Question

Prove that 2 linearly independent vectors in \(\mathbb{R}^2\) is a basis for \(\mathbb{R}^2\).

Answer 1

I think you need to show the vectors span \(\mathbb R^2\)

Answer 2

How about this:
\[
\mathbf{u}=
\begin{bmatrix}
u_1\\
u_2
\end{bmatrix}
, \, \mathbf{v}=
\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}
,\, \mathbf{b}=
\begin{bmatrix}
b_1\\
b_2
\end{bmatrix}\\
\text{We have to prove that for all }\mathbf{b}\in\mathbb{R}^2,
\begin{bmatrix}
\mathbf{u}&\mathbf{v}
\end{bmatrix}
\begin{bmatrix}
x_1\\
x_2
\end{bmatrix}
=\mathbf{b}
\text{ has a unique solution.}\\
\]
The matrix \(\begin{bmatrix}\mathbf{u}&\mathbf{v}\end{bmatrix}\) has full rank. Therefore, it is invertible. Thus, the equation has a unique solution.

Am I doing circular reasoning?

Answer 3

i think so, the statement we're trying to prove is equivalent to below :
prove that a basis of "n" dimensional vector space contains exactly "n" vectors

Answer 4

which is equivalent to proving :
prove that the vectors in a full rank nxn matrix span the n dimensional space

Answer 5

so i don't think we can use rank here..

Answer 6

But the definition of dimension of a vector space is the number of vectors in the basis of that vector space...

Answer 7

yes, so is it satisfying to conclude like this : since basis of n dimensional vector space contains 2 independent vectors, the 2 given independent vectors represent a basis. QED.

? i thought we need to say something about why "n" given vectors represent a basis for "n" dimensional vector space