Showing a set is a basis for a vector space

[penroseandpaper]

If I'm given a set of four vectors, such as A={(0,1,4,2),(1,0,0,1)...} and am given another set B, whose vectors are given as a form such as (x, y, z, x+y-z) all in ℝ, what steps are needed to show A is a basis of B?

I have calculated another basis of B, and found I can use linear combinations of the vectors in this basis to make each of the four vectors in A. But I'm not sure if I can use that as proof or if it means anything.

No answers being sought, simply a checklist of steps to take. The set notation including (x, y, z, x+y-z) has thrown me.

Penn

 

[fresh_42]

$A$ is a basis of $B$ makes only sense, if $B$ is a linear space. So I assume we have to decide whether $A$ is a basis for the linear span $\operatorname{lin}B$ of the vectors in $B$. Let's call this vector space $\mathcal{B}=\operatorname{lin}B$ in order to distinguish it from the set of vectors $B$ which you used.

The necessary steps are as follows:

1. $A \subseteq \mathcal{B}$, i.e. each vector $\vec{a}\in A$ must have a linear combination of vectors from $B$, i.e. $\vec{a}=\sum_{i=1}^k \lambda_i\vec{b}_i$ with $\lambda_i\in \mathbb{R},\vec{b}_i\in B$.
2. $A$ must span $B$, i.e. the other way around must also hold: each vector $\vec{b}\in B$ must have a linear combination of vectors from $A$, i.e. $\vec{b}=\sum_{i=1}^k \mu_i\vec{a}_i$ with $\mu_i\in \mathbb{R},\vec{a}_i\in A$.
3. $A=\{\vec{a}_1,\ldots,\vec{a}_n\}$ must be linearly independent, i.e. from $\vec{0}=\sum_{i=1}^n x_i\vec{a}_i$ must follow, that the equation can only hold if $x_1=\ldots=x_n=0.$

 

[martinbn]

Are you sure about the problem? B is 3 dimensional, 4 vectors cannot be a basis.