8

Homework 8.pdf

1. (20 pts) Let \(B = \{\alpha_1, \alpha_2, \alpha_3\}\) be the basis for \(\mathbb{C}^3\) defined by \(\alpha_1 = (1, 0, -1)\), \(\alpha_2 = (1, 1, 1)\), \(\alpha_3 = (2, 2, 0)\). Find the dual basis of \(B\).

   We know any \((x,y,z)\in \mathbb{C}^3\) can be written \((x,y,z)=a_1\alpha_1+a_2\alpha_2+a_3\alpha_3\) since \(\alpha_1,\alpha_2,\alpha_3\) is the basis of \(B\)​

   Then we have \(\begin{pmatrix}\begin{array}{ccc|c}1 & 1 & 2 & x\\ 0 & 1 & 2 & y\\ -1 & 1 & 0 & z\end{array}\end{pmatrix}\Rightarrow\begin{pmatrix}\begin{array}{ccc|c}0 & 1 & 0 & x+z-y\\ 0 & 0 & 1 & \frac{2y-x-z}{2}\\ 1 & 0 & 0 & x-y\end{array}\end{pmatrix}\Rightarrow\begin{cases}\alpha_1=x-y\\ \alpha_2=x+z-y\\ \alpha_3=\frac{2y-x-z}{2}\end{cases}\)​

   Since \(\varphi_1\left(x,y,z\right)=a_1,\varphi_2\left(x,y,z\right)=a_2,\varphi_3\left(x,y,z\right)=a_3\), then \(\varphi_1\left(x,y,z\right)=x-y,\varphi_2\left(x,y,z\right)=x+z-y,\varphi_3\left(x,y,z\right)=\frac{2y-x-z}{2}\)​

2. (20 pts) Suppose \(V_1, \dots, V_n\) are vector spaces. Prove that \((V_1 \times \cdots \times V_n)^*\) and \(V_1^* \times \cdots \times V_n^*\) are isomorphic vector spaces.

   First we prove both are vector spaces

   Since \(V_1, \dots, V_n\) are vector spaces, then the product space \(V_1 \times \cdots \times V_n\) is a vector space

   Since \((V_1 \times \cdots \times V_n)^*\) is the dual space of \(V_1 \times \cdots \times V_n\), then it is also a vector space

   Similarly, since \(V_1, \dots, V_n\) are vector spaces, then the dual space \(V_1^{*},\cdots,V_{n}^{*}\) is a vector space

   Since \(V_1^{*},\cdots,V_{n}^{*}\) are vector spaces, then the product space \(V_1^* \times \cdots \times V_n^*\) is a vector space

   Then

   Use Classification Theorem, \(\dim(V_1\times\cdots\times V_{n})^{*}=\dim\left(V_1\times\cdots\times V_{n}\right)=\dim V_1+\cdots+\dim V_{n}\)​

   Also \(\dim\left(V_1^{*}\times\cdots\times V_{n}^{*}\right)=\dim V_1^{*}+\cdots+\dim V_{n}^{*}=\dim V_1+\cdots+\dim V_{n}\)

   Thus \(\dim\left(V_1^{*}\times\cdots\times V_{n}^{*}\right)=\dim(V_1\times\cdots\times V_{n})^{*}\), then \((V_1 \times \cdots \times V_n)^*\) and \(V_1^* \times \cdots \times V_n^*\) are isomorphic vector spaces.

3. (20 pts) Suppose \(U\) is a subspace of \(V\) such that \(V/U\) is finite-dimensional. Prove that there exists a subspace \(W\) of \(V\) such that \(\dim(W) = \dim(V/U)\) and \(V = U \oplus W\).

   Since \(U\) is a subspace of \(V\), by theorem there exists a subspace \(W\) such that \(V = U \oplus W\)​

   Then we need to prove \(\dim(W) = \dim(V/U)\)

   We know \(\dim(V/U)=\dim(V)-\dim(U)\).

   Also we have \(\dim(U\oplus W)=\dim(U)+\dim(W)-\dim\left(U\cap W\right)=\dim(U)+\dim(W)\) since \(U+W\) is a direct sum and \(U\cap W=\{0\}\)

   Thus \(\dim\left(V\right)=\dim(U)+\dim(W)\Rightarrow\dim\left(W\right)=\dim(V)-\dim(U)\)

   Thus \(\dim(W) = \dim(V/U)\), then there exists a subspace \(W\) of \(V\) such that \(\dim(W) = \dim(V/U)\) and \(V = U \oplus W\).

4. (10 pts) Let \(F\) be a field and let \(f : F^2 \to F\) be the linear functional defined by \(f(x_1, x_2) = ax_1 + bx_2\). For each of the following linear operators \(T\), find \(g(x_1, x_2)\) where \(g = T^*f\).

