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!