Basis and Dimension

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 combinations
    • \(a_1v_1+a_2v_2+\cdots\) for \(v_i\in V\) and \(a_i\in F\)
  • Span\((v_1,\ldots,v_n)\)
    • All linear combinations
  • Subspace
    • Subset closed under linear combinations
  • Linearly independent set of vectors
    • No non-trivial linear combination is zero
  • Question: How to establish linear independence or dependence?

Finite dimensional vector spaces

Spanning set of vectors: \(\{v_1, v_2, \ldots, v_n\}\) is a spanning set if \(V=\text{span}(v_1, v_2, \ldots, v_n)\)

  • Every vector \(v\in V\) can be expressed as a linear combination of vectors of a spanning set.

  • Example: \(F^n\), \(F=\mathbb{R}\) or \(\mathbb{C}\) (denote vectors as rows with comma-separated coordinates)

    • \(F^2\): (1, 0), (0, 1); \(F^3\): (1, 0, 0), (0, 1, 0), (0, 0, 1)

    • \(F^n\): (1, 0,…, 0), (0, 1, 0,…, 0),…, (0,…, 0, 1)

Finite dimensional vector space: \(V\) is said to be finite-dimensional if it has a finite spanning set.

\(F^n\) is finite-dimensional.

  • Polynomials (degree \(\le n\)): Finite-dimensional
    • Spanning set: \(1,x,x^2,\ldots,x^n\)
  • Matrices (\(m\times n\)): Finite-dimensional
    • Spanning set: \(E_{ij}\), matrix with 1 in \((i,j)\)-th position and zero elsewhere
  • Functions (from \(\mathbb{R}\) to \(\mathbb{R}\)): Not finite-dimensional
    • No finite spanning set (can you prove this?)

Linear Dependence Lemma

Suppose \(\{v_1,v_2,\ldots,v_m\}\) is a linearly dependent list in a vector space \(V\).

  1. There exists \(j\) such that \(v_j\in\text{span}(v_1,\ldots,v_{j-1})\).
    • For \(j=1\), we interpret as \(v_1=0\)
  2. \(\text{span}(v_1,\ldots,v_m)=\text{span}(v_1,\ldots,v_{j-1},v_{j+1},\ldots,v_m)\)
  • Very useful result to manipulate lists of vectors and understanding linear dependence properties.

Linearly independent set vs Spanning set

In a finite-dimensional vector space \(V\), size of any linearly independent set is lesser than or equal to the size of any spanning set.

  • Example: \(F^3\)
    • (1, 2, 3), (-2, 5, 1), (4, -5, 3), (9, 8, -7) are linearly dependent
  • Example: \(F^{10}\)
    • Any set of 9 vectors is not a spanning set.
  • Exercise: In a finite-dimensional vector space, every subspace is finite-dimensional.

Bases

A basis of \(V\) is a set of vectors satisfying the following two properties:

  1. The set of vectors is linearly independent.
  2. The span of the set is equal to \(V\).
  • Standard basis: (1,0,…,0), (0,1,0,…,0),…,(0,…,0,1) in \(F^n\)
  • (1,2), (2,3) in \(F^2\): basis; infinitely many bases…
  • (1,2), (2,3), (3,4) in \(F^2\): not a basis
  • (1,2,3), (2,3,4) in \(F^3\): not a basis

Unique basis representation: Every vector can be represented uniquely as a linear combination of vectors in a basis. The coefficients of the linear combination represent the vector and become its coordinates with respect to the basis.

  • (3,5) = 3 (1,0) + 5 (0,1). So, (3,5) are coordinates in the standard basis.
  • (3,5) = 1 (1,2) + 1 (2,3). So, (3,5) in standard basis is (1,1) in the basis \(\{(1,2),(2,3)\}\).
  • Infinitely many bases => infinitely many coordinates for the same vector over the different bases.

Some results on bases

  • Every spanning set can be reduced to a basis.

    • Proof: Use linear dependence lemma to find linearly dependent vectors and reduce.

  • Every finite-dimensional vector space has a basis.

    • Proof: Start with a finite spanning set and reduce to get a basis.

  • Every linearly independent set can be extended to a basis.

    • Proof: Append spanning set and use linear independence lemma to reduce.

  • Any two bases have the same number of vectors.

    • Proof: size of linearly independent set \(\le\) size of spanning set.

Dimension

The dimension of a finite-dimensional vector space is the length of any basis.

Notation: dim \(V\)

  • Examples
    • dim \(F^n=n\)
    • dim (Polynomials of degree \(\le n\)) \(=\) \(n+1\)
    • dim (\(m\times n\) matrices) \(=\) \(mn\)

Dimension of a subspace: Let \(U\subseteq V\) be a subspace of a finite-dimensional vector space \(V\). \(U\) is finite-dimensional and has a basis. dim \(U\) is the size of a basis of \(U\).

Some results on dimension and bases

  • If \(V\) is finite-dimensional and \(U\) is a subspace of \(V\), dim \(U\) \(\le\) dim \(V\).

    • Proof: size of linearly independent set \(\le\) size of spanning set.

  • A set of linearly independent vectors of size dim \(V\) is a basis for \(V\).

    • Proof: Extend linearly independent set to get the same set as basis.

  • A spanning set of size dim \(V\) is a basis for \(V\).

    • Proof: Reduce spanning set to get the same set as basis.

Problems

  1. Find a basis for the subspace \(\{(x,x,y)\in F^3: x,y\in F\}\).

  2. Find a basis for the subspace \(\{(x,y,z)\in F^3: x+y+z=0\}\).

  3. Find a basis for the subspace \(\{(x_1,x_2,x_3,x_4,x_5)\in F^5: 2x_1=5x_3,x_4+x_5=0\}\).