In [math]\mathbb{R}^2[/math], we consider two special vectors [math]\hat{\mathbf{i}}=\begin{pmatrix}1 \\ 0 \end{pmatrix}, \hat{\mathbf{j}}=\begin{pmatrix} 0 \\ 1 \end{pmatrix} [/math]. Then it is easy to see that for any vector [math]v=\begin{pmatrix} a \\ b \end{pmatrix}[/math], [math] v = a\begin{pmatrix}1 \\ 0 \end{pmatrix}+b\begin{pmatrix} 0 \\ 1 \end{pmatrix}=a\hat{\mathbf{i}}+b\hat{\mathbf{j}}[/math] i.e. v is a linear combination of [math]\hat{\mathbf{i}}[/math] and [math] \hat{\mathbf{j}}[/math] with weights [math] a [/math] and [math] b [/math]. Therefore, [math]\text{Span}\{\hat{\mathbf{i}},\hat{\mathbf{j}}\}=\mathbb{R}^2[/math]. And [math]\{\hat{\mathbf{i}},\hat{\mathbf{j}}\}[/math] is called the [b]standard basis [/b] for [math]\mathbb{R}^2[/math].[br][br]In [math]\mathbb{R}^3[/math], we have three special vectors [math]\hat{\mathbf{i}}=\begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}, \hat{\mathbf{j}}=\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, \hat{\mathbf{k}}=\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} [/math]. Similarly, for any vector [math]v=\begin{pmatrix} a \\ b \\ c \end{pmatrix}[/math], [math] v = a\begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}+b\begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix}+c\begin{pmatrix} 0 \\ 0 \\ 1\end{pmatrix}=a\hat{\mathbf{i}}+b\hat{\mathbf{j}}+c\hat{\mathbf{k}}[/math]. Therefore, [math]\text{Span}\{\hat{\mathbf{i}},\hat{\mathbf{j}},\hat{\mathbf{k}}\}=\mathbb{R}^3[/math]. And [math]\{\hat{\mathbf{i}},\hat{\mathbf{j}},\hat{\mathbf{k}}\}[/math] is called the [b]standard basis [/b] for [math]\mathbb{R}^3[/math].[br][br]More generally, in [math]\mathbb{R}^n[/math], we can define the following standard basis in a similar way:[br][br][math]e_1=\begin{pmatrix}1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, e_2=\begin{pmatrix} 0 \\ 1 \\ \vdots \\ 0\end{pmatrix}, \ldots , e_n=\begin{pmatrix} 0 \\ \vdots \\ 0 \\ 1 \end{pmatrix}[/math].[br][br]The following is an illustration of a vector as a linear combination of the standard basis in [math]\mathbb{R}^3[/math]:[br]

There are two good features of the standard basis for [math]\mathbb{R}^n[/math]:[br][br][list][*]Any vector in [math]\mathbb{R}^n[/math] can be written as a linear combination of the standard basis.[/*][*]There is no "redundant vector" in the standard basis i.e. the standard basis is linearly independent.[/*][/list][br]In fact, any set of vectors that satisfies the above two features is called a [b]basis[/b] for [math]\mathbb{R}^n[/math] i.e. for any set of vectors [math]S[/math], [math]S[/math] is a basis if [math]\text{Span} S=\mathbb{R}^n[/math] and [math]S[/math] is linearly independent.[br][br]Writing a vector as a linear combination of a basis is like resolving a force into components in physics. Sometimes this can make vector calculation much easier. Moreover, there are many different choices of basis that can suit different purposes.[br][br]The standard basis is called “standard” because when a vector is written as a linear combination of the standard basis, the weights are exactly the coordinates of the arrowhead. However, if a different basis is used, the weights of the corresponding linear combination will obviously be different. In fact, such weights can be regarded as the “new coordinates” with respect to this new basis. It can be shown that[br][br][list][*]Each vector is a linear combination of the basis. (Because by definition, any basis spans the whole space)[/*][*]Each vector has a unique set of weights when it is written as a linear combination of the basis. (Why?)[br][/*][/list]

In [math]\mathbb{R}^n[/math], we have a standard basis which consists of n vectors. Now suppose we consider any other basis for [math]\mathbb{R}^n[/math], you may guess that there should also be n vectors in the basis. And the following theorem confirms this:[br][br][b]Theorem[/b]: Any two basis in [math]\mathbb{R}^n[/math] have the same number of vectors.[br][br]Here is a nice [url=https://www.khanacademy.org/math/linear-algebra/vectors-and-spaces/null-column-space/v/proof-any-subspace-basis-has-same-number-of-elements?modal=1]video[/url] on the proof of this theorem in Khan Academy.[br][br]Therefore, we define [b]dimension[/b] of the space to be the number of vectors in a basis. Obviously, for any natural number n, [math]\mathbb{R}^n[/math] has dimension n.[br][br][br]The following are some T/F questions that test your understanding of the basis: