I am reading C*-Algebras and Operator Theory – Gerald J. Murphy. Here is where I'm stack
$U $ is invertible if and only if $\inf_n|\lambda_n|>0$
1.4.3. Example. One should not be misled by Theorem 1.4.11—the spectral behaviour of compact operators is not typical of all operators. To illustrate this, let $H$ be a separable Hilbert space with an orthonormal basis $(e_n)_{n=1}^\infty$. If $(\lambda_n)$ is a bounded sequence of scalars, define $u\in B(H)$ by setting $u(x)=\sum_{n=1}^\infty\lambda_n\alpha_ne_n$ when $x=\sum_{n=1}^\infty\alpha_ne_n$. We call $u$ the diagonal operator with diagonal $(\lambda_n)$ with respect to the basis $(e_n)$. It is readily verified that $\|u\|=\sup_n|\lambda_n|$, and that $u$ is invertible if and only if $\inf_n|\lambda_n|>0$, and in this case $u^{-1}$ is the diagonal operator with respect to $(e_n)$ with diagonal $(\lambda_n^{-1})$. These observations imply that $\sigma(u)$ is the closure of the set $\{\lambda_n\mid n=1,2,\ldots\}$.
$u$ is diagonal operator on Hilbert space, then $u$ is invertible iff $\inf \lambda >0$.
Best Answer
$u(e_n)=\lambda_ne_n$. So, if $u$ is invertible, then $\lambda_n \neq 0$, $u^{-1}(e_n)=\frac 1 {\lambda_n}e_n$ and $\frac 1 {|\lambda_n|} \leq \|u^{-1}\|$ proving that $\inf_n |\lambda_n| >0$.
Conversely, if $\inf_n |\lambda_n| >0$ then $v(\sum \alpha_n e_n)=\sum \frac1 {\lambda_n} \alpha_n e_n$ defines a bounded operator and $u\circ v=v\circ u=I$.