1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Decomposition of a complex vector space into 2 T-invariant subspaces

  1. Nov 22, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose V is a complex vector space and [tex]T \in L(V)[/tex]. Prove that
    there does not exist a direct sum decomposition of V into two
    proper subspaces invariant under T if and only if the minimal
    polynomial of T is of the form [tex](z - \lambda)^{dim V}[/tex] for some [tex]\lambda \in C[/tex].

    2. Relevant equations



    3. The attempt at a solution

    First suppose that the minimal polynomial of T is [tex]p(z) = (z - \lambda)^{n}[/tex] where n = dim V. Suppose further that [tex]V = U \oplus W[/tex] where U and W are T-invariant proper subspaces of V. So [tex]dim U \geq 1[/tex] and [tex]dim W \geq 1[/tex]. Since n = dim U + dim W, we have that [tex]dim U \leq n-1[/tex] and [tex]dim W \leq n-1[/tex].
    Now the minimal polynomial of T is [tex]p(z) = (z - \lambda)^{n}[/tex], therefore [tex](T - \lambda I)^{n}v = 0[/tex] for all v in V but there is at least one v in V such that [tex](T - \lambda I)^{n-1}v \neq 0[/tex]. Because of the decomposition of V, we can write: v = u + w for some u in U and some w in W. Applying [tex](T - \lambda I)^{n-1} [/tex] to both sides we get:

    [tex](T - \lambda I)^{n-1}v = (T - \lambda I)^{n-1}u + (T - \lambda I)^{n-1}w \neq 0[/tex]

    since both U and W are invariant under T, we have Tu in U and and [tex] -\lambda u \in U [/tex] so [tex](T - \lambda I)u \in U [/tex]. Thefore U (and similarly W) are both invariant under [tex](T-\lambda I) [/tex], so invariant under [tex](T-\lambda I)^{n-1} [/tex]. [tex](T - \lambda I)^{n-1}v \neq 0 [/tex], so either one of [tex](T - \lambda I)^{n-1}u [/tex] or [tex](T - \lambda I)^{n-1}w [/tex] is not zero, because they're in different subspaces. Assume it's the first one. So there exists u in U such that

    [tex](T - \lambda I)^{n-1}u \neq 0[/tex]. But [tex](T - \lambda I)^{n}u = 0 [/tex] (for all u in U). Constraining T to U, we see that the minimal polynomial of T on U is [tex](z - \lambda I)^{n} [/tex]. But [tex] dim U \leq n-1 < n [/tex], a contradiction.

    I hope the reasoning above is correct; if it is, it proves that minimal polyomial of T is [tex](z-\lambda I)^{n} [/tex] implies that T cannot be decomposed into the direct sum of two proper T-invariant subspaces.

    However I'm stumped about how to prove the other direction.
    Also, I would like to know what's the mistake in the following reasoning:

    Since V is a complex vector space, then T in L(V) has an eigenvalue [tex]\lambda[/tex], so V has an T-invariant subspace of dimension 1. Then there exists a subspace W of V such that [tex]V = U \oplus W[/tex]. It follows that W is T-invariant and of dimension dim V - 1[/tex]. So it's always possible to decompose V into two proper T-invariant proper subspaces :S

    Thank you :)
     
    Last edited: Nov 22, 2008
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you help with the solution or looking for help too?



Similar Discussions: Decomposition of a complex vector space into 2 T-invariant subspaces
Loading...