Question about diagonalization and projections

Question

Let a finite dimensional vector space $V$ above $\mathbb{F}$. Let $T:V\to V$ a diagonlizable transformation. We denote $a_1 \ldots a_r$ the $r$ different eigenvalues of $T$. By diagonalization, we have $V = \oplus_{i=1}^r W_i$ where $W_i = \ker (T-a_i Id)$. We denote $E_i:V\to V$, the $i$-th projection:

$${E_i}(v) = \left\{ {\matrix{ {v,v \in {W_i}} \cr {0,Otherwise} \cr } } \right.$$

Show that:

$${E_i} = {1 \over {\prod\limits_{j \ne i} {({a_i} - {a_j})} }}\prod\limits_{j \ne i} {(T - {a_j} \cdot Id)} $$

I don't even know how to approach this..

Answer 1

It's enough to show the equality for each $\;w_i\in W_i\;,\;\;i=1,...,r\;$ :

$$\frac1{\prod\limits_{j\neq i}(a_i-a_j)}\prod_{j\neq i}\left(T-a_j I\right)(w_i)=\frac1{\prod\limits_{j\neq i}(a_i-a_j)}\prod_{j\neq i}\left(Tw_i-a_jw_i\right)=$$

$$=\frac1{\prod\limits_{j\neq i}(a_i-a_j)}\prod_{j\neq i}(a_iw_i-a_jw_i)=\frac1{\prod\limits_{j\neq i}(a_i-a_j)}\prod_{j\neq i}(a_i-a_j)w_i=w_i=E_i(w_i)$$

and if $\;m\neq i\;$ :

$$\frac1{\prod\limits_{j\neq i}(a_i-a_j)}\prod_{j\neq i}\left(T-a_j I\right)(w_m)=\frac1{\prod\limits_{j\neq i}(a_i-a_j)}\prod_{j\neq i}\left(Tw_m-a_jw_m\right)=$$

$$=\frac1{\prod\limits_{j\neq i}(a_i-a_j)}\prod_{j\neq i}(a_mw_m-a_jw_m)=\frac1{\prod\limits_{j\neq i}(a_i-a_j)}\prod_{j\neq i}(a_m-a_j)w_m=0\cdot w_m=0=E_i(w_m)$$

Answer 2

Hints: The elements of the $W_i$ and $W_j$, $j \neq i$, span $v$ so you only have to show that those maps are equal in two cases: 1) when $v \in W_i$ and 2) when $v \in W_j$ for $j \neq i$.

In both those cases you know exactly what $T$ and $E_i$ does to $v$ so you should be able to just plug into that formula and observe that both sides are equal.