9

Homework 9.pdf

1. (30pts) Suppose \(V\) is finite-dimensional and \(\Gamma\) is a subspace of \(V^*\).

   Show that \(\Gamma=\{v\in V:\varphi(v)=0\text{ for every }\varphi\in\Gamma\}^0.\)

   Let \(W=\{v\in V:\varphi(v)=0\text{ for every }\varphi\in\Gamma\}\), then we need to prove \(\Gamma=W^\circ\)

   \(\subseteq\)) Suppose \(\Gamma\cancel{\subseteq}W^{\circ}\), then \(\exists \varphi \in \Gamma,\) but not in \(W^{\circ}\)

   Then \(\exists \varphi \in \Gamma\) and \(\varphi(v)\neq0,\exists v\in W\) by definition of annihilator

   But we know if \(v\in W\), then \(\varphi(v)=0,\forall \varphi\in \Gamma\), which is a contradiction

   Thus \(\Gamma \subseteq W^{\circ}\)

   \(\supseteq\)) Suppose \(\Gamma\cancel{\supseteq}W^{\circ}\), then \(\exists \varphi \in W^\circ\) but not in \(\Gamma\)

   Then \(\exists\varphi(v)=0,\forall v\in W\) and \(\varphi \notin \Gamma\)

   Then \(\exists\varphi(v)=0,\forall v\in W\) and \(\varphi\left(v\right)\neq a,\forall v\in V\) where \(a\in\mathbb{F}\)

   In particular, \(\varphi(v)\neq 0,\forall v\in V\), which is a contradiction

   Thus \(\Gamma \supseteq W^{\circ}\)

2. (20pts) Suppose \(V\) and \(W\) are finite-dimensional. Let \(T \in L(V, W)\) and suppose there exists \(\varphi\in V^{*}\) such that \(\text{Null}(T^{*})=\langle\varphi\rangle\)

   Prove that \(\text{Range}(T)=\text{Null}(\varphi)\)

   By the theorem, we know that \(\text{Null}(T^*) = (\text{Range}(T))^\circ \implies (\text{Range}(T))^\circ = \langle \varphi \rangle = \{ \alpha \varphi : \alpha \in \mathbb{R} \}\).

   By a exercise in tutorial, we know if \(V\) is finite-dimensional and \(U\) is a subspace \(\Rightarrow U = \{v \in V : \varphi(v) = 0 \; \forall \varphi \in U^\circ \}\)​

   Thus \(\text{Range}T=\{w\in W:\varphi(w)=0\;\forall\varphi\in\left(\text{Range}T\right)^{\circ}\}=\{w\in W:\alpha\varphi(w)=0\;\forall\alpha\in\mathbb{R}\}=\{w\in W:\varphi(w)=0\}=\text{Null}(\varphi)\)​

3. (30pts) Let \(T : \mathbb{R}^3 \to \mathbb{R}^3\) be the linear map defined by \(T(x, y, z) = (3x + 4y - z, x + y + z, -3x + 6y)\) and consider the bases of \(\mathbb{R}^3\) given by \(B = \{(1, 1, 0), (0, 1, 1), (1, 0, 0)\}\) and \(B' = \{(1, -2, 1), (2, -3, 3), (-2, 2, -3)\}\).

   1. Find the matrix \([T]_{B'}\) of \(T\) in the basis \(B'\).

      First, let's find \([T]_{C}\) where \(C\) is the canonical basis of \(\R^3\)​

      Then \(T(1,0,0)=\left(3,1,-3\right)=3e_1+e_2-3e_3\)

      \(T(0,1,0)=\left(4,1,6\right)=4e_1+e_2+6e_3\)

      \(T(0,0,1)=\left(-1,1,0\right)=-e_1+e_2\)

      Thus \([T]_{C}=\begin{pmatrix}3 & 4 & -1\\ 1 & 1 & 1\\ -3 & 6 & 0\end{pmatrix}\)

      Then we write \(P_{B^{\prime}}^{C}=\begin{pmatrix}1 & 2 & -2\\ -2 & -3 & 2\\ 1 & 3 & -3\end{pmatrix}\), then \(\left(P_{B^{\prime}}^{C}\right)^{-1}=\begin{pmatrix}3 & 0 & -2\\ -4 & -1 & 2\\ -3 & -1 & 1\end{pmatrix}\)​

      Then we can get \([T]_{B^{\prime}}=\left(P_{B^{\prime}}^{C}\right)^{-1}\left\lbrack T\right\rbrack_{C}P_{B^{\prime}}^{C}=\begin{pmatrix}3 & 0 & -2\\ -4 & -1 & 2\\ -3 & -1 & 1\end{pmatrix}\begin{pmatrix}3 & 4 & -1\\ 1 & 1 & 1\\ -3 & 6 & 0\end{pmatrix}\begin{pmatrix}1 & 2 & -2\\ -2 & -3 & 2\\ 1 & 3 & -3\end{pmatrix}=\begin{pmatrix}12 & 21 & -21\\ -6 & -14 & 19\\ 3 & 1 & 6\end{pmatrix}\)​ 2. Write \([T]_{B'} = P^{-1}[T]_B P\) where \(P\) is an invertible matrix.

