Find basis and dimension in $\mathbb{R}^6$

Question

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.

Gram-Schmidt allows you to find the maximal set of orthogonal vectors that span a space. Try it on $A$ to see if you get $6$. (Evidently, you will not.). Then move to $5$. — David G. Stork, Dec 19 '20 at 02:55
@David Gram-Schmidt is used to orthogonalize a known basis, not for finding the basis. How do you suggest we "try it on$A$"? Am I missing something? — Shubham Johri, Dec 19 '20 at 03:03
Take two (distinct) elements from the set of $A$. Form an orthogonal basis (using GS). Add a third element from $A$ to your two; use GS to construct an orthogonal 3-basis. Continue until it cannot be done. — David G. Stork, Dec 19 '20 at 03:15

score 2 · Accepted Answer · edited Dec 19 '20 at 02:53

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!

Find basis and dimension in $\mathbb{R}^6$

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