Characterization of Triangularizability and Diagonalizability in Terms of the Minimal Polynomial

Theorem

1. Let $T\in \mathcal{L}(V)$
2. Let $dim(V)=n<\infty$ Then,
3. $T$ is Triangularizable $⟺$ the Prime Factorization of the Minimal Polynomial for $T$ is a product of linear factors
4. $T$ is Diagonalizable $⟺$ the Prime Factorization of the Minimal Polynomial for $T$ is a product of distinct linear factors

Proof

Proving Triangular Case

1. Suppose $T$ is Triangular with basis $\beta$
2. Then, the Characteristic Polynomial of $T$ is $\det (xI-[T{]}_{\beta })=f(x)=(x-a_{11})(x-a_{22})\dots (x-a_{nn})=(x-c_1{)}^{d_1}\dots (x-c_k{)}^{d_k}$ Where $a_i$ are distinct characteristic values. Therefore, the minimal polynomial for $T$ has similar form since it divides $f(x)$ by Cayley Hamilton Theorem

Proving $⟹$Triangularizability

1. If $[T{]}_{\beta }$ is diagonal, then $p(x)=(x-c_1)\dots (x-c_k)$ is the minimal polynomial for $T$
2. Then, $p(T)=(T-c_1)\dots (T-c_k)=0$ as each diagonal entry will be multiplied by $0$ at some point

Proving $⟸$ Triangularizability

1. Suppose the minimal polynomial is a product of linear factors. Apply the previous lemma to $W=\{0\}$ to get ${\alpha }_1$
2. Then, $(T-eI){\alpha }_1=0⟹T{\alpha }_1=c{\alpha }_1\in span(\{{\alpha }_1\})$
3. Let $W_1=span(\{{\alpha }_1\})$. Note that $W_1$ is $T$-invariant.
4. We can apply the previous lemma to $W_1$ to get ${\alpha }_2$.
5. Then, $(T-e){\alpha }_2=0⟹T{\alpha }_2=c_1{\alpha }_1+c_2{\alpha }_2\in span(\{\alpha .{\alpha }_2\})$
6. Continue this way until you have $W_n=V$ (Linear Independence Extension Lemma)
7. Let $\beta =\{{\alpha }_1,\dots ,{\alpha }_n\}$. Then, $[T{]}_{\beta }$ is Upper Triangular

Proving Diagonizable Case

Proving $⟸$ Diagonalizability

1. Suppose that $T$ is not diagonal for sake of contradiction
2. Suppose $W$ is the subspace spanned by all Characteristic Vector of $T$.
3. If $W=V$, then $T$ is diagonizable, as proven earlier.
4. This contradicts our assumption
5. Instead, consider $W=V$
6. By the previous lemma, there exists a vector $\alpha$ not in $W$ and a characteristic value $c_i$ of $T$ such that the vector $\beta =(T-c_{i}I)\alpha \in W^{\ast }$
7. Let $h(T)$ be some polynomial
8. We will find $h(T)\in W$
9. Now, $p=(x-c_i)(q_i)$ for some polynomial $q$
10. We pick $h$ s.t $q-q(ci)=(x-c_i)h$
11. Thus, we have that $q(T)\alpha =q(c_i)\alpha =h(T)(T-c_i)\alpha =h(T)\beta$
12. As $0=p(T)\alpha =(T-cI)q(T)\alpha$, the vector $q(T)\alpha$ is an Eigenvector as long as $q(T)\alpha \ne 0$. This implies that $q(T)\alpha =\lambda \alpha ,\lambda \in \mathbb{F}$
13. This implies that $q(T)\alpha -q(c_i)\alpha =k\alpha =h(T)\beta \in W$
14. Thus, $\alpha \in W$
15. As $\alpha \in W$ , as $q(c_i)\alpha =q(T)\alpha \in W$, then $q(c_i)=0$
16. By Fundamental Remainder Theorem, $c_i$ is a root of $q$. Thus, $p$ has a repeated root in $c_i$ for the minimal polynomial has a non-linear factor.

Proving $⟹$ Diagonalizability