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 \(T:V\to W\) preserves linear combinations
- null \(T=\{v\in V:Tv=0\}\), range \(T=\{Tv:v\in V\}\)

- Fundamental theorem of linear maps
- dim null \(T\) + dim range \(T\) = dim \(V\)

- \(m\times n\) matrix A represents a linear map \(T:F^n\to F^m\)
- colspace(A) = range T, null(A) = null(T)

- Matrix-vector multiplication, matrix multiplication

A linear map \(T:V\to W\) is said to be *invertible* if there exists a linear map \(S:W\to V\) (called *inverse* of \(T\)) such that \(ST:V\to V\) is the identity map on \(V\) and \(TS:W\to W\) is the identity map on \(W\).

*Properties*

An invertible linear map has a unique inverse.

Suppose \(S_1,S_2\) are two inverses. \[S_1=S_1I=S_1(TS_2)=(S_1T)S_2=IS_2=S_2\]

\(T^{-1}\): denotes unique inverse when it exists

\(T\) is invertible iff it is injective and surjective.

This is a classic general result about maps.

See book for proof.

Maps that are not injective are non-invertible

- Pick any \(T\) with null \(T\ne \{0\}\)

Maps that are not surjective are non-invertible

- Pick any \(T\) with range \(T\ne W\)

Polynomials: multiplication by \(x^2\)

- Injective, not surjective, non-invertible

Polynomials: left shift

- Not injective, surjective, non-invertible

An invertible linear map is called an *isomorphism*. Two vector spaces are called *isomorphic* if there is an isomorphism between them.

Two finite-dimensional vector spaces over the same field \(F\) are isomorphic iff they have the same dimension.

- Proof
- \(V,W\): isomorphic
- There exists an invertible \(T:V\to W\)
- dim null \(T=0\), dim range \(T=\) dim \(W\)
- Fundamental theorem: dim \(V=\) dim \(W\)

- dim \(V=\) dim \(W\)
- Define \(T\) by mapping basis to basis
- \(T\) is invertible

- \(V,W\): isomorphic

A finite-dimensional vector space \(V\) is isomorphic to \(F^n\), where \(n=\) dim \(V\).

dim \(V=n\), dim \(W=m\), \(\mathcal{L}(V,W)\): vector space of linear maps, \(F^{m,n}\): vector space of \(m\times n\) matrices

\(\mathcal{L}(V,W)\) and \(F^{m,n}\) are isomorphic.

Fix bases for \(V\) and \(W\)

- Define map \(\mathcal{M}:\mathcal{L}(V,W)\to F^{m,n}\) as follows.
- \(T:V\to W\) mapped to matrix with respect to chosen bases.

\(\mathcal{M}\): linear, injective, surjective

dim \(\mathcal{L}(V,W)=\text{dim}(V)\text{dim}(W)\)

A linear map from a vector space to itself is called an *operator*.

\(\mathcal{L}(V)\): set of all operators on \(V\).

Operators are the most important linear maps.

Invertible operators or invertible square matrices are an important class.

When are operators injective, surjective and invertible?

- Polynomials: multiplication by \(x^2\)
- Injective, not surjective, non-invertible

- Polynomials: left shift
- Not injective, surjective, non-invertible

- Finite dimensions?

- Polynomials: multiplication by \(x^2\)

Let \(T:V\to V\) be an operator and let \(V\) be finite-dimensional. Then the following are equivalent: (a) \(T\) is invertible; (b) \(T\) is injective; (c) \(T\) is surjective.

*Proof*

a implies b: by definition

b implies c: Fundamental theorem

- dim null \(T=0\); so, dim \(V=\) dim range \(T\)

c implies a: Fundamental theorem

- dim range \(T=\) dim \(V\); so, dim null \(T=0\)

How to find if a matrix is invertible?

Matrix has to be square

\(m>n\): not surjective

\(m<n\): not injective

If square, the column rank has to be full

Find dimension of column space

Gaussian elimination