Vector space/Basis/Polynomials/Introduction/Section

Example

The standard vectors in ${}K^{n}$ form a basis. The linear independence was shown in example. To show that they also form a generating system, let

{}v={\begin{pmatrix}b_{1}\\b_{2}\\\vdots \\b_{n}\end{pmatrix}}\in K^{n}\,

be an arbitrary vector. Then we have immediately

{}v=\sum _{i=1}^{n}b_{i}e_{i}\,.

Hence, we have a basis, which is called the standard basis of ${}K^{n}$ .

Example

We consider the ${}K$ -linear subspace ${}U\subset K^{n}$ given by

{}U={\left\{v\in K^{n}\mid \sum _{i=1}^{n}v_{i}=0\right\}}\,.

A basis for ${}U$ is given by the ${}n-1$ vectors

u_{1}=(1,-1,0,0,\ldots ,0),\,u_{2}=(0,1,-1,0,\ldots ,0),\,,\ldots ,u_{n-1}=(0,0,\ldots ,0,1,-1),\,.

These vectors belong evidently to ${}U$ . The linear independence can be checked in ${}K^{n}$ . From an equation

{}\sum _{i=1}^{n-1}a_{i}u_{i}=0\,

we can deduce step by step ${}a_{1}=0$ , ${}a_{2}=0$ , etc. That the system is a generating system follows from

{}{\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\\\vdots \\v_{n-1}\\v_{n}\end{pmatrix}}=v_{1}{\begin{pmatrix}1\\-1\\0\\\vdots \\0\\0\end{pmatrix}}+(v_{1}+v_{2}){\begin{pmatrix}0\\1\\-1\\\vdots \\0\\0\end{pmatrix}}+(v_{1}+v_{2}+v_{3}){\begin{pmatrix}0\\0\\1\\-1\\\vdots \\0\end{pmatrix}}+\cdots +(v_{1}+v_{2}+v_{3}+\cdots +v_{n-1}){\begin{pmatrix}0\\0\\0\\\vdots \\1\\-1\end{pmatrix}}\,,

where the equality in the last row rests on the condition

{}\sum _{i=1}^{n}v_{i}=0\,.

For the complex numbers, the elements ${}1,{\mathrm {i} }$ form a real basis. In the space of all ${}m\times n$ -matrices, that is, ${}\operatorname {Mat} _{m\times n}(K)$ , those matrices where exactly one entry is ${}1$ , and all other entries are ${}0$ , form a basis, see exercise.