$Av=λv$ basis proof problems So the question asks: Let A be a $2\times 2$ matrix, and suppose that there exist scalars $λ_1,λ_2$ in$\mathbb{R}$ and nonzero vectors $v_1,v_2$ in $\mathbb{R}$ such that $Av_1=λ_1v_1, Av_2=λ_2v_2$, and $λ_1$ does not equal to $λ_2$.Prove that the vectors $v_1,v_2$ form a basis of $R^2$. 
So so far I got: 
As the eigenvector $v_1,v_2$ are corresponding to two different eigen values$λ_1,λ_2$ respectfully. 
Therefore the eigenvectors $v_1,v_2$ are linearly independent. 
And since $v_1,v_2$ spans A in $\mathbb{R}$. 
Hence,  $v_1,v_2$ forms a basis of $\mathbb{R}$. 
Does this proof look right? Do I need to write a linear combination of $v_1,v_2$ to show they span A? 

Establishing the following facts would suffice:


*

*Different eigenvalues correspond to linearly independent vectors.

*Any linearly independent list whose length is equal to dimension is a basis.
You need to explain why "Therefore the eigenvectors $v_1,v_2$ are linearly independent." and "And since $v_1,v_2$ spans $A$ in $\mathbb{R}^2$"
