I have this question that has me pretty stumped. It is as follows:
Assume $A$ is $3 \times 5, v_1, v_2 \in N(A)$ and are not multiples of each other. Assume $(Av_3, Av_4, Av_5)$ is linearly independent. Prove that $(v_1, v_2, v_3, v_4, v_5)$ is a basis of $\mathbb{R}^5.$"
I understand that $v_1$ and $v_2$ are not linear combinations of each other (right?) and $v_3,v_4$ and $v_5$ form a basis, but I don't know how to "word" this very well. Is it right to say that $v_1$ and $v_2$ add to the span of the sequence of vectors? Or that $v_1$ and $v_2$ can also form basis vectors?
I'm still pretty new to Linear and a little lost when it comes to this stuff. So please, please correct me if anything I said doesn't make sense or is incorrect.
You just need to show that $(v_1, v_2, v_3, v_4, v_5)$ are linearly independent.
Since you have information about $Av_i$, the key will be to multiply a linear relation on $v_i$ by $A$.
Now suppose you had a relation $$c_1v_1 + c_2v_2 + c_3v_3 + c_4v_4 + c_5v_5 = 0$$ You want to show all the $c_i$ must be $0$. Multiplying through by $A$ you get a new dependence $$c_3Av_3 + c_4Av_4 + c_5Av_5 = 0$$ since $v_1, v_2$ are in the nullspace of $A$.
From the definition of linear independence, what can you say about $c_3, c_4, c_5$? Can you use this to show all $c_i$ are $0$?