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Thread: Find a T invariant subspace if it exists

  1. #1
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    Find a T invariant subspace if it exists

    Hello,

    I want to prove,

    Let the minimal polynomial of $T$ on a finite dimensional vectore space $V$ be p^2 where $p$ is irreducible. Is it true that $V$ contains a proper $T$ invariant subspace?

    (I'm using theorems out of Hoffman and Kunze. $\S$ 6.1)

    So the minimal polynomial divides the characteristic polynomial $\det (\lambda - T) = c(\lambda)$ so in particular $c(p) = 0$ and therefore, $p$ is an eigenvalue of $T$.

    I can let $E_p$ be a projection onto the subspace $p v$ where $v$ is the eigenvector associated with $p$. We have $E_p w = pw`$ for all $w \in V$.

    To show that $T$ is invariant on $p v$ I can check that $T$ commutes with $E_p$.

    I tried writing:

    $$(E_p T) y = E_p (T y) = y` = p (\alpha v) = p E_p y = T (E_p y) = (TE_p)y.$$

    Here $y`$ is the projection of $Ty$ onto $ v$. Since $y`$ is colinear to $v$ I should be able to find a scalar such that $p(\alpha v ) = y`$.

    I'm not super confidant with this: is there any advice?
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  2. #2
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    Re: Find a T invariant subspace if it exists

    Let the minimal polynomial of T on a finite dimensional vectore space V be p^2 where p is irreducible
    .....
    therefore, p is an eigenvalue of T


    p is a polynomial. How is it an eigenvalue of T?
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  3. #3
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    Re: Find a T invariant subspace if it exists

    Quote Originally Posted by Idea View Post
    Let the minimal polynomial of T on a finite dimensional vectore space V be p^2 where p is irreducible
    .....
    therefore, p is an eigenvalue of T


    p is a polynomial. How is it an eigenvalue of T?
    Ah ya, that's pretty bad. I guess I was thinking of $p$ as a root of the minimal polynomial which doesn't exist in the field. So ya that's pretty bad.

    So how does it go then? The eigenvalues are associated with linear subsets of $V$ such that $T$ is invariant on them so I'm pretty sure I need to use that. I also know the eigenvalues are roots of the minimal polynomial. So then what (?) $V$ has no $T$ invariant subspaces because $p$ is irreducible?

    If the minimal polynomial splits into $\dim V$-many (distinct) linear factors then $T$ will have a basis in which the matrix of $T$ is diagonal.

    Here the minimal polynomial doesn't split, but all that says is that there's no basis in which a matrix representing $T$ is diagonal so it doesn't say anything about the non-existance of $T$ invariant subspaces?

    It sounds like there is a $T$ invariant subspace but I don't know how to calculate it.

    I haven't tried using this Jordan normal form is that related? It's the part of the book next to the eigenvector/eigenvalue section so I assume it's related - I'm just guessing based on proximity.

    Any suggestions?
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