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4.9 Jordan Basis

Find Jordan basis

Recall General case, then we call \(f\) be \(N\) since it is nilpotent

Here \(N\neq 0,s\geq 2\)

If \(s-2=0\), then we are done. \(V=\ker N^{s-1}\oplus \lang C_{s}\rang\)​

If \(s-2>0\), then let us choose a basis \(B_{s-2}\) of \(\ker N^{s-2}\), we extend \(B_{s-2}\cup N(C_s)\) which is linear independent in \(\ker N^{s-1}\) since Remark of last lecture to a basis of \(\ker N^{s-1}\), that is we choose a subset \(\tilde C_{s-1}\) of \(\ker N^{s-1}\) such that \(B_{s-2}\cup\underbrace{ N(C_{s})\cup \tilde C_{s-1}}_{C_{s-1}}\) this is a basis of \(\ker N^{s-1}\)

\(\tilde C_{s-1}=\{v_{1}^{(s-1)},v_{2}^{(s-1)},...,v_{r_{s-1}}^{(s-1)}\}\). Let \(C_{s-1}=N(C_{s})\cup \tilde C_{s-1}\), now \(V=\ker N^{s}=\ker N^{s-2}\oplus \lang C_{s-1}\rang\oplus \lang C_{s}\rang\)

If \(s-3=0\), we stop here

If \(s-3>0\), we continue the procedure in a similar way we find a set \(C_{s-3}\) such that \(V=\ker N^s=\ker N^{s-3}\oplus \lang C_{s-2}\rang \oplus \lang C_{s-1}\rang \oplus\lang C_{s}\rang\)

Example

\(N:\mathbb{R}^{6}\to\mathbb{R}^{6},\left\lbrack N\right\rbrack_{E}= \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ -1 & -1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & -1 & 0 \end{pmatrix}\). Find \(J_N\) and a Jordan basis

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We know \(\chi_N(x)=x^6\Rightarrow N\) is nilpotent and we know \(\dim \ker N=3\), then we can deduce that there are 3 elementary Jordan blocks, because each zero column separate the Jordan Block, then \(\deg m_{N}(x)\leq3\)image​

Computing we get \(A^2\neq 0\) and \(A^3=0\), so \(m_A(x)=x^3\). Here \(s=3\)

Since \(\{0_{V}\}\subsetneq\ker N\subsetneq\ker N^{2}\subsetneq\ker N^{3}=\mathbb{R}^{6}\) and \(\dim \ker N=3\), \(\dim\ker N^{3}=6\), then \(\dim \ker N^2=4\text{ or }5\)

A basis for \(\ker N\) is \(\{e_3,e_4,e_6\}\) and a basis for \(\ker N^2\) is \(\underbrace{\{e_{3},e_{4},e_{6},e_{2},e_{5}\}}_{B_{s-1}=B_2}\) since \(N(e_2)=-e_3+e_4\) and \(N(e_5)=e_4-e_6\) both are in \(\ker N\)

Thus \(\dim \ker N^2=5\), we extend it to a basis of \(\ker N^3=\R^3\) by adding \(C_s=C_3=\{e_1\}\)

\(N(e_{1})=e_{2}-e_{3}-e_{5}+e_{6}\), then (check) \(N(C_{s})=\{e_{2}-e_{3}-e_{5}+e_{6}\}\) this is in the kernel of \(N^2\) and \(\ker N^{s-1}\cap\langle N(C_{3})\rangle=\ker N\cap\langle N(e_{1})\rangle=\{0_{V} \}\)

For \(\{0_{V}\}\subsetneq\ker N\subsetneq\ker N^{2}\subsetneq\ker N^{3}=\mathbb{R}^{6}\), \(e_1\in \ker N^3\) and \(e_1\notin \ker N^2\) and \(N(e_1)\in \ker N^2,\notin \ker N\)

Then since \(s-2=1>0\), then we choose \(B_{s-2}=B_{1}=\{e_3,e_4,e_6\}\) be basis of \(\ker N^{s-2}=\ker N\)

Then \(B_{1}\cup N(C_{3})\cup \tilde C_2\) is basis of \(\ker N^2\) is equal to \(\left\lbrace e_{3},e_{4},e_{6}\right\rbrace\cup\left\lbrace e_{2}-e_{3}-e_{5}+e_{6} \right\rbrace\cup\left\lbrace\text{choose for example }e_2\text{ or }e_5\right\rbrace\)

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Finally, we extend the linear independent set \(\underbrace{\{N^{2}(e_{1}),N(e_{5})\}}_{N^2(C_3)\cup N(C_2)}\) to a basis of \(\ker N=\ker N^{s-2}\)

