12

Homework 12.pdf

1. (30pts) Let \(T:\mathbb{R}^4\to\mathbb{R}^4\) be given by \(T(x,y,z,w)=(4x+7z+w,7x+2y+11z+3w,x+2z+w,-4x-8z-w)\)​
   Find a basis \(\mathcal{B}\) of \(\mathbb{R}^4\) such that \([T]_{\mathcal{B}}\) is upper triangular.

   Step 1: Find the eigenvalues of \(T\).

   So, consider \([T]_{C}=\begin{pmatrix}4 & 0 & 7 & 1\\ 7 & 2 & 11 & 3\\ 1 & 0 & 2 & 1\\ -4 & 0 & -8 & -1\end{pmatrix}\), where \(C\) = canonical basis.

   Then consider \(\text{Null}(T-\lambda I)\;\;\iff\;\;\det\left(T-\lambda I\right)=0\)

   Then \(\det\left(T-\lambda I\right)=\det\begin{pmatrix}4-\lambda & 0 & 7 & 1\\ 7 & 2-\lambda & 11 & 3\\ 1 & 0 & 2-\lambda & 1\\ -4 & 0 & -8 & -1-\lambda\end{pmatrix}=\left(-1\right)^{2+2}\cdot\left(2-\lambda\right)\cdot\det\begin{pmatrix}4-\lambda & 7 & 1\\ 1 & 2-\lambda & 1\\ -4 & -8 & -1-\lambda\end{pmatrix}\)​

   \(=\left(2-\lambda\right)\cdot\left\lbrack\left(4-\lambda\right)\left(2-\lambda\right)\left(-1-\lambda\right)+\left(-8\right)+\left(-4\right)\cdot7-\left(2-\lambda\right)\cdot\left(-4\right)-\left(-8\right)\cdot\left(4-\lambda\right)-\left(-1-\lambda\right)\cdot7\right\rbrack\)

   \(=\left(2-\lambda\right)\cdot\left(-1\lambda^3+5\lambda^2-7\lambda+3\right)=0\Rightarrow\lambda=1,2,3\)​

   Step 2: Choose an eigenvalue \(\lambda\) and find \(U=\text{Range}(T-\lambda I)\) (invariant subspace).

   We choose \(\lambda=1\), and we have that \(T-I=\begin{pmatrix}3 & 0 & 7 & 1\\ 7 & 1 & 11 & 3\\ 1 & 0 & 1 & 1\\ -4 & 0 & -8 & -2\end{pmatrix}\).

   Then we reduce it, we get \(\begin{pmatrix}0 & 0 & 4 & -2\\ 0 & 1 & 4 & -4\\ 1 & 0 & 1 & 1\\ 0 & 0 & 0 & 0\end{pmatrix}\). Then we can choose the first three columns as range

   Thus \(\text{Range}(T-I)=\langle(3,7,1,-4),(0,1,0,0),\left(7,11,1,-8\right)\rangle=U\)​

   Basis \(B_{U}=\{(3,7,1,-4),(0,1,0,0),\left(7,11,1,-8\right)\}=\{u_1,u_2,u_3\}\).

   Step 3: Find \(T|_U : U \to U\) (the representation of \(T\) restricted to \(U\) in the basis \(B_U\)).

   Note that: \(T|_{U}(u_1)=(15,34,1,-16)=-2u_1+15u_2+3u_3\), \(T|_{U}(u_2)=\left(0,2,0,0\right)=2u_2\). \(T|_{U}(u_3)=\left(27,58,1,-28\right)=-5u_1+27u_2+6u_3\).

   Thus \(\left\lbrack T|_{U}\right\rbrack_{B_{U}}=\begin{pmatrix}-2 & 0 & -5\\ 15 & 2 & 27\\ 3 & 0 & 6\end{pmatrix}\).

   Step 4: Triangulate \(\big[T|_U\big]_{B_U}\) (so we need to find an eigenvalue of \(T|_U\)).

   \(\det(T|_{U}-\mu I)=\det\begin{pmatrix}-2-\mu & 0 & -5\\ 15 & 2-\mu & 27\\ 3 & 0 & 6-\mu\end{pmatrix}=\left(\mu+2\right)\left(\mu-2\right)\left(6-\mu\right)+5\cdot\left(2-\mu\right)\cdot3=0\) Then \(\mu =1,2,3\)​

   Thus we can diagonalize this matrix.

   For \(\mu=1\): \(\text{Null}(T|_{U}-I)=\text{Null}\begin{pmatrix}-3 & 0 & -5\\ 15 & 1 & 27\\ 3 & 0 & 5\end{pmatrix}=\text{Null}\begin{pmatrix}1 & 0 & \frac53\\ 0 & 1 & 2\\ 0 & 0 & 0\end{pmatrix}=\langle\left(-5,-6,3\right)\rangle\).

