Andrew Thangaraj

Aug-Nov 2020

- Vector space \(V\) over a scalar field \(F\)
- \(F\): real field \(\mathbb{R}\) or complex field \(\mathbb{C}\) in this course

- Linear map \(T:V\to W\)
- \(T(au+bv)=aT(u)+bT(v)\)

- Matrix of linear map with respect to bases for \(V\) and \(W\)
- Basis for \(V\): \(\{v_1,\ldots,v_n\}\)
- Column \(j\): coordinates of \(T(v_j)\) with respect to basis of \(W\)

- We will start studying some important aspects of linear maps in this lecture

\(T:V\to W\) is a linear map. The null space of \(T\), denoted null \(T\), is defined as \[\text{null } T=\{v\in V:Tv=0\}.\] null \(T\): subset of vectors that get mapped to 0 by \(T\)

*Examples*

- \(T=0\), null \(T=V\)
- \(T=1\), null \(T=\{0\}\)
- \(D\): Differentiation of polynomials
- null \(D\) = constant polynomials

- Multiplication of polynomials by \(x^2\)
- null space \(= \{0\}\)

- \(T(x,y)=x+2y\)
- null \(T=\{(x,y):x=-2y\}=\text{span}\{(-2,1)\}\)

- \(T(x,y,z)=(x,x+y)\)
- null \(T=\{(0,0,z):z\in\mathbb{R}\}\), \(z\) axis

\(T:V\to W\) is a linear map. null \(T\) is a subspace of \(V\).

*Proof*: If \(u,v\in\text{null }T\), \(au+bv\in\text{null }T\)

*Corollary*: A linear map \(T\) always maps \(0\) to \(0\).

A map \(T:V\to W\) is said to be *injective* if \(Tu=Tv\) implies \(u=v\).

*Examples*: identity, multiplication by \(x^2\) are injective; zero, differentiation are not injective

A linear map \(T:V\to W\) is injective iff null \(T=\{0\}\).

- \(T\): injective
- if \(v\in\text{null }T\), \(Tv=0=T0\). So, \(v=0\).

- null \(T=\{0\}\)
- \(Tu=Tv\) implies \(T(u-v)=0\). So, \(u-v\in\text{null }T\) or \(u-v=0\).

\(T:V\to W\) is a map. The range of \(T\), denoted range \(T\), is defined as \[\text{range } T=\{Tv:v\in V\}.\] range \(T\): set of outputs of \(T\)

*Examples*

- \(T=0\), range \(T=\{0\}\)
- \(T=1\), range \(T=V\)
- \(D\): Differentiation of polynomials
- range \(D\) = all polynomials

- Multiplication of polynomials by \(x^2\)
- range = polynomials with constant zero and coefficient of \(x\) zero

- \(T(x,y)=x+2y\)
- range \(T=\mathbb{R}\)

- \(T(x,y,z)=(x,x+y)\)
- range \(T=\mathbb{R}^2\), \(x{-}y\) plane

\(T:V\to W\) is a linear map. range \(T\) is a subspace of \(W\).

*Proof*

If \(w_1,w_2\in\text{range }T\), we have \(Tv_1=w_1\) and \(Tv_2=w_2\)

\(aw_1+bw_2=T(av_1+bv_2)\in\text{range }T\)

A map \(T:V\to W\) is said to be *surjective* if range \(T=W\).

*Examples*

- identity, differentiation are surjective
- zero, multiplication by \(x^2\) are not surjective
- \(T(x,y)=x+2y\), \(T(x,y,z)=(x,x+y)\) are surjective

Suppose \(V\) is finite-dimensional and \(T:V\to W\) is a linear map. Then, range \(T\) is finite-dimensional and \[\text{dim } V = \text{dim null } T + \text{dim range } T\]

*Proof sketch*

\(\{u_1,\ldots,u_k\}\): basis for null \(T\)

\(\{u_1,\ldots,u_k,v_1,\ldots,v_l\}\): extension of above basis to basis of \(V\)

Show \(\{Tv_1,\ldots,Tv_l\}\) is a basis for range \(T\)

\(T:V\to W\) is a linear map, \(V\): finite-dimensional

null \(T\): mapped to \(0\)

\(\{u_1,\ldots,u_k\}\): basis for null \(T\)

\(\{u_1,\ldots,u_k,v_1,\ldots,v_l\}\): extension

Vectors of the form: \(av_1+\) vector from null \(T\)

- mapped to \(a\;Tv_1\)

Vectors of the form: \(bv_2+\) vector from null \(T\)

- mapped to \(b\;Tv_2\)

and so onâ€¦

Each mapping above is to linearly independent vectors