Proving T is Scalar Multiple of Identity Operator: Invariant Subspace

[cubixguy77]

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.

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?

 

[Pere Callahan]

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 $\{e_i\}$. From what you have shown you know $T e_i =\lambda_i e_i$.

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