Invertible maps, Isomorphism, Operators

Andrew Thangaraj

Aug-Nov 2020

Recap

  • 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

Invertible linear maps

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.

Examples

  • 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

Isomorphism

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

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

Isomorphism of linear maps and matrices

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

Operators

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?

Invertible operators in finite dimensions

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 \(T=0\)

Invertibility of matrices

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