Homework 7

7-HW-AlgebraB.pdf

1. Consider a real number \(a\) and let \(A = \begin{pmatrix} a & 0 & 1 & 0 \\ 0 & -1 & 3a+3 & 0 \\ 0 & 0 & a & 0 \\ 0 & 1 & -3 & -1 \end{pmatrix}\)

   1. Determine the Jordan form \(J\) of \(A\).

      \(\chi_{A}(x)=\det \begin{pmatrix} x-a & 0 & -1 & 0 \\ 0 & x+1 & -3a-3 & 0 \\ 0 & 0 & x-a & 0 \\ 0 & -1 & 3 & x+1 \end{pmatrix}=(x-a)^{2}(x+1)^{2}\)

      1. \(a \neq -1\)
         ​\(E_{a}= \ker \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & a+1 & -3a-3 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & 3 & a+1 \end{pmatrix} = \ker \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix} = \text{span }\{(1,0,0,0)\}\)
         ​\(\dim E_{a}\neq\text{mult}(\chi_{A},a)\)
         Then \(E_{a}^{2}= \ker \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & a+1 & -3a-3 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & 3 & a+1 \end{pmatrix} \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & a+1 & -3a-3 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & 3 & a+1 \end{pmatrix} = \ker \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & (a+1)^{2} & -3(a+1)^2 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & -2(a+1) & 6(a+1) & (a+1)^{2} \end{pmatrix}\)
         ​\(= \ker \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & -3 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} = \text{span }\{(1,0,0,0), (0,3,1,0)\}\)
         ​\(E_{-1}= \ker \begin{pmatrix} -1-a & 0 & -1 & 0 \\ 0 & 0 & -3a-3 & 0 \\ 0 & 0 & -1-a & 0 \\ 0 & -1 & 3 & 0 \end{pmatrix} = \ker \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} = \text{span }\{(0,0,0,1)\}\)
         ​\(\dim E_{-1}\neq\text{mult}(\chi_{A},-1)\)
         Then \(E_{-1}^{2}= \ker \begin{pmatrix} -1-a & 0 & -1 & 0 \\ 0 & 0 & -3a-3 & 0 \\ 0 & 0 & -1-a & 0 \\ 0 & -1 & 3 & 0 \end{pmatrix} \begin{pmatrix} -1-a & 0 & -1 & 0 \\ 0 & 0 & -3a-3 & 0 \\ 0 & 0 & -1-a & 0 \\ 0 & -1 & 3 & 0 \end{pmatrix} = \ker \begin{pmatrix} (a+1)^{2} & 0 & 2(a+1) & 0 \\ 0 & 0 & 3(a+1)^2 & 0 \\ 0 & 0 & (a+1)^{2} & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}\)
         ​\(= \ker \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} = \text{span }\{(0,1,0,0), (0,0,0,1)\}\)

         Thus, \(m_A(x) = (x-a)^2 (x+1)^2\) when \(a \neq -1\)
         ​\(J_A(x) = \begin{pmatrix} a & & & \\ 1 & a & & \\ & & -1 & \\ & & 1 & -1 \end{pmatrix}\)​ 2. \(a = -1\). \(\chi_A(x) = (x+1)^4\)
         ​\(E_{-1}=\ker \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & 3 & 0 \end{pmatrix}=\ker \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}=\text{span }\{e_{1},e_{4}\}\)
         ​\(E_{-1}^{2} = \ker \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & 3 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & 3 & 0 \end{pmatrix} = \ker \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}\)
         ​\(= \text{span }\{e_{1}, e_{2}, e_{3}, e_{4}\}\)
         Thus, \(m_A(x) = (x+1)^2\)
         Then \(J_A = \begin{pmatrix} -1 & & & \\ 1 & -1 & & \\ & & -1 & \\ & & 1 & -1 \end{pmatrix}\) since \(\dim \ker (A+I) = 2 =\) number of Jordan blocks 2. Find an invertible matrix \(P\) such that \(A = PJP^{-1}\).

      2. When \(a\neq -1\)​

         \(E_a = \text{span } \{(1,0,0,0)\}\), \(E_a^2 = \text{span } \{(1,0,0,0), (0,3,1,0)\}\)
         ​\(\{0\}\subsetneq\ker(f-aI_{d})\subsetneq\ker^{2}(f-aI_{d})\)
         \(e_{1}\leftarrow3e_{2}+e_{3}\)
         Thus \(B_a = \{3e_2+e_3, e_1\}\)
         ​\(E_{-1} = \text{span } \{(0,0,0,1)\}\), \(E_{-1}^2 = \text{span } \{(0,1,0,0), (0,0,0,1)\}\)
         ​\(\{0\}\subsetneq\ker(f+I_{d})\subsetneq\ker^{2}(f+I_{d})\)
         \(e_{4} \leftarrow e_{2}\)
         Thus \(B_{-1} = \{e_2, e_4\}\)
         Thus when \(a \neq -1\), \(B = \{3e_2+e_3, e_1, e_2, e_4\}\)
         ​\(J = PAP^{-1}\) where \(P\) is \(P_E^B\)
         Then \(P = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 3 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\) 2. When \(a=-1\), \(E_{-1} = \text{span } \{e_1, e_4\}\), \(E_{-1}^2 = \text{span } \{e_1, e_2, e_3, e_4\}\)
         ​\(A+I_d = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & -3 & 0 \end{pmatrix}\)
         ​\(Ae_2 = e_4\)
         ​\(Ae_3 = e_1 - 3e_4\)
         ​\(\{0\} \subsetneq \ker(A+I_d) \subsetneq \ker^2(A+I_d)\)
         \(e_1 \leftarrow e_3 + 3e_2\)
         \(e_4 \leftarrow e_2\)
         Thus when \(a=-1\), \(B = \{e_3+3e_2, e_1, e_2, e_4\}\)
         ​\(J = PAP^{-1}\) where \(P\) is \(P_E^B\)
         Then \(P = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 3 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\) 3. Find \(J^n\) for all \(n \in \mathbb{N}\).

      Since we have \(J_{A}(x) = \begin{pmatrix} a & & & \\ 1 & a & & \\ & & -1 & \\ & & 1 & -1 \end{pmatrix}\), then \(J^{n} = \begin{pmatrix} a^{n} & & & \\ na^{n-1} & a^{n} & & \\ & & (-1)^{n} & \\ & & n(-1)^{n-1} & (-1)^{n} \end{pmatrix}\)

      In particular, if \(a=-1\), \(J^{n}=\begin{pmatrix} (-1)^{n} & & & \\ n(-1)^{n-1} & (-1)^{n} & & \\ & & (-1)^{n} & \\ & & n(-1)^{n-1} & (-1)^{n} \end{pmatrix}\)​

2. Let \(A \in \mathbb{R}^{5 \times 5}\) with characteristic polynomial \(\chi_A(x) = (x - 1)^2 x^3\). Find all the Jordan forms of \(A\) if \(A \sim A^2\)

   Since we know \(A = P^{-1} J_A P\), then \(A^2 = P^{-1} J_A^2 P\).
   And \(A \sim A^2\), thus if \(A \sim A^2\), then \(J_A \sim J_A^2\).
   Since \(\chi_A(x) = (x-1)^2 x^3\), then \(m_A(x) = (x-1)x\), \((x-1)^2x\), \((x-1)x^2\), \((x-1)x^3\), \((x-1)^2x^2\), \((x-1)^2x^3\).

