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Homework 11

  1. (10 pts) Let \(A \in M_n(\mathbb{F})\). Prove that \(A\) and \(A^T\) have the same eigenvalues. Give an example where the eigenvectors are different.

    Assume \(A, A^T \in M_n(\mathbb{F})\). Then \(\chi_{A}(x) = det(xI{d} - A)\)

    Then \(\chi_{A^{T}}(x)=\det(xId-A^{T})=\det(xId^{T}-A^{T})=\det(xId-A)=\chi_{A}(x)\).

    Thus \(A\) and \(A^T\) have same eigenvalues.
    Let \(A = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}\), \(A^T = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\) and \(\chi_A(x) = (x-1)^2 = \chi_{A^T}(x)\), then \(x=1\).
    \(E_1\) of \(A\) is \(\text{ker}(A - Id) = \text{ker} \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} = \langle (0,1) \rangle\)
    \(E_1\) of \(A^T\) is \(\text{ker}(A^T - Id) = \text{ker} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \langle (1,0) \rangle\)
    Thus they have different eigenvectors.

  2. (20 pts) Suppose \(S, T: V \to V\) are linear maps such that \(S\) is an isomorphism.
    (a) Prove that \(T\) and \(S^{-1}TS\) have the same eigenvalues.
    (b) What is the relationship between the eigenvectors of \(T\) and the eigenvectors of \(S^{-1}TS\)?

    (a)
    Assume \(T\) has an eigenvalue \(\lambda\). Then \(T(v) = \lambda v\).
    Since \(S\) is an isomorphism, then \(\ker S = \{0\}\), \(\text{Im } S = V\). Then \(\exists v' : S(v') = v\). \(S^{-1}(v) = v'\).
    Then \(S^{-1}TS(v') = S^{-1}T(v) = S^{-1}\lambda v = \lambda S^{-1}(v) = \lambda v'\).
    Thus they have same eigenvalues.

    (b)

    For \(T\):
    \(E_\lambda = \ker(T - \lambda Id) = \ker([T - \lambda Id]_E)\)
    For \(S^{-1}TS\):
    \(E_\lambda = \ker(S^{-1}TS - \lambda Id) = \ker(S^{-1}(T - \lambda Id)S)\) \(= \ker([T - \lambda Id]_{B})\) where \(S\) is change basis.
    Thus \([T - \lambda Id]_E \cdot [v]_E = 0\) iff \(S^{-1}[T - \lambda Id]_E \cdot [v]_E \cdot S = 0\) iff \([T - \lambda Id]_B \cdot [v]_B = 0\).
    Thus \([v]_E\) is eigenvector of \(T \iff [v]_B\) is eigenvector of \(S^{-1}TS\) where \(S\) is change of basis

  3. (50 pts) Decide whether or not the following linear maps \(T\) are diagonalizable and if so, find an invertible matrix \(P\) such that \(P^{-1}[T]_C P\) is a diagonal matrix. Consider them first as linear maps in \(\mathbb{R}\) and then in \(\mathbb{C}\).

    (a) \(T(x, y) = (2x + 3y, -x + y)\).

    In \(\R\)

    First, we need \([T]_{C}=\begin{pmatrix}2&3\\-1&1\end{pmatrix}.\)

    Then we calculate eigenvalues, let \(\det(T-\lambda I)=\det\begin{pmatrix}2-\lambda&3\\-1&1-\lambda\end{pmatrix}=(2-\lambda)(1-\lambda)+3=\lambda^2-3\lambda+5=0\;\Rightarrow\;\lambda=\frac{3\pm i\sqrt{11}}2.\)

    Thus there is no eigenvalue in \(\R\), thus \(T\) is not diagonalizable in \(\R\)

    In \(\mathbb{C}\)

    First, we need \([T]_{C}=\begin{pmatrix}2&3\\-1&1\end{pmatrix}.\)

    Then we calculate eigenvalues, let \(\det(T-\lambda I)=\det\begin{pmatrix}2-\lambda&3\\-1&1-\lambda\end{pmatrix}=(2-\lambda)(1-\lambda)+3=\lambda^2-3\lambda+5=0\;\Rightarrow\;\lambda=\frac{3\pm i\sqrt{11}}2.\)

    Thus we find the basis of eigenvectors

    For \(\displaystyle\lambda=\frac{3+i\sqrt{11}}2\): \(\text{Null}(T-\lambda I)=\text{Null}\begin{pmatrix}\tfrac{1-i\sqrt{11}}2&3\\-1&\tfrac{-1-i\sqrt{11}}2\end{pmatrix}=\bigl\langle\bigl(-\tfrac{1+i\sqrt{11}}2,\,1\bigr)\bigr\rangle.\)

    For \(\displaystyle\lambda=\frac{3-i\sqrt{11}}2\): \(\text{Null}(T-\lambda I)=\text{Null}\begin{pmatrix}\tfrac{1+i\sqrt{11}}2&3\\-1&\tfrac{-1+i\sqrt{11}}2\end{pmatrix}=\bigl\langle\bigl(-\tfrac{1-i\sqrt{11}}2,\,1\bigr)\bigr\rangle.\)

    Now, \(\mathcal{B}=\{(-\tfrac{1+i\sqrt{11}}2,1),\,(-\tfrac{1-i\sqrt{11}}2,1)\}\) is my basis of eigenvectors of \(T\).

