06. Jordan Normal Form

Nilpotent

Let $V$ be finite-dimensional
Let $T:V\to V$ be a linear transformation

If

$$T^n=0\text{ for some }n>0$$

Then

$$T\text{ is nilpotent}$$

Link to original

Jordan Block

Let $J_n$ be a $(n\times n)-$matrix then

$$J_i:=\left(\begin{array}{cccc}0& 1& & 0\\ & \ddots & \ddots & \\ & & \ddots & 1\\ 0& & & 0\end{array}\right)$$

Link to original

15 - Jordan Canonical Form for Nilpotent Operators

Jordan Canonical Form for Nilpotent Operators

If $T$ is nilpotent, then the minimal polynomial has form

$$m_T(x)=x^m\text{ for some }m$$

and there exists basis $\mathcal{B}$ of $V$ such that

$${}_{\mathcal{B}}[T{]}_{\mathcal{B}}=\left(\begin{array}{cccc}0& \ast & & 0\\ & \ddots & \ddots & \\ & & \ddots & \ast \\ 0& & & 0\end{array}\right)\text{ with each }\ast =0\text{ or }1$$

Proof

As $T$ is nilpotent then $T^n=0$ for some $n$ and hence $m_T(x)\mid x^n$
Hence

$$m_T(x)=x^m\text{ for some }m$$

As we have

$$\{0\}⫋\mathrm{Ker}(T)⫋\mathrm{Ker}(T^2)⫋\cdots ⫋\mathrm{Ker}(T^m)=V$$

As $m$ is minimal then the inclusions are strict as if

$$\mathrm{Ker}(T^k)=\mathrm{Ker}(T^{k+1})⟹\mathrm{Ker}(T^k)=\mathrm{Ker}(T^{k+s})\text{ for all }s\ge 0$$

For $1\le i\le m$, let ${\mathcal{B}}_i\subset \mathrm{Ker}(T^i)$ such that

$$\{w+\mathrm{Ker}(T^{i-1})\mid w\in {\mathcal{B}}_i\}\text{ is a basis for }\mathrm{Ker}(T^i)/\mathrm{Ker}(T^{i-1})$$

Hence $|{\mathcal{B}}_i|$ must then be $\dim (\mathrm{Ker}(T^i))-\dim (\mathrm{Ker}(T^{i-1}))$

By Basis Extension through Quotient Spaces and induction then

$$\mathcal{B}=\bigcup _{i=1}^m{\mathcal{B}}_i\text{ is a basis for }V$$

By considering $T{\mid }_{\mathrm{Ker}(T^{m-1})}$ and by induction then

$$\bigcup _{i=1}^{m-1}{\mathcal{B}}_i\text{ is a basis for }\mathrm{Ker}(T^{m-1})$$

As $\overline{{\mathcal{B}}_m}$ is a basis for quotient space $V/\mathrm{Ker}(T^{m-1})$ so using Basis Extension then

$$\bigcup _{i=1}^{m-1}{\mathcal{B}}_i\cup {\mathcal{B}}_m\text{ is a basis for }V$$

For $1\le i\le m-1$ the set

$$\{T(w)+\mathrm{Ker}(T^{i-1})\mid w\in {\mathcal{B}}_{i+1}\}$$

is linearly independent in $\mathrm{Ker}(T^i)/\mathrm{Ker}(T^{i-1})$ (with ${\mathcal{B}}_{i+1}\subset \mathrm{Ker}(T^{i+1})$)
Suppose there exists $a_1,\cdots ,a_t\in \mathbf{F}$ with

$$\begin{array}{rl}\sum _{s=1}^t{a}_s(T(w_s)+\mathrm{Ker}(T^{i-1}))& =0_{\mathrm{Ker}(T^i)/\mathrm{Ker}(T^{i-1})}\\ T\left(\sum _{s=1}^t{a}_s{w}_s\right)& \in \mathrm{Ker}(T^{i-1})\\ \sum _{s=1}^t{a}_s{w}_s& \in \mathrm{Ker}(T^i)\end{array}$$

Hence

$$\sum _{s=1}^t{a}_s(w_s+\mathrm{Ker}(T^i))=0_{\mathrm{Ker}(T^{i+1})/\mathrm{Ker}((T^i))}$$

Which is a contradiction of the choice of ${\mathcal{B}}_{i+1}$ unless all coefficients $a_1=\cdots =a_t=0$

As $|\mathcal{B}{|}_{i+1}$ has size $\dim (\mathrm{Ker}(T^{i+1}))-\dim (\mathrm{Ker}(T^i))$ so we get

$$|{\mathcal{B}}_{i+1}|=\dim (\mathrm{Ker}(T^{i+1}))-\dim (\mathrm{Ker}(T^i))\le \dim (\mathrm{Ker}(T^i))-\dim (\mathrm{Ker}(T^{i-1}))=|{\mathcal{B}}_i|$$

for $1\le i\le m-1$

Constructing the desired basis $\mathcal{B}$ in an inductive manner then

$$\{w+\mathrm{Ker}(T^{m-1})\mid w\in {\mathcal{B}}_m\}\text{ is a basis for }\mathrm{Ker}(T^m)/\mathrm{Ker}(T^{m-1})$$

