Linear Maps
Linear mapping in linear algebra.
Table of Contents
1. Linear Maps
- Linear Map
- A linear map from \(V\) to \(W\) is a function \(T:V \mapsto W\) with the property of addition \(T(u+v)=Tu+Tv, \forall u,v \in V\) and homogeneity \(T(\lambda u)=\lambda T u, \forall \lambda \in F, u \in V\).
The set of all linear maps from \(V\) to \(W\) is denoted \(\mathcal{L}(V,W)\)
Suppose \(v_{1}, \dots, v_{n}\) is a basis of \(V\) and \(w_{1}, w_{n} \in W\), then there exists a unique linear map \(T: V \mapsto W\) such that
\[ T v_{j} = w_{j}, \forall j=1,\dots,n \]
1.1. Null Spaces
- Null Space
- For \(T \in \mathcal{L}(V,W)\), the null space of \(T\), denoted \(\mathop{\mathrm{null}} T\) is the subset of \(V\) consisting of those vectors that \(T\) maps to \(0\), i.e., \[ \mathop{\mathrm{null}} T = \set{v \in V: Tv = 0} \]
Suppose \(T \in \mathcal{L}(V,W)\), then \(\mathop{\mathrm{null}} T\) is a subspace of \(V\).
Let \(T \in \mathcal{L}(V,W)\), then \(T\) is injective (one-to-one) if and only if \(\mathop{\mathrm{null}} T=\set{0}\)
1.2. Ranges
- Range
- For \(T:V\mapsto W\), the range of \(T\) is the subset of \(W\) consisting of those vectors that are of the form \(Tv\) for some \(v \in V\), i.e., \[ \mathop{\mathrm{range}} T = \set{Tv: v\in V} \]
1.3. Fundamental Theorem of Linear Maps
Fundamental Theorem of Linear Maps
Suppose \(V\) is finite-dimensional and \(T \in \mathcal{L}(V,W)\), then \(\mathop{\mathrm{range}} T\) is finite-dimensional and
\[ \dim V=\dim\mathop{\mathrm{null}} T + \dim \mathop{\mathrm{range}}T \]
Suppose \(\mathop{\mathrm{null}} T\) has basis \(u_{1}, \dots, u_{m}\), therefore, we can find one set of basis of \(V\)
\[ u_{1}, \dots, u_{m}, v_{1}, \dots, v_{n} \]
First, we prove \(\mathop{\mathrm{range}} T\) is finite-dimensional. For any vector \(w \in V\), we have
\[ w = a_{1}v_{1} + \dots + a_{n}v_{n} + b_{1}u_{1} + \dots + b_{m}u_{m} \]
Multiply by \(T\) on both sides, since \(T u_{i} = 0\) as \(u_{i} \in \mathop{\mathrm{null}} T\)
\[ Tw = a_{1}Tv_{1}+\dots+a_{n}Tv_{n} \]
Therefore, \(\mathop{\mathrm{range}} T\) is spanned by \(Tv_{i}\), so it’s finite-dimensional.
Next, we prove that \(Tv_{i}\) forms a basis. Consider
\[ T(c_{1}v_{1}+\dots +c_{n}v_{n})=0 \]
If \(w=c_{1}v_{1}+\dots +c_{n}v_{n} \ne 0\), then \(w\in\mathop{\mathrm{null}}T\), therefore
\[ w = d_{1}u_{1} + \dots + d_{m}u_{m} = c_{1}v_{1} + \dots + c_{n}v_{n} \]
So \(c_{1}v_{1}+\dots +c_{n}v_{n}-d_{1}u_{1} -\dots-d_{m}u_{m}=0\), which says that \(v_{1},\dots,v_{n},u_{1},\dots,u_{m}\) is not basis of \(V\), contradiction!
Therefore \(w=0\), so
\[ c_{1}v_{1}+\dots +c_{n}v_{n}=0 \]
This implies \(c_{i}=0\) as \(v_{i}\) are orthogonal to each other because they are a part of the basis of \(V\). Therefore, we proved that \(Tv_{i}\) is the basis of \(\mathop{\mathrm{range}}T\), so \(\dim\mathop{\mathrm{range}}T=n\)