if $Y$ is eigenspace to an eigenvalue $\lambda$ of an operator $T,$ and if $T_1$ is restriction of $T$ to $Y$. find spectrum of $\sigma(T_1)$

Question

if $Y$ is eigenspace corresponding to an eigenvalue $\lambda$ of an operator $T,$ and if $T_1$ is restriction of $T$ to $Y$. then find spectrum of $\sigma(T_1)$

its easy to see that $\lambda \in \sigma(T_1)$. but why is $\sigma(T_1)=\{\lambda\}$

if $\alpha\neq \lambda$ we need to show $T_1-\alpha I$ is invertible.

$T_1-\alpha I $ is one one over $Y$. Also range $R$ of $T_1-\alpha I $ is closed set and hence banach space.

so by inverse mapping theorem. $T_1-\alpha I $ is invertible

is this approach correct?? or there is some other better way

Answer 1

I think you are making it too complicated. The operator $T_1$ is simply $\lambda I$.

So the question is whether you believe that $\sigma(I)=\{1\}$.

Answer 2

You can compute directly the inverse: $T_1-\alpha I= (\lambda-\alpha)I$ is invertible and its inverse is ${1\over{\lambda-\alpha}}I$.