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10.22 Linear combination

Linear combination

Good list

Introduction

\((V=\R^2\) or \(V=\mathbb{F}^2)\)

  1. Is there an \(m\in \N\) and a list \(v_1,v_2,...,v_m\) such that every \(v\in V\) can be written as a linear combination of them? (\(v=a_1v_1+...+a_mv_m\)) The list of vector is fixed, so the important thing is the coefficients (\(a_1,a_2,...,a_m\)​)

    Yes. \(m=2\) is the minimum possible.

  2. Does every \(2-\)list do the job

    No, \((1,0),(1,0) \times\) \((0,0),(0,0)\times\) \((1,0),(0,1)\checkmark\)​

    If we have \(\left(a,b\right)\neq\left(0,0),(c,d\right)\), then bad case: \((c,d)=\lambda (a,b)=(\lambda a,\lambda b)\)​

    good case: \(\nexists \lambda\) such that \((c,d)=\lambda(a,b)\)​

  3. Does any \(3-\)list do the job?

    No, A good \(3-\)list is that containing a good \(2-\)list. That is if the \(3\) are not colinear

  4. Given a good \(3-\)list, in how many ways a given \(v\) can be written as a linear combination of them?

    At least two

    \(v,w,z\) (\(v,w\) is good \(2-\)list)

    \(z=0v+0w+1z\)​

    \(z=\lambda v+\mu w+0z\)​

Examples of good \(2-\)lists

  1. \(V=\mathbb{C}^2\), \(v_1=(i,1+i),v_2=(-1,?)\)

    If \(\lambda(i,1+i)=(-1,?)\Rightarrow\lambda=i\Rightarrow?=i-1\)

    If \(?\neq i-1\) good

  2. \(V=\mathbb{Z}_7^2,v_1=\left(3,6\right),v_2=\left(1,?\right)\), \(?\neq 2\) good

  3. \(V=\mathbb{Q}[\sqrt2]^2,v_1=(\sqrt2,-\sqrt2),v_2=(-2,?)\) \(?\neq2\) good

Vectors space and subfields

Definition

Let \(\left(\mathbb{F},+,\times\right)\) be a field.

A subset \(\mathbb{E}\) of \(\mathbb{F}\) (\(\mathbb{E}\subseteq\mathbb{F}\)) is a subfield of \(\mathbb{F}\) if \(\mathbb{E}\) is itself a field with the restrictions of \(+\) and \(\times\) to \(\mathbb{E}\).

That is \(^+|_{\mathbb{E}\times\mathbb{E}}:\mathbb{E}\times\mathbb{E}\rightarrow\mathbb{E}\) and \(^\times|_{\mathbb{E}\times\mathbb{E}}:\mathbb{E}\times\mathbb{E}\rightarrow\mathbb{E}\)

Remark

\(0\in\mathbb{F}\) is the neutral element of \(\mathbb{E}\)

\(1\in\mathbb{F}\) is the identity element of \(\mathbb{E}\)

If \(a\in \mathbb{E}\), its opposite in \(\mathbb{E}\) is \(-a\) (the one in \(\mathbb{F}\))

Let \(a'\in \mathbb{E}\) such that \(a+a'=0\Rightarrow(-a)+(a+a')=-a+0\Rightarrow a'=-a\)

If \(a\in \mathbb{E},a\neq 0\), its inverse in \(\mathbb{E}\) is \(a^{-1}\) (the one in \(\mathbb{F}\))

Let \(a'\in \mathbb{E}\) such that \(a\cdot a^{\prime}=1\Rightarrow a^{-1}\cdot(a\cdot a^{\prime})=a^{-1}\cdot1\Rightarrow a^{\prime}=a^{-1}\)

Given a \(\mathbb{F}\)-vector space \(V\), it follows that it is also an \(\mathbb{E}\)-vector space with the following:

\(+:V\times V\rightarrow V\) and \(\cdot :\mathbb{E}\times V\rightarrow V\)

Field can narrow into a subfield, but if a subfield expands to the field, it will have problem.

Example

  1. Every \(\mathbb{C}\)-vector space is an \(\mathbb{R}\)-vector space (Every \(\mathbb{F}\)-vector space is an \(\mathbb{E}\)-vector space)

    That is said, you can use \(\mathbb{C}\)-vector space as \(\mathbb{R}\)-vector space because \(\mathbb{R}\subseteq\mathbb{C}\), but not vice versa

  2. Every \(\R\)-vector space is a \(\mathbb{Q}\)-vector space

  3. \(V=\mathbb{C}\), As a \(\mathbb{C}\)-vector space: \(1\in V=\mathbb{C}\) (vector), given \(v\in V=\mathbb{C}\) (vector).

    Let \(\lambda=v\in\mathbb{C}\) (number), then \(\lambda \cdot 1=v\)

    The list: 1, is a good \(1-\)list

    As an \(\R\)-vector space: \(V=\mathbb{C}\sim\mathbb{R}^2\) (as a set). The sum in \(\mathbb{C}\) is the same as in \(\R^2\)​

    The multiplication by real scalars in \(\mathbb{C}\) is the same as the multiplication by real sacalars in the \(\R\)-vector space \(\R^2\)​

    A good list must have at least \(2\)-vectors

    image​

    \(\mathbb{R}^2\sim\mathbb{C}\)

    \(v,w\) are not \(\R\)-colinear

    The linear combination of the list: \(v\), are just the complex numbers on the line generated by \(v,Lv\)​

    The linear combination of \(2\)-list: \(v,w\) are all the complex numbers

    image​

More General

\(V=\mathbb{R}\begin{cases}as~\R-vector~space:\lambda\cdot 1=\lambda\\as~\mathbb{Q}-vector~space:let~v_1,v_2,...,v_m\text{ be a m-list of vectors in V=}\R\end{cases}\)

The set of \(\mathbb{Q}\)-linear combination of \(v_1,...,v_m\) is \(\{\lambda_1v_1+\cdots+\lambda_{m}v_{m}:\lambda_{i}\in\mathbb{Q}\}\subsetneq\mathbb{R}\)

\(\{\lambda_1v_1+\cdots+\lambda_{m}v_{m}:\lambda_{i}\in\mathbb{Q}\}\sim\left\lbrace\left(\lambda_1,\ldots,\lambda_{m}\right):\lambda_{i}\in\mathbb{Q}\right\rbrace\sim\mathbb{Q}^{m}\)​

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