# Thread: Find a T invariant subspace if it exists

1. ## 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?

2. ## 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?

3. ## Re: Find a T invariant subspace if it exists

Originally Posted by Idea
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?