    Thus \([T]_{\mathcal{B}}=\begin{pmatrix}\tfrac{3+i\sqrt{11}}2&0\\0&\tfrac{3-i\sqrt{11}}2\end{pmatrix}.\)

    Then \(P= \begin{pmatrix} -\tfrac{1+i\sqrt{11}}{2} & -\tfrac{1-i\sqrt{11}}{2} \\ 1 & 1 \end{pmatrix},\quad P^{-1}[T]_{C}P=[T]_{\mathcal{B}}.\)

    (b) \(T(x, y, z) = (-x - 2y + 2z, 4x + 3y - 4z, -2y + z)\).

    In \(\R\)

    First, we need \([T]_{C}=\begin{pmatrix}-1 & -2 & 2\\ 4 & 3 & -4\\ 0 & -2 & 1\end{pmatrix}\)

    Then we calculate eigenvalues, let \(\det(T-\lambda I)=\det\begin{pmatrix}-1-\lambda & -2 & 2\\ 4 & 3-\lambda & -4\\ 0 & -2 & 1-\lambda\end{pmatrix}=0\)

    \(=(-1-\lambda)(3-\lambda)(1-\lambda)+4\cdot (-2)\cdot 2-(-4)\cdot(-2)(-1-\lambda)-(1-\lambda)\cdot(-2)\cdot4\)

    \(=-\lambda^3+3\lambda^2+\lambda-3=0\Rightarrow\lambda=1,-1,3\)

    Thus we find the basis of eigenvectors

    For \(\lambda = -1\): \(\text{Null}\left(T+I\right)=\text{Null}\begin{pmatrix}0 & -2 & 2\\ 4 & 4 & -4\\ 0 & -2 & 2\end{pmatrix}=\text{Null}\begin{pmatrix}0 & 1 & -1\\ 1 & 0 & 0\\ 0 & 0 & 0\end{pmatrix}=\langle(0,1,1)\rangle\).

    For \(\lambda = 1\): \(\text{Null}(T-I)=\text{Null}\begin{pmatrix}-2 & -2 & 2\\ 4 & 2 & -4\\ 0 & -2 & 0\end{pmatrix}=\text{Null}\begin{pmatrix}1 & 0 & -1\\ 0 & 1 & 0\\ 0 & 0 & 0\end{pmatrix}=\langle(1,0,1)\rangle\).

    For \(\lambda = 3\): \(\text{Null}(T-3I)=\text{Null}\begin{pmatrix}-4 & -2 & 2\\ 4 & 0 & -4\\ 0 & -2 & -2\end{pmatrix}=\text{Null}\begin{pmatrix}0 & 1 & 1\\ 1 & 0 & -1\\ 0 & 0 & 0\end{pmatrix}=\langle(1,-1,1)\rangle\).

    Now, \(\mathcal{B}=\{(0,1,1),\left(1,0,1),(1,-1,1\right)\}\) is my basis of eigenvectors of \(T\).

    Thus \([T]_{\mathcal{B}}=\begin{pmatrix}-1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 3\end{pmatrix}\)

    Then \(P= \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & -1 \\ 1 & 1 & 1 \end{pmatrix}\), then \(P^{-1}[T]_CP=[T]_B\)

    Thus \(T\) is diagonalizable in \(\R\), then \(T\) is also diagonalizable in \(\mathbb{C}\)

    (c) \(T(x, y, z) = (6x - 3y - 2z, 4x - y - 2z, 10x - 5y - 3z)\).

    In \(\R\)

    First, we need \([T]_{C}=\begin{pmatrix}6&-3&-2\\4&-1&-2\\10&-5&-3\end{pmatrix}.\)

    Then we calculate eigenvalues, let \(\det(T-\lambda I)=\det\begin{pmatrix}6-\lambda&-3&-2\\4&-1-\lambda&-2\\10&-5&-3-\lambda\end{pmatrix}=0\;\Rightarrow\;- \lambda^3+2\lambda^2-\lambda+2=0\;\Rightarrow\;\lambda^3-2\lambda^2+\lambda-2=(\lambda-2)(\lambda^2+1)=0\) \(\Rightarrow\;\lambda=2,\;i,\;-i.\)

