Let $W$ denote the subspace of $R^5$ consisting of all the vectors having coordinates that sum to zero. The vectors $$u_1 = (2, −3, 4, −5, 2),$$$$ u_2 = (−6, 9, −12, 15, −6),$$$$ u_3 = (3, −2, 7, −9, 1),$$$$ u_4 = (2, −8, 2, −2, 6),$$$$ u_5 = (−1, 1, 2, 1, −3),$$$$ u_6 = (0, −3, −18, 9, 12),$$$$ u_7 = (1, 0, −2, 3, −2),$$$$ u_8 = (2, −1, 1, −9, 7).$$ generate $W.$ Find a subset of the set $\{u_1, u_2,...,u_8\}$ that is a basis for $W.$
I tried solving the problem using the idea to find a linearly independent subset of the set of the given set $\{u_1, u_2,...,u_8\}$ having the same number of elements as the dimension of $W$ so, that subset will be a basis of $W$ as well, from the replacement theorem.
Now, I took an empty set $S$ and then started putting elements from the set given until the set becomes linearly dependent upon addition of any more elements from the given set.
I started by adding $$u_1 = (2, −3, 4, −5, 2).$$ Then I added $$ u_3 = (3, −2, 7, −9, 1),$$ as $u_2$ and $u_3$ aren't multiples of each other. So, $S$ is linearly independent. Then I added $$u_6 = (0, −3, −18, 9, 12),$$ because, I verified that the addition of it, in $S$ does not make it linearly dependent i.e it stays linearly independent. Finally, I tried to check that whether adding $$u_7 = (1, 0, −2, 3, −2)$$ in the set $S$ keeps it linearly independent or not.
To do this, I chose some scalars $a,b,c,d$ such that $au_1+bu_3+cu_6+du_7=0$. In matrix representation this is equivalent to the equation $Ax=b$ where $A=\begin{pmatrix} 2 && 3 && 0 && 1\\ -3 &&-3 && -3 && 0\\ 4 && 7 && -18 && 2\\ -5 && -9 && 9 && 3\end{pmatrix},x=\begin{pmatrix} a \\ b \\ c \\ d\end{pmatrix}$ and $b=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0\end{pmatrix}.$
Now, since, $\text{rank}(A,b)=\text{rank}(A)=4$ so the system has a unique solution and the solution is thus, $a=b=c=d=0$ and so, we can add $$u_7 = (1, 0, −2, 3, −2)$$ in $S.$
Now, I think we need to check that whether any more elements can be added to keep $S$ a linearly independent set and thus a basis for $W,$ as $\dim (W)\leq \dim (R^5)=5.$ But the thing is, the calculation becomes even bigger and larger. I had a hard time calculating the ranks of the matrices, $(A,b)$ and $A$ using elementary row operations as the calculation was bigger. So, I am kind of tired doing the same thing again for a larger matrix now.
Is there any way, I can find the dimension of the subspace $W$ so that I know whether I should stop now or not ? Is there any simpler way to solve this question which requires lesser calculation? Any tips regarding this will also be helpful.