Show that a given linear transformation has no eigenvectors

Let V = \mathbb{R}^\mathbb{N} be a countable-dimensional \mathbb{R} vector space, and define T : V \rightarrow V by T((a_i))_j = 0 if j = 0 and a_{j-i} otherwise. Show that T has no eigenvectors.

Suppose there exists a \in V, r \in \mathbb{R} nonzero, such that T(a) = ra. We claim that a_i = 0 for all i, and prove it by induction. For the base case i = 0, we have ra_0 = 0. Since r \neq 0, a_0 = 0. For the inductive step, if a_i = 0, then ra_{i+1} = a_i = 0. Again since r \neq 0, a_{i+1} = 0. So in fact a = 0.

So T has no eigenvectors.

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