    Thus we find the basis of eigenvectors

    For \(\lambda=2\): \(\text{Null}(T-2I)=\text{Null}\begin{pmatrix}4&-3&-2\\4&-3&-2\\10&-5&-5\end{pmatrix}=\bigl\langle(1,0,2)\bigr\rangle.\)

    Then eigenspace for \(\lambda =2\) is less than \(3\), thus \(T\) is not diagonalizable in \(\R\)

    But if in \(\mathbb{C}\)

    For \(\lambda=i\): \(\text{Null}(T-iI)=\text{Null} \begin{pmatrix} 6-i & -3 & -2 \\ 4 & -1-i & -2 \\ 10 & -5 & -3-i \end{pmatrix}=\lang(3+i,3+i,5){\rangle}.\)

    For \(\lambda=-i\): \(\text{Null}(T+iI)=\text{Null} \begin{pmatrix} 6+i & -3 & -2 \\ 4 & -1+i & -2 \\ 10 & -5 & -3+i \end{pmatrix}=\lang(3-i,3-i,5){\rangle}.\)

    Now, \(\mathcal{B}=\{(1,0,2),\;(3+i,3+i,5),\;(3-i,3-i,5)\}\) is my basis of eigenvectors of \(T\).

    Thus \([T]_{\mathcal{B}}= \begin{pmatrix} 2 & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & -i \end{pmatrix}.\)

    Then \(P_{}= \begin{pmatrix} 1 & 3+i & 3-i \\ 0 & 3+i & 3-i \\ 2 & 5 & 5 \end{pmatrix},\quad P^{-1}[T]_{C}P=[T]_{\mathcal{B}}.\)

    Thus \(T\) is diagonalizable in \(\mathbb{C}\) but not in \(\R\)

    (d) \(T(x, y, z, w) = (-5x - 5y - 9z + 7w, 8x + 9y + 18z - 9w, -2x - 3y - 7z + 4w, 2w)\).

    In \(\R\)

    First, we need \([T]_{C}=\begin{pmatrix}-5&-5&-9&7\\8&9&18&-9\\-2&-3&-7&4\\0&0&0&2\end{pmatrix}.\)

    Then we calculate eigenvalues, let \(\det(T-\lambda I)=(2-\lambda)\det\begin{pmatrix}-5-\lambda&-5&-9\\8&9-\lambda&18\\-2&-3&-7-\lambda\end{pmatrix}=-(\lambda-2)(\lambda+1)^3=0\;\Rightarrow\;\lambda=2,\,-1,-1,-1.\)

    For \(\lambda=2\): \(\text{Null}(T-2I)=\text{Null} \begin{pmatrix} -7 & -5 & -9 & 7 \\ 8 & 7 & 18 & -9 \\ -2 & -3 & -9 & 4 \\ 0 & 0 & 0 & 0 \end{pmatrix}=\bigl\langle(-9,63,-7,27)\bigr\rangle.\)

    For \(\lambda=-1\): \(\text{Null}(T+I)=\text{Null}\begin{pmatrix}-4&-5&-9&7\\8&10&18&-9\\-2&-3&-6&4\\0&0&0&3\end{pmatrix}=\bigl\langle(\tfrac32,-3,1,0)\bigr\rangle.\)

    Since the eigenspace for \(\lambda=-1\) has dimension \(1 < 3\), \(T\) is not diagonalizable.

    Thus \(T\) is not diagonalizable in \(\R\) and in \(\mathbb{C}\)

  4. (20pts) Suppose \(T \in \mathcal{L}(\mathbb{C}^3, \mathbb{C}^3)\) is such that \(6\) and \(7\) are eigenvalues of \(T\). Furthermore, suppose \([T]_{\mathcal{B}}\) is not a diagonal matrix with respect to any basis \(\mathcal{B}\) of \(\mathbb{C}^3\). Prove that there exists \((x, y, z) \in \mathbb{C}^3\) such that \(T(x, y, z) = (17 + 8x, \sqrt{5} + 8y, 2\pi + 8z)\).

    Proof

    We need to prove the existence of \((x, y, z) \in \mathbb{C}^3\) such that \(T(x, y, z) = (17 + 8x, \sqrt{5} + 8y, 2\pi + 8z)\).

    Then we need to prove the existence of \((x, y, z) \in \mathbb{C}^3\) such that \(T(x,y,z)-8\left(x,y,z)=(17,\sqrt5,2\pi\right)\).

    Then we need to prove the existence of \(v=(x, y, z) \in \mathbb{C}^3\) such that \(\left(T-8I\right)\left(v\right)\left.=(17,\sqrt5,2\pi\right)\).

    Since \(6,7\) are eigenvalues and we know \(T\) is not diagonalizable, then \(8\) is not eigenvalue.

    Then \(\dim \text{Null}(T-8I)=0\), then \(T-8I\) is injective.

    Since \(T-8I\) is an endomorphism, then \(T-8I\) is surjective.

    Then there exists such linear map.