Question

determine if a set of vectors forms a basis for Rn.
(Can someone do an example of this please??)

Answer 1

any bunch of n vector independents

Answer 2

for R³ could be (1,0,0);(0,1,0);(0,0,1)

Answer 3

with linear combination of this three vector you can fill the whole R³ space

Answer 4

A basis is a set of vectors which span the space. As @RaphaelFilgueiras said.

Answer 5

Given a set of vectors, you would check to see if they span of the vector space in question (i.e. any vector in the given space can be written as a linear combo of the given vectors), and you would check to see if they're linearly independent.

Continuing with @RaphaelFilgueiras's example, suppose you're told to show that the set \(\left\{(1,0,0)~,~(0,1,0)~,~(0,0,1)\right\}\) is a basis of \(\mathbb{R}^3\).

One of the things you have to do is show \(\text{span}\left\{(1,0,0)~,~(0,1,0)~,~(0,0,1)\right\}=\mathbb{R}^3\).

Let \((x,y,z)\in\mathbb{R}^3\) be some vector. Show that \((x,y,z)=c_1(1,0,0)+c_2(0,1,0)+c_3(0,0,1)\) for any \(x,y,z\in\mathbb{R}\).

Basically, you have the system of equations
\[\begin{cases}
c_1=x\\
c_2=y\\
c_3=z
\end{cases}\]
So, letting \(c_1=x,c_2=y,c_3=z\) means you can indeed write any vector in the space as the linear combination of the given set of vectors. Thus \(\text{span}\left\{(1,0,0)~,~(0,1,0)~,~(0,0,1)\right\}=\mathbb{R}^3\).

Lastly, show that the given set of vectors is linearly independent. This basically means you can write the zero vector as a linear combination of the vectors only if the coefficients in front of them are all zero.
\[c_1(1,0,0)+c_2(0,1,0)+c_3(0,0,1)=(0,0,0) ~ \iff ~c_1=c_2=c_3=0\]
This should be fairly clear for these vectors once you set it up as a system of equations. You'll see that the only possible solutions are \(c_1=c_2=c_3=0\). Thus the set of vectors is linearly independent.

And there you have it: \(\left\{(1,0,0)~,~(0,1,0)~,~(0,0,1)\right\}\) is a basis of \(\mathbb{R}^3\).