Let $W_i$ be the null space of $(T-c_i I)$. If $T$ is diagonalizable, then prove that $\sum_i \dim(W_i) = \dim (V)$

Question

Let $T$ be a linear operator on a finite dimensional vector space $V$ and let $c_1,c_2\ldots ,c_k$ be the distinct eigenvalues of T. Let $W_i$ be the null space of $(T-c_i I)$. If $T$ is diagonalizable, then prove that $$\sum_i \dim(W_i) = \dim(V)$$

I am unable to connect things in order to logically think about this. How do I go about proving this?

Answer 1

Since $T$ is similar to a diagonal matrix, there exists a basis $\mathfrak{B}=\{v_j\}_{j\le \dim V}$ of $V$ such that for each $j$, $v_j \in W_k =\ker (T-c_kI)$ for some $k$. Since $\mathfrak{B} \subset \bigoplus_k W_k$, it follows $$ V =\text{span}\ \mathfrak{B} =\bigoplus_k W_k $$ and $\dim V =\dim \bigoplus_k W_k=\sum_k \dim W_k.$