Let $e_1=(1,1,0,0)$ and $e_2=(1,1,0,1)$ and $e_3=(0,0,0,1)$. Let $E=\operatorname{span}(e_1,e_2,e_3)$. I want to determine the dimension of the vector subspace $E$. And then complete into a basis of $\mathbb R^4$.
My try: We have that $e_3=e_2-e_1$ so the family $e_1,e_2,e_3$ is linearly dependent. The family $e_1,e_2$ is linearly independent, since it generates $E$ then it is a basis of $E$ and hence $\dim(E)=2$.
Now by exchange lemma we know that we this linearly independent family can be extended into a basis of $\mathbb R^4$ and the vectors to be added are be chosen from a generating family of $\mathbb R^4$ for example the canonical basis of $\mathbb R^4$, $u_1=(1,0,0,0)$, $u_2=(0,1,0,0)$, $u_3=(0,0,1,0)$ and $u_4=(0,0,0,1)$. We have to add the vectors and check for linear independence. We can't add $u_4$ since we know that $u_4=e_3$ and we already checked that $e_1,e_2,u_4$ is linearly dependent. We add $u_1$ and we verify that $e_1,e_2,u_1$ is linearly independent. It remains another vector to have a basis. We take $u_2$ but we show that the family $e_1,e_2,u_1,u_2$ is not linearly independent. Hence we can be sure without checking that $ e_1,e_2,u_1,u_3$ is linearly independent and hence it is a basis for $\mathbb R^4$.
Is my reasoning correct and is this the standard method to complete a linearly independent family into a basis? thank you for your clarifications!
Best Answer
You need not choose vectors from the canonical basis for $\mathbb{R}^4$; any basis will do. There may be more convenient bases from which you can add vectors, if it's easier to prove that those vectors are linearly independent with the vectors you already have.
Using the canonical basis is certainly not wrong (and it is an intuitive choice), but you should know that there is no property unique to this basis that makes this process work.