   For \(\mu=2\): \(\text{Null}(T|_{U}-I)=\text{Null}\begin{pmatrix}-4 & 0 & -5\\ 15 & 0 & 27\\ 3 & 0 & 4\end{pmatrix}=\text{Null}\begin{pmatrix}1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\end{pmatrix}=\langle\left(0,1,0\right)\rangle\).

   For \(\mu=3\): \(\text{Null}(T|_{U}-I)=\text{Null}\begin{pmatrix}-5 & 0 & -5\\ 15 & -1 & 27\\ 3 & 0 & 3\end{pmatrix}=\text{Null}\begin{pmatrix}1 & 0 & 1\\ 0 & 1 & -12\\ 0 & 0 & 0\end{pmatrix}=\langle\left(-1,12,1\right)\rangle\).

   Now, \(\{-5u_1-6u_2+3u_3,u_2,-u_1+12u_2+u_3\}\) is my basis of eigenvectors of \(T|_U\).

   Then \(\left\lbrack T|_{U}]_{B_{U}^{\prime}}=\right.\begin{pmatrix}1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 3\end{pmatrix}\)

   Step 4: Extend \(B_U'\) to a basis of \(\mathbb{R}^4\): We take \(B=\{-5u_1-6u_2+3u_3,u_2,-u_1+12u_2+u_3,\left(1,0,0,0\right)\}\)​

   \(T(1,0,0,0)=(4,7,1,-4)=\frac{3}{10}(20,47,0,-20)+\frac{27}{5}(0,1,0,0)-\frac12(6,25,-2,-4)+\left(1,0,0,0\right)\).

   And now \([T]_{B}=\begin{pmatrix}1 & 0 & 0 & -\frac12\\ 0 & 2 & 0 & -21\\ 0 & 0 & 3 & \frac32\\ 0 & 0 & 0 & 1\end{pmatrix}\) where \(B=\{(6,-8,-2,-4),(0,1,0,0),(4,16,0,-4),(1,0,0,0)\}\)​

2. (30pts) Decide whether or not the following linear maps \(T \in \mathcal{L}(\mathbb{R}^n, \mathbb{R}^n)\) are diagonalizable and if so, find a matrix \(P\) such that \(P^{-1}[T]_{\mathcal{C}}P\) is a diagonal matrix.

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

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

   ---

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

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

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

   \(=\left(-1\right)^2\cdot\left(2-\lambda\right)\cdot\det\begin{pmatrix}4-\lambda & 0 & 0\\ 0 & 1-\lambda & 3\\ 0 & 1 & -1-\lambda\end{pmatrix}+\left(-1\right)^3\cdot\left(1\right)\cdot\det\begin{pmatrix}-1 & 0 & 0\\ 0 & 1-\lambda & 3\\ 0 & 1 & -1-\lambda\end{pmatrix}\)​

   \(=\left(2-\lambda\right)\cdot\left(\left(4-\lambda\right)\left(1-\lambda\right)\left(-1-\lambda\right)-3\left(4-\lambda\right)\right)-\left\lbrack\left(-1\right)\cdot\left(1-\lambda\right)\left(-1-\lambda\right)-3\cdot\left(-1\right)\right\rbrack\)

   Then \(\lambda = -2,2,3\)​

   Thus we find the basis of eigenvectors

   For \(\lambda=-2\): \(\text{Null}\left(T+2I\right)=\text{Null}\begin{pmatrix}4 & -1 & 0 & 0\\ 1 & 6 & 0 & 0\\ 0 & 0 & 3 & 3\\ 0 & 0 & 1 & 1\end{pmatrix}=\text{Null}\begin{pmatrix}0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0\end{pmatrix}=\langle(0,0,-1,1)\rangle\).

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

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

   The dimension of eigenvector is 3 less than 4, thus it can not be diagonalized.

   ---

   (c) \(T(x,y,z,w,v)=(3x+2y+4z,2x+2z,4x+2y+3z,3w+v,2w+2v)\).

   First, we need \([T]_{C}=\begin{pmatrix}3 & 2 & 4 & 0 & 0\\ 2 & 0 & 2 & 0 & 0\\ 4 & 2 & 3 & 0 & 0\\ 0 & 0 & 0 & 3 & 1\\ 0 & 0 & 0 & 2 & 2\end{pmatrix}=\begin{pmatrix}A & 0\\ 0 & B\end{pmatrix}\), notice that this is a block matrix.

   Then if \([T_A]_C\) and \([T_B]_C\) is diagonalizable, then \([T]_C\) is also diagonalizable.