   1. \(T(x_1, x_2) = (x_1, 0)\).

      \(g\left(x_1,x_2\right)=T^{*}f\left(x_1,x_2\right)=f\left(T\left(x_1,x_2\right)\right)=f\left(x_1,0\right)=ax_1\) 2. \(T(x_1, x_2) = (-x_2, x_1)\).

      \(g\left(x_1,x_2\right)=T^{*}f\left(x_1,x_2\right)=f\left(T\left(x_1,x_2\right)\right)=f\left(-x_2,x_1\right)=bx_1-ax_2\) 3. \(T(x_1, x_2) = (x_1 - x_2, x_1 + x_2)\).

      \(g\left(x_1,x_2\right)=T^{*}f\left(x_1,x_2\right)=f\left(T\left(x_1,x_2\right)\right)=f\left(x_1-x_2,x_1+x_2\right)=\left(a+b\right)x_1+\left(b-a\right)x_2\)

5. (30 pts) Consider the real vector space of \(n \times n\) matrices \(V = M_n(F)\).

   1. If \(B \in V\) is a fixed matrix, define a function \(f_B\) on \(V\) by \(f_B(A) = \text{trace}(B^T A)\). Show that \(f_B\) is a linear functional on \(V\).

      Since \(f_{B}:V\rightarrow\mathbb{F}\), then we need to prove \(f_B\) is a linear map

      Consider \(f_{B}(A+\lambda X)=\text{trace}\left(B^{T}\left(A+\lambda X\right)\right)=\text{trace}\left(B^{T}\left(A+\lambda X\right)\right)=\text{trace}\left(B^{T}A+B^{T}\lambda X\right)=\text{trace}\left(B^{T}A+\lambda B^{T}X\right)\)

      Since the trace is a linear map, then \(\text{trace}\left(B^{T}A+\lambda B^{T}X\right)=\text{trace}\left(B^{T}A)+\text{trace}(\lambda B^{T}X\right)=\text{trace}\left(B^{T}A)+\lambda\text{trace}(B^{T}X\right)=f_{B}\left(A\right)+\lambda f_{B}\left(X\right)\)​

      Thus it is a linear functional. 2. Show that every linear functional on \(V\) is of the above form, i.e., is \(f_B\) for some \(B\).

      Take any linear functional \(g:V\rightarrow\mathbb{F}\), we consider the basis \(\{E_{ij},\forall i,j\leq n\}\) of \(V\) that is \(ij\) entry is \(1\) and others are \(0\) which is a trivial basis of \(V\)​

      Then \(g(A)=g(\sum_{i=1,j=1}^{n^2}a_{ij}E_{ij})=\sum_{i=1,j=1}^{n^2}a_{ij}\cdot g\left(E_{ij}\right)\)​

      Let's define \(g(E_{ij})=b_{ji}\), then \(g(A)=\sum_{i=1,j=1}^{n^2}a_{ij}\cdot b_{ji}\)​

      Also \(f_{B}(A)=\text{trace}(B^{T}A)\), define \(B=b_{ij},B^{T}=b_{ji}\)​

      Then \(f_{B}(A)=\text{trace}\left(\sum_{i=1,j=1}^{n^2}b_{ji}a_{ij}E_{ij}\right)=\sum_{i=1,j=1}^{n^2}b_{ji}a_{ij}=\sum_{i=1,j=1}^{n^2}a_{ij}b_{ji}\)​

      Thus any \(g(A)\) can be written as \(f_B(A)\)​ 3. Show that \(\Phi: V \to V^*\), \(\Phi(B) = f_B\) is an isomorphism.

      1. linear map

         Take \(\Phi(B+\lambda C)\left(A\right)=f_{\left(B+\lambda C\right)}\left(A)=\text{trace}((B+\lambda C\right)^{T}A)=\text{trace}(B^{T}A+\lambda C^{T}A)=\text{trace}\left(B^{T}A\right)+\lambda\text{trace}\left(C^{T}A\right)=\Phi(B)\left(A\right)+\lambda\Phi(C)\left(A\right)\)​ 2. injective

         We need to prove \(\text{Null}(\Phi)=\left\lbrace0\right\rbrace\), consider \(\Phi(B)\left(A\right)=f_{B}\left(A\right)=0\Rightarrow\text{trace}(B^{T}A)=0\)​

         Since \(A\) can be any, then \(B=0\), then \(\text{Null}(\Phi)=\left\lbrace0\right\rbrace\)​ 3. surjective

         We need to prove \(\text{Range}\Phi=V^*\), since in (2) we have proved any \(f\in V^*\) can be written as\(f_B\) for some \(B\), which satisfy the definition of surjectivity

      Thus \(\Phi: V \to V^*\), \(\Phi(B) = f_B\) is an isomorphism.

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