Diagonalizable operator of a finite vector space

Question

Let $V$ a vector space of finite dimension, $dim (V) = r$, and $T: V \rightarrow V$ a diagonalizable operator with $ \lambda _1,\lambda_ 2,...,\lambda _r$ distincts eigenvalues of $T$ then $ (T- \lambda _1 I) ... (T- \lambda _r I) = 0$.

i stuck with this problem someone can help to understand and prove this, please it is an observation of Linear Algebra, Friedberg.

Answer 1

First note that $(T-\lambda_i I)$ and $(T-\lambda_j I)$ commute: $$(T-\lambda_i I)(T-\lambda_j I) = T^2 - \lambda_i T - \lambda_j T + \lambda_i \lambda_j I = T^2 - \lambda_j T - \lambda_i T + \lambda_j \lambda_i I = (T-\lambda_j I)(T-\lambda_i I)$$ for each $i$ and $j$. For shorthand we will call each $T-\lambda_i I = T_i$.

Then we can write any vector $v \in V$ as a linear combination of the eigenvectors, call them $e_i$: $$v=a_1 e_1 + \cdots + a_r e_r$$

Now $$(\prod_i T_i) v = \sum_{j = 1}^r (\prod_i T_i) a_j e_j = \sum_{j=1}^r (T_1 T_2 \cdots T_{j-1} T_{j+1} \cdots T_r) T_j (a_j e_j)$$

Here we used commutatitivity to move $T_j$ to the front. Now recall that $T_j = (T-\lambda_j I)$ and when you apply this to $a_j e_j$ we find $$(T-\lambda_j I) a_j e_j = a_j (Te_j - \lambda_j e_j) = a_j (\lambda_j e_j - \lambda_j e_j) = 0.$$

Answer 2

Hint

Since $T$ is diagonalizable then

$$V=\ker(T-\lambda_1I)\oplus\cdots \oplus\ker(T-\lambda_rI)$$ which means that we have a basis $\mathcal B$ of eigenvectors of $T$ for $V$. Notice also that $T-\lambda_iI$ commutes with $T-\lambda_jI$. Now let $x\in V$ and write it in $\mathcal B$. How we compute $$(T-\lambda_1I)\cdots (T-\lambda_rI)(x)$$ to get the desired answer?

Answer 3

Try the Cayley-Hamilton Theorem:http://en.wikipedia.org/wiki/Cayley-Hamilton_theorem