Given a set $A=(x_1,x_2,x_3,x_4,x_5,x_6)$ in $\mathbb{R}^6$, that follows
$$5x_1+x_2+x_3+x_4+x_5+5x_6=0$$
How to find a basis and dimension to its subspace?
By now, I only proved that the set is linear dependent, but I have no idea about how to continue it.
HINT
Take a vector $x\in\mathbb{R}^{6}$ and apply the restriction related to the set $A$. After so, you'll get
\begin{align*} x & = (x_{1},x_{2},x_{3},x_{4},x_{5},x_{6})\\\\ & = (x_{1},-5x_{1} - x_{3} - x_{4} - x_{5} - 5x_{6},x_{3},x_{4},x_{5},x_{6})\\\\ & = x_{1}(1,-5,0,0,0,0) + x_{3}(0,-1,1,0,0,0) + x_{4}(0,-1,0,1,0,0) + x_{5}(0,-1,0,0,1,0)\\\\ & + x_{6}(0,-5,0,0,0,1) \end{align*}
Based on such result, we conclude that \begin{align*} A = \text{span}\{(1,-5,0,0,0,0),(0,-1,1,0,0,0),(0,-1,0,1,0,0),(0,-1,0,0,1,0),(0,-5,0,0,0,1)\} \end{align*}
It also can be proven that such set is LI. Consequently, it is also a basis. Finally, we deduce that $\dim A = 5$.
Hopefully this helps!