14 - Diagonalisation Theorem

Diagonalisation Theorem

$$T\text{ is diagonalisable }⟺m_T\text{ factors as a product of distinct linear polynomials}$$

Proof

If $T$ is diagonalisable then there exists a basis $\mathcal{B}$ of eigenvectors for $V$ such that

$${}_{\mathcal{B}}[T{]}_{\mathcal{B}}\text{ is diagonal }$$

with entries from a list of distinct eigenvalues ${\lambda }_1,\cdots {\lambda }_r$ then

$$m(x)=(x-l_1)\cdots (x-{\lambda }_r)$$

is annihilating as

$$m(T)v=0\text{ for all }v\in \mathcal{B}\text{ hence }\forall v\in V$$

It is minimal as every eigenvalue is a root of the minimal polynomial by this theorem

For the converse, assume that $m_T(x)-(x-{\lambda }_1)\cdots (x-{\lambda }_r)$
Then by Primary Decomposition Theorem

$$V=E_{{\lambda }_1}\oplus \cdots \oplus E_{{\lambda }_r}$$

where $E_{{\lambda }_i}:=\mathrm{Ker}(T-{\lambda }_{i}I)$ which is the ${\lambda }_i-$eigenspace

So $V$ is a direct sum decomposition of eigenspaces

Let $\mathcal{B}={\bigcup }_i{\mathcal{B}}_i$ where ${\mathcal{B}}_i$ is the basis for $E_{{\lambda }_i}$
So we have a basis of eigenvectors where ${}_{\mathcal{B}}[T{]}_{\mathcal{B}}$ is diagonal