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@jemoka / Jemoka Knowledge Base / wiki/concepts/linear_map.md
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--- title: "Linear Map" type: concept related: [Zero, Product Of Linear Maps, Subspace, Polynomial, Range] source: https://www.jemoka.com/posts/kbhlinear_map/ confidence: high status: active --- A Linear Map (a.k.a. Linear Transformation) is a function which maps elements between two vector space that follows linear properties. constituents vector spaces \(V\) and \(W\) (they don’t have to be subspaces) A function \(T: V \to W\) (when we put something in, it only goes to one place) requirements \(T\) is considered a Linear Map if it follows… (properties of “linearity”) additivity \begin{equation} T(u+v) = Tu+Tv,\ \forall u,v \in V \end{equation} homogeneity \begin{equation} T(\lambda v) = \lambda (Tv),\ \forall \lambda \in \mathbb{F}, v \in V \end{equation} additional information note on notation The “application” of a Linear Map \(T\) on a vector \(V\) is written as: \begin{equation} \begin{cases} Tv \\ T(v) \end{cases} \end{equation} both are acceptable. \(\mathcal{L}(V,W)\) The set of all Linear Maps from \(V\) to \(W\) is denoted as \(\mathcal{L}(V,W)\). some fun linear maps zero There is, of course, the Linear Map that maps everything to the \(0\) — as the zero exists in all vector spaces. That is: \begin{equation} 0 \in \mathcal{L}(V,W) \implies 0v = 0, v\in V \end{equation} additivity Let \(v_1+v_2= v \in V\). \begin{equation} 0(v_1+v_2) = 0(v) = 0 = 0+0 = 0v_1+0v_2 \end{equation} homogeneity Let \(\lambda v = u \in V\). \begin{equation} 0(\lambda v) = 0(u) = 0 = \lambda 0 = \lambda 0v \end{equation} identity Another classic. \(I\), the identity map, is denoted as (for some \(v \in V\) and \(I \in \mathcal{L}(V,V)\)): \begin{equation} Iv = v \end{equation} i.e. it does nothing additivity Let \(v_1,v_2 \in V\): \begin{equation} I(v_1+v_2) = v_1+v_2 = Iv1+Iv2 \end{equation} homogeneity \begin{equation} I(\lambda v) = \lambda v = \lambda Iv \end{equation} any map from \(\mathbb{F}^{n}\) to \(\mathbb{F}^{m}\) turns out any map that follows a specific pattern of polynomials between two vector spaces are Linear Maps. Define some two vector spaces \(\mathbb{F}^{n}\) and \(\mathbb{F}^{m}\), some set of scalars \(a_{jk} \in \mathbb{F}: j=1, \dots m; k=1, \dots n\). We construct \(T \in \mathcal{L}(\mathbb{F}^{n}, \mathbb{F}^{m})\) by: \(T(x_1, \dots x_{n}) = a_{11} x_1+ \cdots + a_{1n} x_{n}, \dots, a_{m1} x_1 + \cdots + a_{mn} x_{n}\) (i.e. a combination of linear combinations). additivity Let \(x,y \in \mathbb{F}^{n}\), with \(x_{j}\) being each coordinate of \(x\) and the same goes for \(y\). \begin{align} T((x_1, \dots x_{n}) + (y_1, \dots y_{n}))) &= T(x_1+y_1 \dots x_{n}+y_{n}) \\ &= a_{11}(x_1+y_1) + \cdots, \dots, \cdots + a_{mn} (x_{n} + y_{n}) \\ &= (a_{11}x_1 + a_{11}y_{1}) + \cdots, \dots, \cdots + (a_{mn} x_{n} + a_{mn} y_{n}) \\ &= (a_{11}x_1 + \cdots) + (a_{11}y_{1}+ \cdots ), \dots, (\cdots + a_{mn}x_n) + (\cdots + a_{mn}y_{n}) \\ &= ((a_{11}x_1 + \cdots), \dots, (\cdots + a_{mn}x_n)) = ((a_{11}y_{1}+ \cdots ), \dots,(\cdots + a_{mn}y_{n})) \\ &= T (x_1, \dots, x_{n}) + T (y_1, \dots, x_{n}) \end{align} homogeneity Proof left to the reader. Pretty much just expand and more