   ① \(m_A(x) = (x-1)x\)
   ​\(J_A = \left(\begin{array}{c|c} \begin{smallmatrix} 1 & \\ & 1 \end{smallmatrix} & 0 \\ \hline 0 & \begin{smallmatrix} 0 & & \\ & 0 & \\ & & 0 \end{smallmatrix} \end{array}\right)\) \(J_A^2 = \left(\begin{array}{c|c} \begin{smallmatrix} 1 & \\ & 1 \end{smallmatrix} & 0 \\ \hline 0 & \begin{smallmatrix} 0 & & \\ & 0 & \\ & & 0 \end{smallmatrix} \end{array}\right)\) \(\Rightarrow J_A \sim J_A^2 \quad \checkmark\)

   ② \(m_A(x) = (x-1)^2x\)
   ​\(J_{A}=\left( \begin{array}{c|c} \begin{smallmatrix}1 & 0\\ 1 & 1\end{smallmatrix} & 0 \\ \hline 0 & \begin{smallmatrix}0 & & \\ & 0 & \\ & & 0\end{smallmatrix} \end{array}\right)\) \(J_{A}^{2}=\left( \begin{array}{c|c} \begin{smallmatrix}1 & 0\\ 2 & 1\end{smallmatrix} & 0 \\ \hline 0 & \begin{smallmatrix}0 & & \\ & 0 & \\ & & 0\end{smallmatrix} \end{array}\right)\) \(\Rightarrow J_A \sim J_A^2 \quad \checkmark\)
   since Jordan form of \(\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}\), \(\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\) are same, which is \(\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}\)

   ③ \(m_A(x) = (x-1)x^2\)
   ​\(J_{A}=\left( \begin{array}{c|c} \begin{smallmatrix}1 & \\ & 1\end{smallmatrix} & 0 \\ \hline 0 & \begin{smallmatrix}0 & & \\ 1 & 0 & \\ & & 0\end{smallmatrix} \end{array}\right)\) \(J_A^2 = \left(\begin{array}{c|c} \begin{smallmatrix} 1 & \\ & 1 \end{smallmatrix} & 0 \\ \hline 0 & \begin{smallmatrix} 0 & 0 & \\ 0 & 0 & \\ & & 0 \end{smallmatrix} \end{array}\right)\) \(\Rightarrow J_A \not\sim J_A^2\) since rank \(J_A \neq\) rank \(J_A^2\)

   ④ \(m_A(x) = (x-1)x^3\)
   ​\(J_{A}=\left( \begin{array}{c|c} \begin{smallmatrix}1 & \\ & 1\end{smallmatrix} & 0 \\ \hline 0 & \begin{smallmatrix}0 & & \\ 1 & 0 & \\ 0 & 1 & 0\end{smallmatrix} \end{array}\right)\) \(J_{A}^{2}=\left( \begin{array}{c|c} \begin{smallmatrix}1 & \\ & 1\end{smallmatrix} & 0 \\ \hline 0 & \begin{smallmatrix}0 & & \\ 0 & 0 & \\ 1 & 0 & 0\end{smallmatrix} \end{array}\right)\) \(\Rightarrow J_A \not\sim J_A^2\) since rank \(J_A \neq\) rank \(J_A^2\)

   ⑤ \(m_A(x) = (x-1)^2x^2\)
   ​\(J_{A}=\left( \begin{array}{c|c} \begin{smallmatrix}1 & \\ 1 & 1\end{smallmatrix} & 0 \\ \hline 0 & \begin{smallmatrix}0 & & \\ 1 & 0 & \\ & & 0\end{smallmatrix} \end{array}\right)\) \(J_{A}^{2}=\left( \begin{array}{c|c} \begin{smallmatrix}1 & \\ 2 & 1\end{smallmatrix} & 0 \\ \hline 0 & \begin{smallmatrix}0 & & \\ 0 & 0 & \\ & & 0\end{smallmatrix} \end{array}\right)\) \(\Rightarrow J_A \not\sim J_A^2\) since rank \(J_A \neq\) rank \(J_A^2\)

