05. The Primary Decomposition Theorem

Property of Direct Sum

Let $V$ be a vector space
Suppose $V=W_1\oplus \cdots \oplus W_r$ of subspaces $W_1,\cdots ,W_r$ so that for all $v\in V$ then

$$v=w_1+\cdots +w_r\text{ with }{w}_i\in W_i$$

Let the ${\mathcal{B}}_i$ be a basis for $W_i$ so

$$\mathcal{B}=\bigcup _i{\mathcal{B}}_i\text{ is a basis for }V$$

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Block Diagonalisation via Invariant Subspaces

Let $V$ be a finite dimensional vector space
Following the setup from direct sum of vector space

Let $T:V\to V$ be a linear map

If each $W_i$ is $T-$ invariant then $T$ with respect to basis $\mathcal{B}$ is block diagonal

$${}_{\mathcal{B}}[T{]}_{\mathcal{B}}=\left[\begin{array}{ccc}{A}_1& & \\ & \ddots & \\ & & A_r\end{array}\right]\text{ with }{A}_i={}_{{\mathcal{B}}_i}[T\mid W_i{]}_{{\mathcal{B}}_i}$$

Hence

$${\chi }_T(x)={\chi }_{T{\mid }_{W_1}}(x)\cdots {\chi }_{T\mid W_r}(x)$$

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Decomposition Theorem of a Vector Space

Let $V$ be a finite dimensional vector space
Let $T:V\to V$ be a linear map

Assume $f(x)=a(x)b(x)$ with $\gcd (a,b)=1$ and $f(T)=0$ then

$$V=\mathrm{Ker}(a(T))\oplus \mathrm{Ker}(b(T))$$

is a $T-$invariant direct sum decomposition

Especially if $f=m_T$ and $a,b$ are monic then

$$m_{T{\mid }_{\mathrm{Ker}(a(T))}}(x)=a(x)\text{ and }{m}_{T{\mid }_{\mathrm{Ker}(b(T))}}(x)=b(x)$$

Proof

By Bezout’s Identity for Polynomials then there exist $s,t$ such that $as+bt=1$ but then

$$a(T)s(T)+b(T)t(T)={\mathrm{Id}}_v$$

So for all $v\in V$

$$a(T)s(T)v+b(T)t(T)v=v$$

As $f$ is annihilating then

$$a(T)(b(T)t(T)v)=f(T)t(T)v=0\text{and}b(T)(a(T)s(T)v)=0$$

Hence

$$V=\mathrm{Ker}(a(T))+\mathrm{Ker}(b(T))$$

To show that this is a direct sum decomposition, assume that $v\in \mathrm{Ker}(a(T))\cap \mathrm{Ker}(b(T))$
So then we get $v=0+0=0$ hence

$$V=\mathrm{Ker}(a(T))\oplus \mathrm{Ker}(b(T))$$

To show that both factors are $T$-invariant then for $v\in \mathrm{Ker}(a(T))$ then

$$a(T)(T(v))=T(a(T)v)=T(0)=0$$

Similarly $b(T)T(v)=0$ for $v\in \mathrm{Ker}(b(T))$

Assume that $f=m_T$ and let

$$m_1=m_{T{\mid }_{\mathrm{Ker}(a(T))}}\text{ and }{m}_2=m_{T_{\mathrm{Ker}(b(T))}}$$

Then $m_1\mid a$ as

$$m_1=m_{T{\mid }_{\mathrm{Ker}(a(T))}}$$

And similarly $m_2\mid b$ and

$$m_T=ab\mid m_1{m}_2\text{ as }{m}_1(T)m_2(T)=0$$

Any $v\in V$ can be written as $v=w_1+w_2$ wh ere $w_1\in \mathrm{Ker}(a(T))$ and $w_2\in \mathrm{Ker}(b(T))$ so

$$m_1(T)m_2(T)v=m_2(T)(m_1(T)w_1)+m_1(T)(m_2(T)w_2)=0+0=0$$

Hence, for degree reasons and because all these polynomials then $m_1=a,m_2=b$

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13 - Primary Decomposition Theorem

Primary Decomposition Theorem

Let $m_T$ be the minimal polynomial and write it in the form such that

$$m_T(x)=f_1^{q_1}(x)\cdots f_r^{q_r}$$

where $f_i$ are the distinct monic irreducible polynomials

Define $W_i:=\mathrm{Ker}(f_i^{q_i}(T))$ then

$$\begin{array}{rl}(i)& V=W_1\oplus \cdots \oplus W_r\\ (ii)& W_i\text{ is }T-\text{invariant}\\ (iii)& m_{T{\mid }_{W_i}}=f_i^{q_i}\end{array}$$

Proof

Put $a=f_1^{q_1}\cdots f_{r-1}^{q_{r-1}}$ and $b=f_r^{q_r}$
Proceed on induction on $r$ using Decomposition Theorem of a Vector Space

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Invariant Factor Decomposition of the Minimal and Characteristic Polynomials

There exist unique distinct irreducible monic polynomials

$$f_1,\cdots ,f_r\in \mathbf{F}[x]$$

and positive integers

$$n_i\ge q_i>0\text{ for }1\le i\le r$$

such that

$$m_T(x)=f_1^{q_1}\cdots f_r^{q_r}\text{ and }{\chi }_T=\pm f_1^{n_1}\cdots f_r^{n_r}$$

Proof $m_T=f_1^{q_1}\cdots f_r^{q_r}$ into distinct monic irreducibles over $\mathbf{F}[x]$ By Cayley-Hamilton then $m_T\mid {\chi }_T$ so

Factor

$${\chi }_T=f_1^{n_1}\cdots f_r^{n_r}\cdot b(x)\text{ for some }{n}_i\ge q_i\text{ and }b(x)\in \mathbf{F}[x]$$

As $m_T$ and ${\chi }_T$ share roots by eigenvalue-polynomial characterisation then
$b(x)$ has no roots and must be constant so by comparing coefficients then

$$b(x)=(-1{)}^n\text{ where }n=\dim V$$

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Triangularisability Criterion

$$\begin{array}{rl}T& \text{ is triangularisable (over a given field)}\\ & ⟺{\chi }_T\text{ factors as a product of linear polynomials (over that field)}\\ & ⟺\text{each }{f}_i\text{ is linear}\\ & ⟺m_T\text{ factors as a product of linear polynomials}\end{array}$$

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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

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