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12.6 Quotient space

  1. For \(n \in \mathbb{N}\), define \(V^n\) by \(V^n = V \times \cdots \times V\), where the product is done \(n\) times. Prove that \(V^n\) and \(\mathcal{L}(\mathbb{F}^n, V)\) are isomorphic vector spaces.

    Use dimension theorem

  2. Suppose \(V_1, \ldots, V_n\) are vector spaces. Prove that \(\mathcal{L}(V_1 \times \cdots \times V_n, W)\) and \(\mathcal{L}(V_1, W) \times \cdots \times \mathcal{L}(V_n, W)\) are isomorphic vector spaces.

    See here

  3. Let \(U\) be a subspace of \(V\). Suppose \(\{v_1 + U, \ldots, v_n + U\}\) is a basis of \(V / U\) and \(\{u_1, \ldots, u_m\}\) is a basis of \(U\). Prove that \(\{v_1, \ldots, v_n, u_1, \ldots, u_m\}\) is a basis of \(V\).

    Since \(\left\lbrace{u_1,\ldots,u_{m}}\}\right.\) is a basis of \(U\), then \(u=\sum_{i=1}^{m}a_{i}u_{i}\)

    Since \(\{v_1 + U, \ldots, v_n + U\}\) is a basis of \(V / U\), then \(V/U\) can be written as \(v+U=\sum_{i=1}^{n}b_{i}\left(v_{i}+U\right)=U+\sum_{i=1}^{n}b_{i}v_{i}\)

    Then \(v-\sum_{i=1}^{n}b_{i}v_{i}\in U\)​

    Since \(u\in U\), then \(v-\sum_{i=1}^{n}b_{i}v_{i}=\sum_{i=1}^{m}a_{i}u_{i}\Rightarrow v=\sum_{i=1}^{n}b_{i}v_{i}+\sum_{i=1}^{m}a_{i}u_{i}\) which is a spanning list

    Since \(\dim (V/U)=\dim V-\dim U\), then \(\dim V=n+m\)​

    Then it is a basis!

  4. Find the dual basis corresponding to the basis:
    (a) \({(2, 1), (3, 1)}\) of \(\mathbb{R}^2\).

    Since \(\left(x,y\right)=a\left(2,1\right)+b\left(3,1\right)\), then we need to find such \(a,b\)​

    \(\begin{pmatrix}\begin{array}{cc|c}2 & 3 & x\\ 1 & 1 & y\end{array}\end{pmatrix}\Rightarrow\begin{pmatrix}\begin{array}{cc|c}0 & 1 & x-2y\\ 1 & 0 & 3y-x\end{array}\end{pmatrix}\Rightarrow b=x-2y,a=3y-x\)​

    Then \(\varphi_1(x,y)=a=3y-x\) \(\varphi_2(x,y)=b=x-2y\)
    (b) \({(1, 0, 1), (1, 1, 0), (0, 1, 1)}\) of \(\mathbb{R}^3\).

    Since \(\left(x,y,z\right)=a\left(1,0,1\right)+b\left(1,1,0\right)+c\left(0,1,1\right)\), then we need to find such \(a,b,c\)

    \(\begin{pmatrix}\begin{array}{ccc|c}1 & 1 & 0 & x\\ 0 & 1 & 1 & y\\ 1 & 0 & 1 & z\end{array}\end{pmatrix}\Rightarrow\begin{pmatrix}\begin{array}{ccc|c}1 & 0 & 0 & \frac{-y+z+x}{2}\\ 0 & 1 & 0 & \frac{y-z+x}{2}\\ 0 & 0 & 1 & \frac{y+z-x}{2}\end{array}\end{pmatrix}\Rightarrow c=\frac{y+z-x}{2},b=\frac{y-z+x}{2},a=\frac{-y+z+x}{2}\)

    Then \(\varphi_1(x,y,z)=a=\frac{-y+z+x}{2}\) \(\varphi_2(x,y,z)=b=\frac{y-z+x}{2}\)

    \(\varphi_3(x,y,z)=c=\frac{y+z-x}{2}\)​

  5. Suppose \(\{v_1, \ldots, v_n\}\) is a basis of \(V\) and \(\{\varphi_1, \ldots, \varphi_n\}\) is the corresponding dual basis of \(V^{*}\). Let \(\psi \in V^*\), prove that \(\psi = \psi(v_1)\varphi_1 + \cdots + \psi(v_n)\varphi_n\).

    \(\psi=\lambda_1\varphi_1+\cdots+\lambda_{n}\varphi_{n}\Rightarrow\psi\left(v_{i}\right)=\lambda_1\varphi_1\left(v_{i}\right)+\cdots+\lambda_{n}\varphi_{n}\left(v_{i}\right)=\lambda_{i}\)​

  6. Suppose \(\varphi \in \mathcal{L}(V, \mathbb{F})\) and \(\varphi \neq 0\). Prove that \(\dim(V / (\text{Null}(\varphi))) = 1\).

    Since \(\dim(V/\text{Null}\varphi)=\dim(V)-\dim(\text{Null}\varphi)\) and \(\dim(V)=\dim(\text{Range}\varphi)+\dim(\text{Null}\varphi)\), then \(\dim(V/\text{Null}\varphi)=\dim(\text{Range}\varphi)\)​

    Since \(\varphi\neq 0\), then \(\dim(\text{Range}\varphi)=1\)​

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