# Basis and Dimension

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 $\{(x_1,x_2,x_3,x_4,x_5)\in F^5: 2x_1=5x_3,x_4+x_5=0\}$.