Finding spectrum of given operator

Question

Let $X=C[0,1]$ and define $T:X \to X$ by $T(x)=vx$ where $v \ \in X$ is fixed.

Find $\sigma(T)$, where $\sigma(T)$ mean spectrum of $T$.

What I know is that $\sigma(T)$ is compact set in complex plane. and it is complement of $\rho(T)$ resolvent set of $T$. And if $T\in B(X)$ where $X$ is Banach space, and $||T||<1$, then $(I-T)^{-1}$ exists as bounded operator on whole of $X$.

Now from this information how to approach this problem. Any hint.

Thanks in advanced.

Answer 1

It really is just a matter of actually writing out the formula and solving it.

$((T-\lambda )x)(t) = v(t)x(t)-\lambda x(t) = (v(t)-\lambda) x(t)$.

Then $(T-\lambda I) x = y$ has a solution for all $y$ iff $(v(t)-\lambda) x(t) = y(t) $ for all $t$ iff $v(t) \neq \lambda$ for all $t$.

It follows that $\sigma(T) = v([0,1])$.