Basis and dimension

Finite representation of subspaces

Goal: defining basis and dimension

A common theme is the extraction of a set of linearly independent vector from a subspace that contains enough information to represent the subspace as a whole.

That is the job of "basis". We will define this concept more precisely.

A by-product is the concept of "dimension".

A basis of the plane

Verify that the set \[ \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix} \,,\, \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right\} \] is linearly independent. Also verify that this set spans $\mathbb{R}^2$.

That is, this subset of $\mathbb{R}^2$ is both linear independent and spanning.

These properties make it very special. In particular, every vector in $\mathbb{R}^2$ can be written as a linear combination of this set. E.g., \[ \begin{bmatrix} 3 \\ 5 \end{bmatrix} = 3 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + 5 \begin{bmatrix} 0 \\ 1 \end{bmatrix}. \] And this can be done in a unique way. (Must be 3 and 5; no other set of coefficients can make this work)

Exercises

Show the two sets \[ \begin{aligned} \mathcal{B}_1 &= \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix} , \begin{bmatrix} 1 \\ 1 \end{bmatrix} \right\} & \mathcal{B}_2 &= \left\{ \begin{bmatrix} 1 \\ 2 \end{bmatrix} , \begin{bmatrix} 3 \\ 5 \end{bmatrix} \right\}. \end{aligned} \] are both linearly independent, and they are both spanning sets of $\mathbb{R}^2$.

They are examples of basis of $\mathbb{R}^2$.

Basis

For a subspace $V$ of $\mathbb{R}^n$ (including the case where $V = \mathbb{R}^n$), a subset $B$ of $V$ is said to be a basis of $V$ if
  • $B$ is linearly independent, and
  • $B$ spans $V$ (i.e., $V = \operatorname{span}(B)$).

Stated intuitively, a basis of a subspace is a subset of vectors in this subspace that is

  • small enough to be linearly independent, and
  • large enough to span the subspace.

A basis is not too big, not too small

We can push this intuitive idea of bases having just the "right size" even further.

Let $B$ be a basis of a subspace $V$ of $\mathbb{R}^n$, then

  • For any $\mathbf{v} \in V \setminus B$, $B \cup \{ \mathbf{v} \}$ is linearly dependent.
  • For any $\mathbf{v} \in B$, the set $B \setminus \{ \mathbf{v} \}$ is not a spanning set of $V$.

Dimension

The dimension of $V$ is the number of elements in a basis of $V$ and is denoted by $\dim V$.

Implicit in this statement is the fact that any two basis for $V$ must be of the exact same size.

I.e., if $B_1$ and $B_2$ are both bases of a subspace $V$ of $\mathbb{R}^n$, then $| B_1 | = | B_2 |$. This fact is not completely obvious. Indeed, it is a part of the "Basis Theorem". One intuitive explanation of this important fact is that a basis has to be of just the "right size".

Exercise. Can you explain why this has to be true? (Try the special case of $n=2$)