   ⑥ \(m_A(x) = (x-1)^2x^3\)
   ​\(J_{A}=\left( \begin{array}{c|c} \begin{smallmatrix}1 & \\ 1 & 1\end{smallmatrix} & 0 \\ \hline 0 & \begin{smallmatrix}0 & & \\ 1 & 0 & \\ 1 & 1 & 0\end{smallmatrix} \end{array}\right)\) \(J_{A}^{2}=\left( \begin{array}{c|c} \begin{smallmatrix}1 & \\ 2 & 1\end{smallmatrix} & 0 \\ \hline 0 & \begin{smallmatrix}0 & & \\ 0 & 0 & \\ 1 & 0 & 0\end{smallmatrix} \end{array}\right)\) \(\Rightarrow J_A \not\sim J_A^2\) since rank \(J_A \neq\) rank \(J_A^2\)

   Thus \(\left( \begin{array}{c|c} \begin{smallmatrix}1 & \\ & 1\end{smallmatrix} & 0 \\ \hline 0 & \begin{smallmatrix}0 & & \\ & 0 & \\ & & 0\end{smallmatrix} \end{array}\right),\left( \begin{array}{c|c} \begin{smallmatrix}1 & \\ 1 & 1\end{smallmatrix} & 0 \\ \hline 0 & \begin{smallmatrix}0 & & \\ & 0 & \\ & & 0\end{smallmatrix} \end{array}\right)\) are the only possibilities.

3. Let \(A \in \mathbb{C}^{3 \times 3}\) be an invertible matrix such that \(A \sim A^{-1}\) and \(\deg(m_A) = 2\). Find all possible Jordan forms of \(A\).

   Since \(A \cdot v = \lambda v\), then \((A \cdot v)^{-1} = A^{-1} \cdot v^{-1} = \frac{1}{\lambda} v^{-1}\)
   Thus if \(\lambda\) is an eigenvalue of \(A\), then \(\lambda^{-1}\) is an eigenvalue of \(A^{-1}\).
   Since \(A^{-1} \sim A\), then \(A^{-1}=(P^{-1}J_{A}P)^{-1}=P^{-1}J_A^{-1}P^{-1}\).
   Thus \(A^{-1} \sim J_A^{-1}\), then \(A \sim J_A^{-1}\). Then \(J_A \sim J_A^{-1}\).
   ​\(J_{A} (\lambda)^{-1} = P^{-1}J_{A}(s) P\Rightarrow (\lambda I + N)^{-1}= P^{-1} (s I + N') P\)
   ​\(\Rightarrow\lambda^{-1}(I+\lambda^{-1}N)^{-1}=sP^{-1}(I+s^{-1}N)P\Rightarrow\lambda ^{-1}=s\)
   Thus if \(\lambda\) is an eigenvalue of \(A\), \(\lambda^{-1}\) is also an eigenvalue of \(A\).
   Then since \(\deg m_A(x) = 2\), then \(m_A(x) = (x-c_1)(x-c_2)\).
   ① \(c_1 = c_2\). \(m_A(x) = (x-c_1)^2\).
   ​\(c_1 = c_1^{-1} \Rightarrow c_1 = \pm 1\).
   ​\(J_{A} = \begin{pmatrix}\begin{array}{cc|c} 1 & &\\ 1 & 1&\\\hline &&1 \end{array} \end{pmatrix}\) or \(\begin{pmatrix}\begin{array}{cc|c} -1 & &\\ 1 & -1&\\\hline &&-1 \end{array} \end{pmatrix}\).
   ② \(c_1 \neq c_2\). \(m_A(x) = (x-c_1)(x-c_2)\).

   Assume multiplicity of \(c_1\) is \(2\), then the diagonal is \(c_1,c_1,c_2\)​

   And \(\frac{1}{c_{1}},\frac{1}{c_{1}},\frac{1}{c_{2}}\) are also in the diagonal.
   Then \(\frac{1}{c_1} = c_1, \frac{1}{c_2} = c_2\) then \(c_1=\pm 1, c_2=\pm 1\).
   ​\(J_{A}= \begin{pmatrix} -1 & & \\ & -1 & \\ & & 1 \end{pmatrix}\) or \(\begin{pmatrix} 1 & & \\ & 1 & \\ & & -1 \end{pmatrix}\).

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