Since \(N^2(e_1)=N(e_2-e_3-e_5+e_6)=-e_6-e_3\) and \(N(e_5)=e_4-e_6\), then \(N^{2}(C_{3})\cup N(C_{2})=\{e_6-e_3,e_4-e_6\}\)

Then we can add \(e_3,e_4\) or \(e_6\), we add \(e_3\) hereimage​

Set \(B=\{e_1,N(e_1),N^2(e_1),e_5,N(e_5),e_3\}\)

image​

Lemma

Let \(J\in M_m(k)\) be an elementary nilpotent Jordan block, then \(J= \begin{pmatrix} 0 & \placeholder{} & \placeholder{} & 0 \\ 1 & \ddots & \placeholder{} & \placeholder{} \\ \placeholder{} & \ddots & \ddots & \placeholder{} \\ 0 & \placeholder{} & 1 & 0 \end{pmatrix}\)

Then \(\forall i,1\leq i\leq m\), \(\text{rank}(J^i)=m-i\)

Proof: Exercise by induction on \(i\)

Proposition

Let \(A\in M_n(k)\) be a nilpotent Jordan matrix of index of nilpotency \(s\). We have:

  1. The size of biggest block is \(s\times s\)​

  2. \(\forall i,0\leq i\leq s-1\), \(b_{i}=\#\{\text{elementary nilpotent Jordan blocks of size}>i\times i\}=\text{rank}A^i-\text{rank}A^{i+1}\)​

In particular, for \(i=0\), \(r_{0}=\#\{\text{elementary Jordan blocks in }A\}=\text{rank}A^{0}-\text{rank}A^{1} =n-\text{rank}A=\dim\ker A\)​

Corollary

\(\forall 1\leq i\leq s\), let \(c_{i}=\#\{\text{elementary Jordan blocks of size }i\times i\}=b_{i-1}-b_{i}=\text{rank} A^{i-1}-\text{rank}A^{i}-(\text{rank}A^{i}-\text{rank}A^{i+1})=\text{rank}A^{i-1} -2\text{rank}A^{i}+\text{rank}A^{i+1}\)

Example

  1. Let \(A\in M_{15}(\R)\) nilpotent, is it possible to have \(\text{rank}A=10?\,\text{rank}A^4=3?\,\text{rank}A^5=0?\)

    We know \(\chi_A(x)=x^{15},m_A(x)=x^5\). The biggest block has size \(5\times 5\)

    \(\#\{\text{elementary Jordan blocks in }J_{A}\}=15-10=5\)

    image​

    Let's consider \(c_{5}=\#\{\text{elementary Jordan blocks of size }5\times5\}=3-0+0=3\)

    Thus we have no space to contain the rest one block

  2. \(f:\mathbb{R}^{4}\to\mathbb{R}^{4},A=[f]_{E}= \begin{pmatrix} 1 & 0 & 0 & 2 \\ 2 & -1 & 0 & 2 \\ 2 & 0 & -1 & 2 \\ 0 & 0 & 0 & 1 \end{pmatrix}\), find \(J_A\) and a Jordan basis

    \(\chi_A(x)=(x-1)^2(x+1)^2\) and \(m_{A}(x)=\text{LCM}\{m_{A,e_1}(x),m_{A,e_2}(x),\underbrace{m_{A,e_3}(x)}_{x+1},m_{A,e_4}(x)\}\)​

    \(V_{1}=\{\text{eigenvectors with eigenvalue 1}\}=\langle(1,1,1,0\rangle\) and \(\dim V_1=1\)​

    Thus \(m_A(x)=(x-1)^2(x+1)^i\) and \(i=1\) or \(2\)​

    Also, \(V_{-1}=\lang e_2,e_3\rang\), thenimage​

    Let's find Jordan Basis

    1. Eigenvalue 1

      Consider \(\{0_{V}\}\subsetneq\ker\left(f-Id\right)\subsetneq\ker\left(f-Id\right)^{2}\), we need to find \(v\in \ker(f-Id)^2\setminus \ker(f-Id)\)

      Since \(\ker(f-Id)^{2}=\lang \underbrace{(1,1,1,0)}_{\in \ker(f-Id)},\underbrace{(0,0,0,1)}_{\notin \ker(f-Id)}\rang\), then let \(v_1=(0,0,0,1)\)

      Then \(\underbrace{(f-Id)(v_1)}_{\in\ker(f-Id)}=\underbrace{2e_1+2e_2+2e_3}_{v_2}\), then \(B=\{v_{1},v_{2},?,?\}\) note that \(v_1,v_2\) are \(f-Id\) not \(f\)​ 2. Eigenvalue -1

      Then \(\{0_{V}\}\subsetneq \ker(f+Id)=\lang e_{2},e_{3}\rang\), thus \(B=\{v_{1},v_{2},e_{2},e_{3}\}\)​

      In the end, image​

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