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Invariant Subspace

  • Thread starter cubixguy77
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  • #1
1. Homework Statement
Suppose T is a linear operator on a finite dimensional vector space V, such that every subspace of V with dimension dim V-1 is invariant under T. Prove that T is a scalar multiple of the identity operator.

3. The Attempt at a Solution
I'm thinking of starting by letting U and W be subspaces of V with dim U = dim W = dim V-1
This means that U and W are invariant under T. Good start? Where do i go from there to show that T is a scalar multiple of the identity operator?
 

Answers and Replies

  • #2
Let n = dim V, v some element of V, <v> the 1-dim subspace generated by v.

Show that there are n-1 subspaces of dimension n-1 such that their intersection is <v>. Conclude that <v> is T-invariant.
Now you can fix a basis [itex]\{e_i\}[/itex]. From what you have shown you know [itex]T e_i =\lambda_i e_i[/itex].

You want to show that all [itex]\lambda_i[/itex] are equal. For this, consider for example [itex]e_i+e_{i+1}[/itex]. Since T fixes evey one-dimensional subspace it has to act on this vector as scalar multiplication, [itex]T(e_i+e_{i+1})=\nu_i^{i+1}(e_i+e_{i+1})[/itex].By linearity you also know that [itex]T(e_i+e_{i+1})=\lambda_i e_i+\lambda_{i+1}e_{i+1}[/itex]. Using the linear independece of the chosen basis you can conclude [itex]\lambda_i=\nu_i^{i+1}=\lambda_{i+1}[/itex]
 

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