4.10 Jordan Form

Let \(V\) be a finite dimensional vector space over \(\mathbb{K}\) and let \(f:V\to V\) be a linear operator

• \(f\) is nilpotent of index \(r\) iff \(m_f(x)=x^r\)​
• If \(f\) is nilpotent of index \(r\), then \(\{0\}\subsetneq \ker f\subsetneq \ker f^{2}\subsetneq ...\subsetneq \ker f^{r}=V\) since \(f^r=0,\ker\{0\}=V\)​
• If \(r=n=\dim V\), then there exists an ordered basis \(\mathcal{B}\) for \(V\) such that \([f]_\mathcal{B}=\)​

Definition

\(J\in \mathbb{K}^{n\times n}\), \(J\) is called an elementary nilpotent Jordan Matrix or a nilpotent Jordan Block if \(J=\)​

Theorem

If \(f\) is nilpotent of index \(r\), then there exists an ordered basis \(\mathcal{B}\) for \(V\) such that \([f]_{\mathcal{B}}=\) (Nilpotent Jordan form) where \(J_{i}\in \mathbb{K}^{{}n_{i}\times n_{i}}\) is a nilpotent Jordan Block for each \(i=1,...,l\) and \(r=n_1\geq n_2\geq ...\geq n_l\) and \(n=\sum^l_{i=1}n_i\)

Exercise

1. Find the nilpotent Jordan Form and a Jordan basis for \(A\in \R^{5\times 5}\) where \(A= \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 1 \\ -1 & -1 & 0 & -1 & 0 \\ -1 & 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 & 0 \end{pmatrix}\)

   1. \(\chi_{A}(x)=\det \begin{pmatrix} x & 0 & 0 & 0 & 0 \\ 1 & x & 0 & 0 & -1 \\ 1 & 1 & x & 1 & 0 \\ 1 & 0 & 0 & x & -1 \\ 1 & 0 & 0 & 0 & x \end{pmatrix}=x^{5}\), thus \(A\) is nilpotent

   2. \(m_A(x)\) using standard basis

      \(e_{1}\mapsto-e_{2}-e_{3}-e_{4}-e_{5}\mapsto-e_{2}+2e_{3}-e_{4}\mapsto-2e_{3}\mapsto 0\)

      \(e_{2}\mapsto-e_{3}\mapsto0\)

      \(e_{3}\mapsto0\)

      \(e_{4}\mapsto-e_{3}\mapsto0\)

      \(e_{5}\mapsto e_{2}+e_{4}\mapsto-2e_{3}\mapsto0\)

      Then \(m_{e_1}(x)=x^{4},m_{e_2}(x)=x^{2},m_{e_3}(x)=x,m_{e_4}(x)=x^{2},m_{e_5}(x)=x^{3}\)

      Thus \(m_A(x)=x^4\) since the least common multiple

      Thus the index of \(f\) is \(4\), then \(J_A=\)​

      Then we need to find Jordan basis 3. \(\{0\}\subsetneq\ker A\subsetneq \ker A^2\subsetneq \ker A^3\subsetneq \ker A^4=\R^5\) (choose a element here not inside)

      Let \(v\in \ker A^4\setminus \ker A^3\), we know \(v=e_1\) above

      Thus ​

      Thus we need another one to form 5 dimension basis, we only can choose \(v'\in \ker f\) and we know \(f(e_3)=0\) is the only possibility, but \(e_3\) is already in the basis.

      Thus we have to choose \(e_2-e_4\) since \(f(e_2-e_4)=0\)​

      Thus the Jordan basis for \(\R^5\) is \(\mathcal{B}=\{e_1,Ae_1,A^2e_1,A^3e_1,e_2-e_4\}\)

      (If dimension of \(\ker A^{3}\) is 3, then we will have at least \(3\times 2\) vectors in basis)

      The cyclic decomposition

      1. \(W_0=\{0\}\): \(\R^{5}=\lang e_{1}\rang_{A}\oplus \lang e_{2}-e_{4}\rang_{A}\)
      2. \(W_{0}=\langle e_{2}-e_{4}\rangle_{A}\): \(\mathbb{R}^{5}=\langle e_{2}-e_{4}\rangle_{A}\oplus\langle e_{1}\rangle_{A}\)

2. Let \(f:\mathbb{K}^8\to\mathbb{K}^8\) such that. Find a Jordan basis for \(f\)

   \(\chi_f(x)=x^8\)

   Choose standard basis

   \(e_{1}\mapsto e_{6}\mapsto e_{4}\mapsto0\)

   \(e_{2}\mapsto e_{6}\mapsto e_{4}\mapsto0\)

