/sci/ - Science & Math

[Anonymous No.1248853]

Let V be some n-dimensional vectorspace over K and f an endomorphism of V.
Theorem: f is diagonizable iff the minimal polynomial splits over K and has pairwise distinct roots.

Why is that?
I know f is diagonizable iff the characteristic polynomial splits over K and the multiplicities of the roots in the characteristic polynomial coincides with the dimension of the respective eigenspace.

Now I can prove the theorem with the Jordan Normal form, but that's a few calibers too big. I think there's a much more basic proof for this.

Anyone any ideas?

[Anonymous No.1248894]

Hm, if I have pairwise distinct eigenvalues <span class="math">a_0, ..., a_k[/spoiler] and V is the direct sum of their eigenspaces, then every <span class="math">v \in V[/spoiler] can be written as a sum <span class="math">v = v_0 + ... + v_k[/spoiler], where <span class="math">v_j[/spoiler] is an eigenvector to the eigenvalue <span class="math">a_j[/spoiler].
Now, if I look at the polynomial <span class="math">(f-a_0)(f-a_1)...(f-a_k)[/spoiler], then I get:
<div class="math">(f-a_0)...(f-a_k)(v) = (f-a_0)...(f-a_k)(v_0 + ... + v_k)
= (f-a_0)...(f-a_k)(v_0) + ... + (f-a_0)...(f-a_k)(v_k)</div>
and every summand is vanishes because powers of f commute.

I rule.