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Proving a matrix is diagonaziable.

  1. Jul 7, 2007 #1
    im given a matrix H from M_n(C) (the space of nxn matrices above the comples field).
    and we know that for every a in C, dim(ker(H-aI)^2)<=1.
    prove that H is diagonizable.

    obviously if i prove that its characteristic poly exists then bacuase every poly above C can be dissected to linear factors, then also its minimal poly can be dissected to linear factors and thus H is diagonaizable, but how to do it?

    thanks in advance.
     
  2. jcsd
  3. Jul 8, 2007 #2

    matt grime

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    Every matrix has a characteristic poly, so your argument is invalid. The characteristic poly doesn't tell you anything about diagonalizability. The minimal poly may tell you something useful.
     
  4. Jul 8, 2007 #3
    care to elaborate a bit more?
    i mean what exactly the minimial polynomial can help here?
     
  5. Jul 8, 2007 #4

    morphism

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    If the minimal polynomial has no repeated roots, then...
     
  6. Jul 8, 2007 #5

    mathwonk

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    do you know what a jordan form is?
     
  7. Jul 10, 2007 #6
    you mean that it is decomposed of linear polynomials, and thus the matrix is diagonizable.
    but how to translate it to here.
    i mean im given that d=dimKer(H-xI)^2<=1, now if H-xI=0 then we have that minimal poly is linear product, but i dont see how to arrive at this, if i prove that dimKer(H-xI)=n then im done, but how to do it?

    p.s
    mathwonk, i do know what is jordan noraml form is, but i don't see here relevancy?
     
  8. Jul 10, 2007 #7

    matt grime

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    Why don't you see the relevance of JNF? It gives the answer instantly. A matrix is not diagonalizable if and only if there is a jordan block of size greater than 1. But the condition you have precludes that happening.
     
  9. Jul 10, 2007 #8
    but the maximum size of jordan block is determined by the multicipity of the eigen value in the minimal polynomial.
    oh, wait a minute, bacuase it's above C, then it means that the minimal poly is product of linear polys, so obviously we don't have higher degrees, and
    Ker(H-aI) is a subspace of Ker(H-aI)^2, so dim(Ker(H-aI))<=1, so we have just one block at most, but every Ker(H-aI)^2 its dim is <= 1, so obviously we don't have here dome subspace which nullifies the entire C^n, but still we ned to have information on the minimal poly to gain information about the maximal size of the block.
     
  10. Jul 10, 2007 #9

    matt grime

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    yes.

    No. There is nothing that states that the linear factors of the minimal poly of a matrix are distinct, irrespective of the field.


    Absolutely not. If there is one block, then it can't be diagonalizable.

    I have no idea what you mean.

    If we have a Jordan block of size 2, or more, and e-value t, then there are vectors x and y such that Mx=tx, and My=ty+x. That is the definition of a Jordan block. Now show that x and y are killed by (M-t)^2.
     
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