Show that $U$ commutes with $T$ if and only if $U$ commutes with each $T_i$

Question

Let $T$ be a normal operator on a finite-dimensional complex inner product space $V$. Use the spectral decomposition $T=\lambda_1T_1+\dots+\lambda_kT_k$ to prove that if $U$ is a linear operator on $V$, then if $U$ commutes with $T$, $U$ commutes with each $T_i$.

$UT=TU$

$U(\lambda_1T_1+\dots+\lambda_kT_k)=(\lambda_1T_1+\dots+\lambda_kT_k)U$

Couple things I have tried are to plug in an orthonormal basis of eigenvectors and to use the fact that $g(T)=\sum_{i=1}^k g(\lambda_i)T_i$ if $g$ is a polynomial but neither led anywhere.

Other post said that because a polynomial $g_i$ such that $g_i(T)=T_i$ can be defined for each $i$, then $g_i(T)U=Ug_i(T)$ however I fail to see why you can ignore the $U$ when taking the $g_i$ of both sides?

Answer 1

\begin{split} g(T)U&=(\alpha_nT^n+\dots+\alpha_0)U\\ &=U(\alpha_nT^n+\dots+\alpha_0)\\ &=Ug(T) \end{split}

Then let $g(T)=g_i(T)=T_i$ for all $i$ (as each $T_i$ is a polynomial in $T$, $g_i(T)$)