algebra. matricies to encode Linear Map we can use matricies to represent Linear Maps. See matrix of Linear Map “basis of domain” This result tells us that we can find a Linear Map for wherever we want to take the basis of a vector space, and that a Linear Map’s behavior on basis uniquely determines that Linear Map. See basis of domain. addition and scalar multiplication on \(\mathcal{L}(V,W)\) Suppose \(S,T \in \mathcal{L}(V,W); \lambda \in \mathbb{F}\). “Sum” and “Product” are defined in the way that one would expect: \begin{equation} (S+T)(v) = Sv+Tv \end{equation} and \begin{equation} (\lambda T)(v) = \lambda (Tv) \end{equation} for all \(v \in V\). These two operations make \(\mathcal{L}(V,W)\) a vector space (\(1Tv = Tv\), \(0+Tv=Tv\), \(Tv + (-1)Tv = 0\), associativity, commutativity, distributive inherits from \(V\).) linear maps take \(0\) to \(0\) We desire that \(T(0) = 0\) for any linear map \(T\) Proof: \begin{equation} T(0) = T(0+0) \end{equation} Then, by additivity: \begin{equation} T(0) = T (0 + 0) = T (0) + T (0) \end{equation} Given \(\mathcal{L}(V,W)\) is a vector space for any \(V,W\), \(\exists -T(0)\) such that \(T(0)+(-T(0)) = 0\). Applying that here: \begin{equation} T(0) = T(0)+T(0) \implies T(0) -T(0) = T(0)+T(0)-T(0) \implies 0 = T(0) \end{equation} Product of Linear Maps See Product of Linear Maps “sizes” of maps map to smaller space is not injective Suppose \(V,W\) are finite-dimensional vector spaces, and \(\dim V > \dim W\). Then, all \(T \in \mathcal{L}(V,W)\) are not injective. We first have that: \begin{align} &\dim V = \dim null\ T + \dim range\ T \\ \Rightarrow\ & \dim null\ T = \dim V - \dim range\ T \end{align} recall at this point that \(\dim range\ T \leq \dim W\) (the range is a subspace of the codomain.) Therefore, subtracting a bigger value means that the value will be smaller. So we have that: \begin{align} & \dim null\ T = \dim V - \dim range\ T \\ \Rightarrow\ & \dim null\ T \geq \dim V - \dim W \end{align} Now, recall that \(\dim V > \dim W\). Therefore, \(\dim V - \dim W\) is strictly bigger than \(0\). So: \begin{align} \dim null\ T &\geq \dim V - \dim W \\ &> 0 \end{align} And so, the dimension of the null space of \(T\) is not \(0\). Therefore, the null space of \(T\) can’t have been \(\{0\}\) because that does have dimension \(0\). This makes the map not injective because injectivity implies that null space is \(\{0\}\) map to bigger space is not surjective Its basically the same thing as the one above. Suppose \(V,W\) are finite-dimensional vector spaces, and \(\dim V < \dim W\). Then, all \(T \in \mathcal{L}(V,W)\) are not injective. We first have that: \begin{align} &\dim V = \dim null\ T + \dim range\ T \\ \Rightarrow\ & \dim range\ T = \dim V - \dim null\ T \end{align} Because the dimension of \(null\ T\) is larger than \(0\) (or, for that matter, the dimension of anything), \(\dim V - \dim\ null\ T \leq \dim\ V\). Hence: \begin{align} & \dim range\ T = \dim V - \dim null\ T \\ \Rightarrow\ & \dim range\ T \leq \dim V \end{align} Now, recall that \(\dim V < \dim W\). \begin{align} & \dim range\ T = \dim V - \dim null\ T \\ \Rightarrow\ & \dim range\ T \leq \dim V < \dim W \end{align} Given the range of \(T\) is smaller than the codomain of \(T\), they cannot be equal spaces. So, \(T\) is not surjective.