## 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.