   \(e_{3}\mapsto e_{5}\mapsto0\)

   \(e_4\mapsto 0\)

   \(e_{5}\mapsto 0\)

   \(e_{6}\mapsto e_{4}\mapsto0\)

   \(e_7\mapsto 0\)

   \(e_{8}\mapsto e_{7}\mapsto0\)

   Then \(m_f(x)=x^3\) since LCM of minimal polynomial of vector

   Since the \(\text{rank}\) is \(4\), then \(\ker f=4\), thus \(\ker f=\{e_{4},e_{5},e_{7},e_{1}-e_{2}\}\)​

   And we have \(\{0\}\subsetneq\ker A\subsetneq\ker A^{2}\subsetneq\ker A^{3}=\mathbb{K}^{8}\)

   Then we possibilities are 4 2 2 or 4 3 1

   But \(\ker f^{2}=\{e_{4},e_{5},e_{7},e_{1}-e_{2},e_{6},e_{8},e_{3}\}\), thus the possibility is 4 2 2

   Thus \(\ker f^{3}=\{e_{4},e_{5},e_{7},e_{1}-e_{2},e_{6},e_{8},e_{3},e_{1}\}\)​

   ​

   

   Note the basis is ordered basis

3. Lemma

   Let \(J\in \mathbb{K}^{m\times m}\) be a nilpotent Jordan block, then \(\text{rank}J^i=m-i\) where \(i=1,...,m\)​

   By induction

   If \(i=1\), \(\text{rank}J=m-1\)​

   Suppose that \(\text{rank}J^i=m-i\), then the nilpotent of \(J^i\) by \(J\) shifts \(i\)s further, reducing the number of non-zero rows by \(1\)​

   Then \(\text{rank}(J^{i+1})=\text{rank}(J^i)-1=m-i-1=m-(i+1)\)​

   Proposition

   Let \(A\in \mathbb{K}^{n\times n}\) be a nilpotent Jordan form of index \(r\), then

   1. The largest Jordan block appearing in \(A\) has size \(r\times r\)​
   2. For each \(0\leq i\leq r-1\) the number of nilpotent Jordan blocks of size greater than \(i\) in \(A\) is given by \(b_i=\text{rank}A^i-\text{rank}A^{i+1}\)​
   3. Then total number of Jordan blocks in \(A\) is \(b_0=n-\text{rank}=\dim\ker A\)​

   Proof

   Since \(A\) has index \(r\), then \(m_A(x)=x^r\), then let \(J_1,...,J_l\) be Jordan blocks in \(A\) with \(J_{i}\in \mathbb{K}^{n_{i}\times n_{i}}\) for each \(1\leq j\leq l\)​

   Then \(m_{A}(x)=\text{LCM}\{m_{J_1}(x),...,m_{J_{l}}(x)\}=\text{LCM}\{x^{n_1},...,x^{n_l} \}=x^{n_1}\) by theorem of the size of block is decreasing

   Thus \(n_1=r\)

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   For each \(1\leq i\leq v\), let \(c_{i}=\# N.J.B.\) of size \(i\) in \(A\), let \(b_i=\#N.J.B.\) of size \(>i\) in \(A\)​

   Now if \(A\) has \(l\) nilpotent Jordan Blocks, then \(\text{rank}A=\text{rank}J_1+\text{rank}J_2+\text{rank}J_l\)​

   \(=n_1-1+n_2-1+...+n_l-1=n-l\)​

   Then \(l=n-\text{rank}A=\dim\ker A\)​

   By definition, \(b_0=l=\dim\ker A\)​

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   Let \(1\leq i\leq r-1\). By the lemma, if \(J\) is a N.J.B in \(\mathbb{K}^{i\times i}\), then \(J^i=0\) if \(j\leq i\) and \(\text{rank}J^i=j-i\) if \(j>i\) by lemma

   And \(\text{rank}(A^{i})=\sum_{s=1}^{l}\text{rank}(J_{s}^{i})\), then \(\text{rank}(A^{i})-\text{rank}(A^{i+1})=\sum_{s=1}^{l}\text{rank}(J_{s}^{i})-\sum _{s=1}^{l}\text{rank}(J_{s}^{i+1})=\sum_{j=i+1}^{r}c_{j}(j-i)-\sum_{j=i+2}^{r}c_{j} \left(j-\left(i+1\right)\right)=\sum_{j=i+1}^{r}c_{j}=b_{i}\)

   This is because the rank of \(J_{s}^{i}\) is equal to the number of size \(j\) times the number of column that are not zero, that is \((j-i)\) since we have \(i\) power, then the non-zero column will decrease \(i\)

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   Example

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