From the above then the set

$$\{T(w)+\mathrm{Ker}(T^{m-2})\mid w\in {\mathcal{B}}_m\}$$

is linearly independent in

$$\mathrm{Ker}(T^{m-1})/\mathrm{Ker}(T^{m-2})$$

By extending the set to a basis for quotient $\mathrm{Ker}(T^{m-1})/\mathrm{Ker}(T^{m-2})$ then
Extend the set $T({\mathcal{B}}_m)$ working in $\mathrm{Ker}(T^{m-1})$ to

$${\mathcal{B}}_{m-1}:=T({\mathcal{B}}_m)\cup {\mathcal{E}}_{m-1}\subset \mathrm{Ker}(T^{m-1})$$

where the image in $\mathrm{Ker}(T^{i-1})/\mathrm{Ker}(T^{i-2})$ is a basis

So we end up with a basis for $V$

$$\mathcal{B}:=\bigcup _{i=1}^m{\mathcal{B}}_i$$

By defining ${\mathcal{E}}_m={\mathcal{B}}_m$ then this is re-ordered as

$$\bigcup _{i=m}^1\left(\bigcup _{v\in {\mathcal{E}}_i}\{T^{i-1}(v),T^{i-2}(v)\cdots ,T(v),v\}\right)$$

Where respect to the basis in that order we get block diagonal block matrix

$${}_{\mathcal{B}}[T{]}_{\mathcal{B}}=\left(\begin{array}{ccc}{A}_m& & \\ & \ddots & \\ & & A_1\end{array}\right)$$

where each $A_i$ ($m\ge i\ge 1$), itself is a block diagonal matrix consisting of
$|{\mathcal{E}}_i|$ many Jordan Blocks

Link to original

Jordan Canonical Form for Nilpotent Operators corollary

Let $V$ be finite-dimensional and $T:V\to V$ be a linear transformation
Assume $m_{T}x)=(x-\lambda {)}^m$ then

There exists basis $\mathcal{B}$ such that ${}_{\mathcal{B}}[T{]}_{\mathcal{B}}$ is block diagonal with blocks of form

$$J_i(\lambda ):=\lambda I_i+J_i=\left(\begin{array}{cccc}\lambda & 1& & 0\\ & \ddots & \ddots & \\ & & \ddots & 1\\ 0& & & \lambda \end{array}\right)$$

Proof

As $m_T(x)=(x-\lambda {)}^m$ then $T-\lambda I$ is Nilpotent with minimal polynomial $x^m$

Applying Jordan Canonical Form for Nilpotent Operators then
So there exists basis of $\mathcal{B}$ such that ${}_{\mathcal{B}}[T-\lambda I{]}_{\mathcal{B}}$ is block diagonal with blocks $J_i$ hence

$${}_{\mathcal{B}}[T{]}_{\mathcal{B}}=\lambda I+{}_{\mathcal{B}}[T-\lambda I{]}_{\mathcal{B}}$$

is of the desired form

Link to original

16 - Jordan Canonical Form Theorem

Jordan Canonical Form Theorem

Let $V$ be finite dimensional
Let $T:V\to V$ be a linear map with minimal polynomial

$$m_T(x)=(x-{\lambda }_1{)}^{m_1}\cdots (x-{\lambda }_r{)}^{m_r}$$

Then there exists basis $\mathcal{B}$ of $V$ such that ${}_{\mathcal{B}}[T{]}_{\mathcal{B}}$ is block diagonal
where each diagonal block is of the form

$$J_i({\lambda }_j)\text{ for one }1\le i\le m_j\text{ and }1\le j\le r$$

Notes

If $\mathbf{F}$ is an algebraically closed field then the minimal polynomial always will be in the form above

There could be several $J_i({\lambda }_j)$ for each pair $(i,j)$ or none
but there is at least one block for $i=m_j$ for $1\le j\le r$

For each $1\le j\le r$ then the number of Jordan Blocks for $1\le i\le m_j$
determines and is determined by the sequence of dimensions

$$\dim \mathrm{Ker}((T-{\lambda }_{j}I{)}^i)\text{ for }1\le i\le m_j$$

As the sequence of dimensions depends only upon $T$ then

$$\text{Jordan Form is unique up to the ordering of blocks}$$

Proof

By Primary Decomposition Theorem then

$$V=\mathrm{Ker}(T-{\lambda }_{1}I{)}^{m_1}\oplus \cdots \oplus \mathrm{Ker}(T-{\lambda }_{r}I{)}^{m_r}$$

Furthermore, $T$ restricted to the $j-$th summand has minimal polynomial $(x-{\lambda }_j{)}^{m_j}$
Hence apply Jordan Canonical Form for Nilpotent Operators

Link to original