11.29 Represent linear maps

Let \(T: \mathbb{R}^3 \to \mathbb{R}^2\) be the linear map defined by \(T(x_1, x_2, x_3) = (x_1 + x_2, 2x_3 - x_1)\) and consider \(\mathcal{B} = \{(1, 0, -1), (1, 1, 1), (1, 0, 0)\}\) and \(\mathcal{B}' = \{(1, 2), (1, -1)\}\), bases for \(\mathbb{R}^3\) and \(\mathbb{R}^2\) respectively. Find the matrix \([T]_{\mathcal{B}, \mathcal{B}'}\) of \(T\) with respect to \(\mathcal{B}\) and \(\mathcal{B}'\).

\(\begin{pmatrix}-\frac23 & 1 & 0\\ \frac53 & 1 & 1\end{pmatrix}\)
Consider the linear map \(T: \mathbb{R}^3 \to \mathbb{R}^3\) given by \(T(x, y, z) = (x - y + 2z, 3x + y + 4z, 5x - y + 8z)\). (a) What is the matrix of \(T\) with respect to the canonical basis of \(\mathbb{R}^3\)? (b) What is the matrix \([T]_{\mathcal{B}, \mathcal{B}'}\) of \(T\) with respect to the bases of \(\mathbb{R}^3\) given by
\(\mathcal{B} = \{(1, 1, 0), (0, 1, 1), (1, 0, 0)\}\) and \(\mathcal{B}' = \{(1, -2, 1), (2, -3, 3), (-2, 2, -3)\}\)? (c) Is \(T\) an isomorphism?
Let \(T: \mathbb{C}^3 \to \mathbb{C}^3\) be the linear map defined by \(T(1, 0, 0) = (1, 0, i)\), \(T(0, 1, 0) = (0, 1, 1)\), \(T(0, 0, 1) = (i, 1, 1)\). Decide whether or not \(T\) is invertible. If invertible, describe \(T^{-1}(x, y, z)\).

Injectivity: Use the property of linear Yes

Surjectivity: Yes, since \((1,0,i),(0,1,1),(i,1,1)\) is lienarly independent, then any \(w\in W\) can be written as unique linear combination Yes

Or use matrix to solve
Let \(V\) and \(W\) be two \(\mathbb{F}\)-vector spaces of dimension \(n\) and \(m\) respectively. Let \(\mathcal{B}\) be a basis of \(V\) and \(\mathcal{B}'\) be a basis of \(W\). Define the following function: \(\Phi: \mathcal{L}(V, W) \to M_{m \times n}(\mathbb{F}), \Phi(T) = [T]_{\mathcal{B} \mathcal{B}'}\). Prove that \(\Phi\) is an isomorphism and conclude that \(\dim(\mathcal{L}(V, W)) = mn\).

First we know this function is a linear map, then we prove injective and surjective
1. Injective
  
  Take \(\Phi(T_1)=\Phi(T_2)\), then \([T_1]_{\mathcal{B} \mathcal{B}'}=[T_2]_{\mathcal{B} \mathcal{B}'}\)
  
  Then every column of matrix must be same, thus \(T_1(v_i)=T_2(v_i)\)
  
  Since the linear map is uniquely decided by basis, then \(T_1=T_2\)
  
  Or to prove \(\text{Null}T=\left\lbrace0\right\rbrace\)
  
  \(\Phi(T)=[T]_{\mathcal{BB}^{\prime}}=0\Rightarrow\) each column of matrix is zero \(\Rightarrow T\left(v_{i}\right)=0,\forall i\in n\Rightarrow T\left(\lambda_1v_1+\cdots+\lambda_{n}v_{n}\right)=0\Rightarrow T\left(v\right)=0,\forall v\in V\Rightarrow\text{Null}T=\left\lbrace0\right\rbrace\) 2. surjective
  
  Take any \([T]_{\mathcal{BB}^{\prime}}\), N.T.P. \(\exists T:\Phi(T)=[T]_{\mathcal{BB}^{\prime}}\)
  
  Let \([T]_{\mathcal{BB}^{\prime}}=\begin{pmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \vdots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn}\end{pmatrix}\), then we let the basis of \(V\) is \(\{v_1,...,v_n\}\) and the basis of \(W\) is \(\{w_1,...,w_{m}\}\)
  
  Then \(T(v_{i})=\sum_{k=1}^{m}a_{ki}w_{k}\), then \(T(v)=\sum_{j=1}^{n}\lambda_{j}\left(\sum_{k=1}^{m}a_{kj}w_{k}\right)=\sum_{k=1}^{m}\left(\sum_{j=1}^{n}\lambda_{j}a_{kj}\right)w_{k}\)
  
  Then \(\Phi(T)=[T]_{\mathcal{BB}^{\prime}}\)
Let \(V\) and \(W\) be vector spaces over the field \(\mathbb{F}\), and let \(U: V \to W\) be an isomorphism.

Prove that \(\Phi(T) = UTU^{-1}\) is an isomorphism from \(\mathcal{L}(V, V)\) onto \(\mathcal{L}(W, W)\).
1. linear
  
  \(\Phi(T+\lambda S)=U(T+\lambda S)U^{-1}=UTU^{-1}+U\lambda SU^{-1}=UTU^{-1}+\lambda USU^{-1}=\Phi(T)+\lambda\Phi(S)\) 2. injective
  
  Prove \(\text{Null}T=\left\lbrace0\}:\right.\) \(\Phi(T)=UTU^{-1}=0\), then \(U^{-1}UTU^{-1}=0\Rightarrow TU^{-1}=0\Rightarrow T=0\) 3. surjective
  
  We need to prove \(\forall S\in\mathcal{L}(W,W),\exists T\in\mathcal{L}(V,V):\Phi(T)=S\)
  
  Consider \(T=U^{-1}SU\), then we have \(\Phi(T)=S\)
Suppose \(\mathcal{B} = \{v_1, \dots, v_n\}\) is a basis of \(V\). Prove that the map \(T: V \to M_{n \times 1}(\mathbb{F})\) defined by \(T(v)=C\) is an isomorphism of \(V\). Here \([T]_{\mathcal{B}}(v)\) is the matrix of \(v \in V\) with respect to the basis \(\mathcal{B}\) (the coordinates of \(v\) with respect to \(\mathcal{B}\)).
1. linear
  
  \(T(v+\lambda w)=\left\lbrack T\right\rbrack_{B}\left(v+\lambda w\right)=\left\lbrack T\right\rbrack_{B}v+\left\lbrack T]_{B}\lambda w\right.=T\left(v\right)+\lambda T\left(w\right)\) 2. Injective
  
  \(T(v)=[T]_{\mathcal{B}}(v)=0\Rightarrow\left\lbrack T\right\rbrack_{B}\left(v\right)=0\Rightarrow v=0\cdot v_1+\cdots+0\cdot v_{n}=0\Rightarrow\text{Null}T=0\) 3. surjective
  
  We need to prove \(\forall A=\begin{pmatrix}a_1\\...\\a_n\end{pmatrix}\in M_{n\times1}(\mathbb{F}),\exists v\in V:T(v)=A\)
  
  Consider \(v=\sum_{i=1}^{n}a_{i}v_{i}\), then \(T(v)=[T]_{\mathcal{B}}(v)=A\)

Actually we can use dimension theorem directly if the problem is about finite dimensional vector space

But if not, then we need to prove by definition.

Note that we need to prove injective and surjective by definition we cannot use the equivalence because that is about finite dimension as well.