> [!NOTE] Definition (Linear Map) > Let $V,W$ be [[Vector Space|vector spaces]] over a *[[Field (Algebra)|field]]* $\mathbb{F}$. Then a [[Function|map]] $\varphi : V \to W$ is a *linear map* iff for all $\lambda\in \mathbb{F},$ and $v,w\in V,$ $\varphi(\lambda v+\mu w)= \lambda \varphi (v) + \mu\varphi(w).$ > [!Example] > Let $\mathbb{C}^{\infty}(\mathbb{R})$ denote the vector space of infinitely [[Real Differentiablity|differentiable functions]] $f:\mathbb{R}$. Differentiation $\begin{align} \mathbb{C}^{\infty}(\mathbb{R}) \to \mathbb{C}^{\infty} (\mathbb{R}) \\ f \mapsto \frac{df}{dx} \end{align}$is a linear map since $\frac{d(f+g)}{dx} = \frac{df}{dg} + \frac{dg}{dx} \quad \text{and} \quad \frac{d(\lambda f)}{dx}= \lambda\frac{ df}{dg}.$ # Properties > [!NOTE] Lemma (Kernel and Image are subspaces) > 1. $\ker \varphi$ is a [[Vector Subspace|subspace]] of $V$; > 2. $\text{Im}\, \varphi$ is a *subspace* of $W$. >Furthermore, if $V$ is finite dimensional then so are the kernel and image of $\varphi$. See [[Kernel and Image of a Linear Map are Subspaces|proof]]. The dimensions of its image and kernel are known the [[Rank-Nullity Formula|rank and nullity]] of $\varphi$ respectively. ### Properties of Linear Map from Finite Dimensional Vector Space > [!NOTE] Theorem (Linear maps are equal if they agree on basis elements/ linear map are uniquely defined by their values on a basis) > Let $V$ be a [[Finite Dimensional Vector Space|FDVS]] with *basis* elements $v_{1},\dots,v_{n}$ and $W$ any *vector space* (not necessarily finite dimensional). > 1. Suppose $\varphi_{1}: V \to W$ and $\varphi_{2}:V\to W$ are two linear maps such that $\forall i = 1,\dots n, \quad \varphi_{1}(v_{i})=\varphi_{2}(v_{i}) ,$Then $\varphi_{1} = \varphi_{2}$. > 2. For any choice of vectors $u_{1},u_{2},\dots,u_{n}\in W$, there is a unique linear map $\varphi:V \to W$ such that $\varphi(v_{i}) = u_{i}$ for $i=1,\dots,n$. ^a8ff22 >*Proof*. Consider any $v\in V$. Then $\exists!\lambda_{i}$ such that $v=\lambda_{1}v_{1}+\dots+\lambda_{n}v_{n}$. >1. STP $\varphi_{1}(v) = \varphi_{2}(v)$. Using *linearity* of the two maps and the fact they agree on the basis elements $\begin{aligned} \varphi_{1}(v)& =\quad\varphi_1(\lambda_1v_1+\ldots+\lambda_nv_n) \\ &=\quad\lambda_1\varphi_1(v_1)+\ldots+\lambda_n\varphi_1(v_n) \\ &=\quad\lambda_1\varphi_2(v_1)+\ldots+\lambda_n\varphi_2(v_n) \\ &=\quad\varphi_2(\lambda_1v_1+\ldots+\lambda_nv_n) \\ &=\quad\varphi_2(v) \end{aligned}$as required. >2. Define $\varphi(v) = \lambda_{1}\varphi(v_{1})+\dots+\lambda_{n}\varphi(v_{n})\in W$Must check that $\varphi$ is linear: $\forall v, w \in V$, $\forall \alpha, \beta \in \mathbb{R}$ $\begin{aligned} \varphi(\alpha v+\beta w)& =\quad\varphi(\alpha(\lambda_1v_1+\ldots+\lambda_nv_n)+\beta(\mu_1v_1+\ldots,\mu_nv_n)) \\ &=\quad\varphi((\alpha\lambda_1+\beta\mu_1)v_1+\ldots+(\alpha\lambda_n+\beta\mu_n)v_n) \\ &\begin{aligned}=&(\alpha\lambda_1+\beta\mu_1)u_1+\ldots+(\alpha\lambda_n+\beta\mu_n)u_n\end{aligned} \\ &=\quad\alpha(\lambda_1u_1+\ldots+\lambda_nu_n)+\beta(\mu_1u_1+\ldots+\mu_nu_n) \\ &=\quad\alpha\varphi(v)+\beta\varphi(w) \end{aligned}$as required. The uniqueness of $\varphi$ follows from $(1)$. > [!NOTE] Theorem/ Definition (The matrix of the linear map $\varphi$ with respect to the given bases of $V$ and $W$) > Let $V$ be an [[Finite Dimensional Vector Space|FDVS]] with basis $v_{1},\dots,v_{n}$ and $W$ be vector space with basis $w_{1},\dots,w_{m}.$ If $\varphi:V \to W$ is a linear map, then there is a matrix $A\in \mathbb{R}^{mn}$ so that $\varphi = \varphi_{A}$ is the [[Left Multiplication Linear Map of Real Matrix#^2b50ef|left multiplication linear map]] of $A$ with respect to these specified bases. ^467233 >Proof. Define the entries $a_{ij}$ of a matrix $\varphi(v_{i})$ in the basis $w_{1},\dots, w_{m}$: $\begin{array}{rcl}\varphi(v_1)&=&a_{11}w_1+a_{21}w_2+\ldots+a_{m1}w_m\\\varphi(v_2)&=&a_{12}w_1+a_{22}w_2+\ldots+a_{m2}w_m\\&\vdots&\\\varphi(v_n)&=&a_{1n}w_1+a_{2n}w_2+\ldots+a_{mn}w_m\end{array}$If we set $A=(a_{ij})\in \mathbb{R}^{mn}$, then $\varphi(v_{i})=\varphi_{A}(v_{i})$. Therefore $\varphi=\varphi_{A}$ as claimed. # Applications - [[Change of Basis Formula For Linear Maps]]. - [[Hom(U,V)]]. - [[Linear Operator]].