Induction on $g(T)=\sum_{i=1}^k g(\lambda_i)T_i$.

Question

Let T be a normal operator on a finite-dimensional complex inner product space V. If g is a polynomial, then $g(T)=\sum_{i=1}^k g(\lambda_i)T_i$.

I know that $T_i T_j =0$ for $ i \not= j$ $(T_i$ is the orthogonal projection of $V$ on a subspace $W_i$).

Suppose that $g(x) = a_1 x + a_0$. Then, $g(\lambda_1)= a_1 \lambda_1 +a_0$, so $g(T^1)=(a_1 \lambda_1 + a_0)T_1$. Assume it holds true, for $n=k-1$, I want to prove it holds true for n=k. Can I just say for linearity, that $g(T)=a_nT^n+...+a_oI \rightarrow T_1 g(\lambda_1)+...+T_k g(\lambda_k)$?

Answer 1

Actually take a normalized basis of $V$

$$ g(T)=\sum_k a_kT^k =\sum_k a_k(\sum_i \lambda_iT_i)^k$$

Because of normality $T_iT_j=0$ when $i \neq j$ and $T^n=T$:

$$(\sum_i \lambda_iT_i)^k=\sum_i\lambda_i^kT_i $$

$$\sum_k\sum_ia_k \lambda_i^k T_i=\sum_iT_i\sum_k a_k\lambda_i^k=\sum_ig(\lambda_i)T_i $$

Hence your result.