11 - Triangularisation of a Linear Operator

Triangularisation of a Linear Operator

Let $V$ be a finite-dimensional vector space
Let $T:V\to V$ be a linear map such that the characteristic polynomial is a product of linear factors then

$$\exists \text{basis }\mathcal{B}\text{ of }V\text{ such that }{}_{\mathcal{B}}[T{]}_{\mathcal{B}}\text{ is upper triangular}$$

Note if $\mathbf{F}$ is an algebraically closed field then the characteristic polynomial always satisfies the hypothesis

Proof

By induction on the dimension of $V$

If $V$ is one dimensional, the nothing to prove

In general assume ${\chi }_T$ has a root $\lambda$ and hence there exists $v_1\ne 0$ such that $T{v}_1=\lambda v_1$
Let $U=⟨v_1⟩$ then $U$ is $T-$invariant

So consider induced map on quotients

$$\overline{T}:=V/U\to V/U$$

By characteristic polynomial of an invariant subspace then

$${\chi }_{\overline{T}}(x)={\chi }_T(x)/(\lambda -x)$$

Hence it is also a product of linear factors so $\dim V/U=\dim (V)-1$

So by induction hypothesis then there exists $\overline{\mathcal{B}}=\{v_2+U,\cdots ,v_n+U\}$ such that

$${}_{\overline{\mathcal{B}}}[T{]}_{\overline{\mathcal{B}}}\text{ is upper triangular}$$

Let $\mathcal{B}=\{v_1,\cdots ,v_n\}$ so $\mathcal{B}$ is a basis for $V$ by basis extension of a quotient space

$${}_{\mathcal{B}}[T{]}_{\mathcal{B}}=\left(\begin{array}{cc}\lambda & \ast \\ 0& {}_{\overline{{\mathcal{B}}^{\prime }}}[\overline{T}{]}_{\overline{\mathcal{B}}}\end{array}\right)$$

which is upper triangular