04. Triangular Form and the Cayley-Hamilton Theorem

Invariant under a linear transformation

Let $T:V\to V$ be a linear transformation
Let subspace $U\subseteq V$ then

$$U\text{ is }T\text{-invariant}\text{ if }T(U)\subseteq U$$

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Properties of Invariant Linear Transformations

Let $T:V\to V,S:V\to V$ be a linear transformation
Let subspace $U\subseteq V$ and $U$ is $T$- and $S$- invariant then

$U$ is invariant under

1. Zero Map
2. Identity Map
3. $aT$, $\forall a\in \mathbf{F}$
4. $S+T$
5. $S\circ T$
6. $p(T)$ for any polynomial $p(x)\in \mathbf{F}[x]$ (combining 2, 3, 4, 5)

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Map of Quotients from Invariant Polynomials

Let $T:V\to V$ be a linear transformation and $U$ is invariant under $t$
So $U$ is also invariant under polynomial $p(x)$ evaluated at $T$ then

$$\overline{p(T)}:V/U\to V/U$$

is a map of quotients

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Characteristic Polynomial of an Invariant Subspace

Let $T:V\to V$ be a linear transformation on a finite dimensional space
Assume $U\subseteq V$ is $T-$invariant then

$${\chi }_T(x)={\chi }_{T{\mid }_U}(x)\times {\chi }_{\overline{T}}(x)$$

Proof

Extend a basis $\mathcal{E}$ for $U$ to a basis $\mathcal{B}$ of $V$
Let $\overline{\mathcal{B}}$ be the associated basis for $V/U$
By Block Matrix Decomposition of a Linear Map then

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

Then the determinant of a triangular block matrix is the product of the determinants of diagonal blocks

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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

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Upper Triangularisation of a Matrix corollary

Let $A$ be a $n\times n$ matrix
If ${\chi }_A(x)$ is a product of linear factors then there

$$\exists P\in M_n(\mathbf{F})\text{ such that }{P}^{-1}AP\text{ is upper triangular}$$

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Annihilating Polynomial of an Upper Triangular Matrix

Let $A$ be an upper triangular $(n\times n)-$matrix with diagonal entries ${\lambda }_1,\cdots ,{\lambda }_n$ then

$$\prod _{i=1}^n(A-{\lambda }_{i}I)=0$$

Proof

Let $e_1,\cdots ,e_n$ be the standard basis vectors for ${\mathbf{F}}^n$ then

$$(A-{\lambda }_{n}I)v\in ⟨e_1,\cdots ,e_{n-1}⟩\text{ for all }v\in {\mathbf{F}}^n$$

So generally

$$(A-{\lambda }_{i}I)w\in ⟨e_1,\cdots ,e_{i-1}⟩\text{ for all }w\in ⟨e_1,\cdots ,e_i⟩$$

Since

$$\begin{array}{rl}\mathrm{Im}(A-{\lambda }_{n}I)& \subseteq ⟨e_1,\cdots ,e_{n-1}⟩\\ \mathrm{Im}(A-{\lambda }_{n-1}I)(A-{\lambda }_{n}I)& \subseteq ⟨e_1,\cdots ,e_{n-2}⟩\end{array}$$

Then we get

$$\prod _{i=1}^n(A-{\lambda }_{i}I)=0$$

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12 - Cayley-Hamilton

Cayley-Hamilton

Let $V$ be a finite dimensional vector space

If $T:V\to V$ is a linear transformation then

$${\chi }_T(T)=0$$

In particular

$$m_T(x)\mid {\chi }_T(x)$$

Proof

Let $A$ be the matrix of $T$ with respect to some basis for $V$
Let $\overline{\mathbf{F}}\subseteq \mathbf{F}$ be the algebraic closure of $\mathbf{F}$

So in $\overline{\mathbf{F}}[x]$ then every polynomial factors into linear terms
So by upper triangularisation of a matrix there exists $P\in M_n(\overline{\mathbf{F}})$ such that

$$P^{-1}AP\text{ is upper triangular }\text{ with }\text{ diagonal entries }{\lambda }_1,\cdots ,{\lambda }_n$$

Hence

$${\chi }_{P^{-1}AP}(x)=(-1{)}^{\dim V}\prod _{k=1}^n(x-{\lambda }_k)$$

By Annihilating Polynomial of an Upper Triangular Matrix then

$${\chi }_{P^{-1}AP}(P^{-1}AP)=0$$

As ${\chi }_T(x):={\chi }_{P^{-1}AP}(x)$ then

$${\chi }_T(T)=0$$

As Minimal Polynomial divides Annihiliating Polynomials then

$$m_T(x)\mid {\chi }_T(x)$$

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