      In the exact same way as (a) we have \([T]_{B}=(P_{B}^{C})^{-1}[T]_{C}P_{B}^{C}\), then \([T]_{C}=P_{B}^{C}\left\lbrack T\right\rbrack_{B}\left(P_{B}^{C}\right)^{-1}(\star1)\)

      By (a), we have \([T]_{B^{\prime}}=(P_{B^{\prime}}^{C})^{-1}[T]_{C}P_{B^{\prime}}^{C}(\star2)\)​

      Then we combine \(\star 1\) and \(\star 2\), we get \([T]_{B^{\prime}}=(P_{B^{\prime}}^{C})^{-1}P_{B}^{C}\left\lbrack T\right\rbrack_{B}\left(P_{B}^{C}\right)^{-1}P_{B^{\prime}}^{C}=P\left\lbrack T\right\rbrack_{B}P^{-1}\) where \(P=(P_{B^{\prime}}^{C})^{-1}P_{B}^{C}\)​

      Since in(a) we already know \(\left(P_{B^{\prime}}^{C}\right)^{-1}=\begin{pmatrix}3 & 0 & -2\\ -4 & -1 & 2\\ -3 & -1 & 1\end{pmatrix}\)​

      Thus we only need to find \(P_B^C\)

      Since \(B = \{(1, 1, 0), (0, 1, 1), (1, 0, 0)\}\), then \(P_{B}^{C}=\begin{pmatrix}1 & 0 & 1\\ 1 & 1 & 0\\ 0 & 1 & 0\end{pmatrix}\)

      Thus \(P=(P_{B^{\prime}}^{C})^{-1}P_{B}^{C}=\begin{pmatrix}3 & 0 & -2\\ -4 & -1 & 2\\ -3 & -1 & 1\end{pmatrix}\begin{pmatrix}1 & 0 & 1\\ 1 & 1 & 0\\ 0 & 1 & 0\end{pmatrix}=\begin{pmatrix}3 & -2 & 3\\ -5 & 1 & -4\\ -4 & 0 & -3\end{pmatrix}\)

4. (20pts) Let \(V\) be finite-dimensional, \(B\) a basis of \(V\), and \(A \in M_n(F)\) an invertible matrix. Prove that there exist bases \(B_1\) and \(B_2\) of \(V\) such that:

   1. \(A\) is the change of basis matrix from \(B_1\) to \(B\).

      Let's \(B=\{v_1,\ldots,v_{n}\}\) be the bases of \(V\) and an inverse matrix \(A^{-1}=\begin{pmatrix}\lambda_{11} & \ldots & \lambda_{1n}\\ \ldots & \ldots & \ldots\\ \lambda_{n1} & \ldots & \lambda_{nn}\end{pmatrix}\)

      Then consider \(B_1=\{w_1,\ldots,w_{n}\}\) be a list, we need to prove it is a basis.

      Define \(w_{j}=\sum_{i=1}^{n}\lambda_{ij}v_{i}\) which \([w_{j}]_{B}=\left(\lambda_{1j},\ldots,\lambda_{nj}\right)\). Then \(([w_1]_{B}^{T},\ldots,[w_{n}]_{B}^{T})=A^{-1}\)

      Let \(v=a_1w_1+\ldots+a_{n}w_{n}=a_1\sum_{i=1}^{n}\lambda_{i1}v_{i}+\cdots+a_{n}\sum_{i=1}^{n}\lambda_{in}v_{i}=v_1\left(a_1\lambda_{11}+\ldots+a_{n}\lambda_{1n}\right)+\ldots+v_{n}\left(a_1\lambda_{n1}+\cdots+a_{n}\lambda_{nn}\right)\)

      Since \(B\) is the basis of \(V\), then \(\forall v\in V\), the coefficient is unique.

      Thus \(B_1\) is a basis and \(A\) is a change of basis matrix from \(B_1\) to \(B\). 2. \(A\) is the change of basis matrix from \(B\) to \(B_2\).

      Let's \(B=\{v_1,\ldots,v_{n}\}\) be the bases of \(V\) and an inverse matrix \(A=\begin{pmatrix}\lambda_{11} & \ldots & \lambda_{1n}\\ \ldots & \ldots & \ldots\\ \lambda_{n1} & \ldots & \lambda_{nn}\end{pmatrix}\)

      Then consider \(B_2=\{w_1,\ldots,w_{n}\}\) be a list, we need to prove it is a basis.

      Define \(w_{j}=\sum_{i=1}^{n}\lambda_{ij}v_{i}\) which \([w_{j}]_{B}=\left(\lambda_{1j},\ldots,\lambda_{nj}\right)\). Then \(([w_1]_{B}^{T},\ldots,[w_{n}]_{B}^{T})=A\)

      Let \(v=a_1w_1+\ldots+a_{n}w_{n}=a_1\sum_{i=1}^{n}\lambda_{i1}v_{i}+\cdots+a_{n}\sum_{i=1}^{n}\lambda_{in}v_{i}=v_1\left(a_1\lambda_{11}+\ldots+a_{n}\lambda_{1n}\right)+\ldots+v_{n}\left(a_1\lambda_{n1}+\cdots+a_{n}\lambda_{nn}\right)\)

      Since \(B\) is the basis of \(V\), then \(\forall v\in V\), the coefficient is unique.

      Thus \(B_2\) is a basis and \(A\) is a change of basis matrix from \(B_2\) to \(B\).