Theroem 6: Let V be a finite-dimensional vector space over the field F
and let T be a linear operator on V. Then T is diagonalizable if and only if the
minimal polynomial for T has the form
$$ p = (x-c_1)…(x-c_n) $$
where $c_1,…c_n$ are distinct elements of $\mathbb{F}$.Proof from Hoffman, Kunze:
We have noted earlier that, if $T$ is diagonalizable, its minimal
polynomial is a product of distinct linear factors (see the discussion
prior to Example 4).To prove the converse, let $W$ be the
subspace spanned by all of the characteristic vectors of $T$, and
suppose $W \ne V$.By the lemma used in the proof of Theorem 5,
there is a vector $\alpha$ not in $W$ and a characteristic value $c_j$
of $T$ such that the vector $\beta= (T – c_jI)\alpha$ lies in W.Since $\beta$ is in $W$, $\beta = \beta_1+\dots\beta_k$ where
$T\beta_i = c_i\beta_i$, $1\le i\le k$, and therefore the vector
$h(T)\beta = h(c_1)\beta_1+\dots+h(c_k)\beta_k$ is in $W$, for every
polynomial $h$.Now $p = (x-c_j)q$, for some polynomial $q$.
Also $q- q(c_j) = (x – c_j)h$.
We have $q(T)\alpha – q(c_j)\alpha = h(T)(T – c_jI)\alpha = h(T)\beta$.
But $h(T)\beta$
is in $W$ and, since $0 = p(T)\alpha = (T – c_jI)q(T)\alpha$, the
vector $q(T)\alpha$ is in $W$.Therefore, $q(c_j)\alpha$ is in
$W$.Since $\alpha$ is not in $W$, we have $q(c_j) = 0$.
That contradicts the fact that $p$ has distinct roots.
Can you explain why it is given that $q(x) – q(c_j) = (x-c_j)h(x)$? I don't understand this part so I don't quite get the remaining proof of the theorem.
Best Answer
Note that $c_j$ is a root of $q(x)-q(c_j)$. Hence $q(x)-q(c_j)=(x-c_j)h(x)$ for some polynomial $h$.