Basis

A basis (plural: bases) is the smallest number of Vectors that span a Vector Space. For this to happen, all the vectors forming the basis must be linearly independent.

Definition §

A basis for the vector space $V_{\mathbb{F}}$ is a set of vectors, $S_k=\{{\vec{v}}_1,{\vec{v}}_2,\dots {\vec{v}}_k\}|{\vec{v}}_{1\dots k}\in V$ . Then:

$$V=\text{span}(S_k)$$

$$S_k\text{ contains linearly independant vectors only}$$

Bases are non-unique for some vector spaces. For example, any two linearly independent vectors in ${\mathbb{R}}^2$ can form the basis for ${\mathbb{R}}_{\mathbb{R}}^2$

The cardinality of a basis ^theorem $V_{\mathbb{F}}$ be a vector space (fixed over field $\mathbb{F}$) and let $B_k=\{{\vec{v}}_1,{\vec{v}}_2,\dots ,{\vec{v}}_k\}$ be the basis of $V_{\mathbb{F}}$. In other words, the dimension of $V_{\mathbb{F}}$ is $k$

Let

• Any subset of $V_{\mathbb{F}}$ with less than $k$ vectors cannot span $V_{\mathbb{F}}$.
• Any subset of $V_{\mathbb{F}}$ with more than $k$ vectors is linearly dependent
• Any bases of $V_{\mathbb{F}}$ must contain exactly $k$ vectors.

Because every spanning set is a subspace, a basis can be extracted from a spanning set, by finding only the linearly independent vectors in it:

Orthonormal Basis §

An orthonormal basis is a set of vectors that has every vector being a unit vector (having a norm of 1) and being orthogonal to every other vector:

Finding a Basis §

Extracting a Basis from a Spanning Set

Standard Basis §

For ${\mathbb{R}}^n$ §

The standard bases for ${\mathbb{R}}^n$ is given by:

$$\mathcal{S}=\{(1,0,0,0,\dots ),(0,1,0,0\dots ),\dots (\dots ,0,0,1)\}$$

The vectors are usually denoted as:

$$\mathcal{S}=\{{\vec{e}}_1,{\vec{e}}_2,{\vec{e}}_3,\dots {\vec{e}}_n\}$$

In ${\mathbb{R}}^3$, they are denoted by $\{\vec{i},\vec{j},\vec{k}\}$

The Dimension of ${\mathbb{R}}^n$ is $n$

For Matrices §

The standard bases for $M_{m,n}(\mathbb{F})$ is given by:

$$\mathcal{S}=\{\left[\begin{array}{ccc}1& 0& \dots \\ 0& 0& \dots \\ \dots & \dots & \dots \end{array}\right],\left[\begin{array}{ccc}0& 1& \dots \\ 0& 0& \dots \\ \dots & \dots & \dots \end{array}\right],\dots ,\left[\begin{array}{ccc}0& 0& \dots \\ 0& 0& \dots \\ \dots & \dots & 1\end{array}\right]\}$$

The dimension of $M_{m,n}(\mathbb{F})$ is $m\times n$

For Polynomials §

The standard bases for ${\mathcal{P}}_n(\mathbb{F})$ is given by:

$$\mathcal{S}=\{1,x,x^2,\dots x^n\}$$

The dimension of ${\mathcal{P}}_n(\mathbb{F})$ is $n+1$