The question I'm trying to answer says:
Suppose the T is an element of L(V), is such that every non-zero vector in V is an eigenvector of T. Prove that T is scalar multiplication of the identity operator
The way I look at it is that for T to be able to take on every vector of V as an eigenvector T must be the identity operator itself, or some multiple of it, otherwise it would not be able to take on these exact values of V. I'm wondering how I would go about proving this or if I am on the wrong track completely.
any help is Greatly appreciated!!
Since and are non-zero eigenvectors corresponding to distinct eigenvalues, they cannot be multiples of one another. But that means we must have and leading to the conclusion that , a contradiction. That is, if T has every vector as eigenvector, it must have only one eigenvalue.