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5.8 Jordan Form

  1. Let \(f:\R^6\to \R^6\) be a linear operator such that

    1. \(\ker (f-3Id)^2\subsetneq \ker (f-3Id)^3\)
    2. \(\dim\ker(f-Id)^2=1\)
    3. \((f-3Id)^4(f-Id)^2=0\)

    Find the Jordan form of \(f\)​

    Remark

    • If \(\dim\ker(f-\lambda d)^{n}\neq 0\) for some \(n\in \N\), then there exists \(v\neq 0\) in \(\ker (f-\lambda Id)\). i.e. \(\lambda\) is an eigenvalue of \(f\)
    • If \(\dim\ker(f-Id)^n\neq 0\), then there exists \(w\in \ker(f-\lambda Id)^n\) such that \((f-\lambda Id)(w)=0\), then \(m_w(x)\mid (x-\lambda)^n\)​

    Conclusion: \((x-\lambda)\) is a factor of \(m_f(x)\) that is, \((x-\lambda)\mid m_{f}(x)\)

    \(\lambda\) is an eigenvalue of \(f\), \(\exists v\in \ker(f-\lambda Id)\setminus \{0\}\)

    Since (a), then \(\dim\ker(f-3Id)^{3}\neq 0\), by remark, we get \(3\) is an eigenvalue, then \((x-3Id)^3\mid m_{v_3}\)​

    Since (b), then \(1\) is an eigenvalue and \(\dim\ker(f-Id)=1\), then \(m_{v_1}=x-1\)​

    Since (c), then \(p(x)=(x-3)^4(x-1)^2\), then \(m_f(x)\mid p(x)\), then \(m_f(x)=(x-3)^3(x-1)\) or \((x-3)^4(x-1)\)​

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  2. Let \(A \in \mathbb{C}^{8 \times 8}\) s.t. \(\chi_A(x) = x^2 (x-1)^3 (x+1)^3\) and \(m_A(x) = x (x-1)^2 (x+1)^2\), and let \(B \in \mathbb{C}^{8 \times 8}\) s.t.

    1. \(\chi_B(x) = \chi_A(x)\)​
    2. \(\text{rank}(B) = 6\)​
    3. \(\text{rank}(I_8 - B) = 6\).

    i) Give all the Jordan forms of B.

    Since \(\chi_B(x) = x^2 (x-1)^3 (x+1)^3\), 0, 1 and -1 are the eigenvalues of B with multiplicities 2, 3 and 3, resp.

    Using b), \(\dim(\text{Ker } B) = 2\), then there are two el. Jordan blocks relative to the eigenvalue 0. \(\Rightarrow \begin{bmatrix} 0&\\ &0 \end{bmatrix}\)​

    Using c), \(6 = \text{rank}(I - B) = \text{rank}(B - I)\), then \(\dim(\text{Ker } (B - I)) = 8 - \text{rank}(B-I) = 8 - 6 = 2\).
    ​\(\therefore\) There are two el. Jordan blocks rel. to the eigenvalue 1.

    \(\text{mult}(1, \chi_B(x)) = 3\), \(3 = 2+1\). \(\Rightarrow \begin{bmatrix} 1 & & \\ 1 & 1&\\ &&1 \end{bmatrix}\).

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    ii) \(A \sim B\)?

  3. \(J= \begin{pmatrix} \lambda & & \\ 1 & \lambda & \\ 0 & 1 & \lambda \end{pmatrix}\in \mathbb{C}^{3\times 3}\). Find \(\cos J\). Ex: \(e^J\)​

    Remark

    \(A=P(J_{A})P^{-1}=P(D+N)P^{-1}\), then \(A^{n}=P(D+N)^{n}P^{-1}=P(\sum^{n}_{i=0}\binom{n}{i}D^{n-i}N^i)P^{-1}\)​

    And we know \(J=D+N= \begin{pmatrix} \lambda & & \\ & \lambda & \\ & & \lambda \end{pmatrix} + \begin{pmatrix} 0 & & \\ 1 & 0 & \\ & 1 & 0 \end{pmatrix}\)​