   • \([T_A]_C\)​

   First, we have \([T_{A}]_{C}=\begin{pmatrix}3 & 2 & 4\\ 2 & 0 & 2\\ 4 & 2 & 3\end{pmatrix}\)

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

   \(=-\lambda\left(\lambda-3\right)^2+16+16+16\lambda+4\left(\lambda-3\right)+4\left(\lambda-3\right)=0\)

   \(=-\lambda^3+6\lambda^2+15\lambda+8=0\Rightarrow\lambda=-1,8\)

   Thus we find the basis of eigenvectors

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

   For \(\lambda=8\): \(\text{Null}(T-8I)=\text{Null}\begin{pmatrix}-5 & 2 & 4\\ 2 & -8 & 2\\ 4 & 2 & -5\end{pmatrix}=\text{Null}\begin{pmatrix}0 & 0 & 0\\ 1 & 0 & -1\\ 0 & 1 & -\frac12\end{pmatrix}=\langle(2,1,2)\rangle\).

   Now, \(\mathcal{B}_{A}=\{(-1,2,0),\left(0,-2,1),(2,1,2\right)\}\) is my basis of eigenvectors of \(T_A\).

   Thus \([T_{A}]_{\mathcal{B}_{A}}=\begin{pmatrix}-1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & 8\end{pmatrix}\)

   Then \(P_{A}=P_{\mathcal{B}_A}^{C}=\begin{pmatrix}-1 & 0 & 2\\ 2 & -2 & 1\\ 0 & 1 & 2\end{pmatrix}\), then \(P_{A}^{-1}[T_{A}]_{C}P_{A}=[T_{A}]_{\mathcal{B}_{A}}\) - \([T_B]_C\)​

   First, we have \([T_{B}]_{C}=\begin{pmatrix}3 & 1\\ 2 & 2\end{pmatrix}\)

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

   \(=(\lambda-3)\left(\lambda-2\right)-2=0\Rightarrow\lambda^2-5\lambda+4=0\Rightarrow\lambda=1,4\)

   Thus we find the basis of eigenvectors

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

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

   Now, \(\mathcal{B}_{B}=\{(1,-2),\left(1,1)\right.\}\) is my basis of eigenvectors of \(T_B\).

   Thus \([T_B]_{\mathcal{B}_{B}}=\begin{pmatrix}1 & 0\\ 0 & 4\end{pmatrix}\)

   Then \(P_{B}=P_{\mathcal{B}_{B}}^{C}=\begin{pmatrix}1 & 1\\ -2 & 1\end{pmatrix}\), then \(P_{B}^{-1}[T_{B}]_{C}P_{B}=[T_{B}]_{\mathcal{B}_{A}}\)

   Thus \(\mathcal{B}=\{(-1,2,0,0,0),\left(0,-2,1,0,0),(2,1,2,0,0\right),(0,0,0,1,-2),\left(0,0,0,1,1)\right.\}\) is my basis of eigenvectors of \(T\).

   Then \([T]_{\mathcal{B}}=\begin{pmatrix}-1 & 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 & 0\\ 0 & 0 & 8 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 4\end{pmatrix}\)​ \(P=P_{\mathcal{B}}^{C}=\begin{pmatrix}-1 & 0 & 2 & 0 & 0\\ 2 & -2 & 1 & 0 & 0\\ 0 & 1 & 2 & 0 & 0\\ 0 & 0 & 0 & 1 & 1\\ 0 & 0 & 0 & -2 & 1\end{pmatrix}\), then \(P^{-1}[T]_{C}P=[T]_{\mathcal{B}}\)

3. (20pts) Let \(A \in M_n(\mathbb{F})\).

   (a) Suppose \(A\) is nilpotent, that is, there exists some \(m \in \mathbb{N}\) such that \(A^m = 0\). Prove that \(0\) is the only eigenvalue of \(A\).

   Proof

   We need to prove \(Av=\lambda v\Rightarrow\lambda=0\)

   Consider \(Av=\lambda v\), then \(A^{m}v=\lambda^{m}v\Rightarrow0=\lambda^{m}v\Rightarrow\lambda=0\) since \(v\neq 0\)​

   (b) Suppose \(A \neq 0\) is idempotent, that is, \(A^2 = A\). Prove that \(0\) and \(1\) are the only eigenvalues of \(A\).

   Proof

   Consider \(Av=\lambda v\Rightarrow A^2v=\lambda^2v\), we also know \(A^2=A\), then \(Av=\lambda^2v\)​

   Thus \(\lambda v=\lambda ^2 v\), then \(\lambda =1,0\) since \(v\neq 0\)

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.

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