
    alright, thank you very much for your help
     
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