    If \(n\geq 2\), then \(J^{n}=\left[ \begin{pmatrix}\lambda & & \\ & \lambda & \\ & & \lambda\end{pmatrix}+\begin{pmatrix}0 & & \\ 1 & 0 & \\ & 1 & 0\end{pmatrix} \right]^n=\sum_{i=0}^{n}\binom{n}{i}\left(\lambda^{n-i}\cdot I\cdot\begin{pmatrix} 0&&\\ 1&0&\\ &1&0 \end{pmatrix}^i\right)\)​

    \(=\lambda^{n}I+n\lambda^{n-1} \begin{pmatrix} 0 & & \\ 0 & 0 & \\ 1 & 0 & 0 \end{pmatrix} + \binom{n}{2}\lambda^{n-2} \begin{pmatrix} 0 & & \\ 0 & 0 & \\ 1 & 0 & 0 \end{pmatrix} +0= \begin{pmatrix} \lambda^{n} & & \\ n\lambda^{n-1} & &\\ \binom{n}{2}\lambda^{n-2}&n\lambda^{n-1}&\lambda ^n \end{pmatrix}=J^n\)​

    Since \(\cos x=\sum^{\infty}_{i=0}\frac{(-1)^{i}}{(2i)!}x^{2i}\), then \(\cos J=\sum^{\infty}_{i=0}\frac{(-1)^{i}}{(2i)!}J^{2i}=I+\frac{(-1)^{1}}{2!}J^{2} +\frac{(-1)^{2}}{4!}J^{4}+...+\frac{(-1)^{i}}{(2i)!}J^{2i}+...\)​

    \(= \begin{bmatrix} 1 & & \\ & 1 & \\ & & 1 \end{bmatrix} + \frac{-1}{2!} \begin{bmatrix} \lambda^{2} & & \\ 2\lambda & \lambda^{2} & \\ 1 & 2\lambda & \lambda^{2} \end{bmatrix} + \frac{(-1)^{2}}{4!} \begin{bmatrix} \lambda^{4} & & \\ 4\lambda^{3} & \lambda^{4} & \\ \binom{4}{2}\lambda^{2} & 4\lambda^{3} & \lambda^{4} \end{bmatrix} + \dots + \frac{(-1)^{i}}{(2i)!} \begin{bmatrix} \lambda^{2i} & & \\ (2i)\lambda^{2i-1} & \lambda^{2i} & \\ \binom{2i}{2}\lambda^{2i-2} & (2i)\lambda^{2i-1} & \lambda^{2i} \end{bmatrix} + \dots\)

    \(= \begin{bmatrix} 1 + (\frac{-1}{2!})\lambda^{2}+ (\frac{-1}{4!})^{2}\lambda^{4}+ \dots + \frac{(-1)^i}{(2i)!}\lambda^{2i}+ \dots =a_{11}| & \,\,\,\,\,\,\,\,| & | \\ \frac{-1}{2!}2\lambda + \frac{(-1)^2}{4!}4\lambda^{3}+ \dots + \frac{(-1)^i}{(2i)!}(2i) \lambda^{2i-1}+ \dots=a_{12}| & a_{11}| & | \\ \frac{-1}{2!}1 + \frac{(-1)^2}{4!}\binom{4}{2}\lambda^{2}+ \dots + \frac{(-1)^i}{(2i)!}\binom{2i}{2}\lambda^{2i-2}+ \dots\quad \quad\,| & a_{12}| & a_{11}| \end{bmatrix}\)

    \(= \begin{bmatrix} \cos\lambda & 0 & 0 \\ -\sin\lambda & \cos\lambda & 0 \\ -\frac{1}{2}\cos\lambda & -\sin\lambda & \cos\lambda \end{bmatrix}\)

  4. Ex 4: Let \(n \in \mathbb{N}\) and \(A \in \mathbb{C}^{n \times n}\) with minimal pol. \(m_A(x) = (x+1)^r x\) for some \(0 \leq r \leq n-1\). Prove that \(A^2\) and \(-A\) are similar.

    Answer:

    • if \(r=0\), then \(m_A(x) = x\). \(\Rightarrow A\) is nilpotent of index 1
      so \(A=0\) but \(-A=0\) and \(A^2=0 \Rightarrow -A \sim A^2\).
    • if \(r>0\), then \(m_A(x) = (x+1)^r x\) and \(\chi_A(x) = (x+1)^a x^b\) with \(a+b=n\). and \(r \leq a\).
      The eigenvalues of A are -1 and 0 then

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