Find dimension and basis of the set of all points in $R^5$ whose coordinates satisfy the relation $x_1+x_2+x_3+x_4=0$ Doesn't a basis in $R^5$ require 5 finite vectors to be a basis? I'm really confused, maybe theres something I am missing on how $R^5$ can have 4 coordinates that can add up to zero? Does that imply that the 5th coordinate is also 0?
 A: Let $V$ be the subset of $\Bbb R^5$ consisting of all points $\vec x$ satisfying
$$
x_1+x_2+x_3+x_4=0
$$
Then every point $\vec x\in V$ may be written as
\begin{align*}
\vec x
&= (x_1,x_2,x_3,x_4,x_5) \\
&= (x_1,x_2,x_3,-x_1-x_2-x_3,x_5) \\
&= x_1\cdot (1,0,0,-1,0)+x_2\cdot (0,1,0,-1,0)+x_3\cdot(0,0,1,-1,0)+x_5\cdot(0,0,0,0,1)
\end{align*}
This proves that $\dim V=4$ and that 
\begin{align*}
(1,0,0,-1,0)&&
(0,1,0,-1,0)&&
(0,0,1,-1,0)&&
(0,0,0,0,1)
\end{align*}
is a basis for $V$. Since $4=5-1$ we say $V$ has codimension one in $\Bbb R^5$. Codimension one subspaces are often referred to as hyperplanes.
A: While it's true that a basis for $\Bbb{R}^5$ requires $5$ linearly independent vectors in order to span the $5$-dimensional space, you are not asking about all of $\Bbb{R}^5$.
This question is about the subspace (let's call it $V$) consisting of vectors whose first four coordinates sum to zero.  In symbols,
$$
V = \left\{
{\scriptsize\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5
\end{bmatrix}} \left|\right. x_1 + x_2 + x_3 + x_4 = 0 \right\}.
$$
This linear relation reduces the dimension of $V$ to $4$ since $x_2$, $x_3$, $x_4$, and $x_5$ are free to take on any real values as long as
$$
x_1 = -x_2 - x_3 - x_4.
$$
The one linear relation forces one coordinate to take a particular value based on the others.  A nice basis is
$$
\left\{
{\scriptsize\begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}}, \quad
{\scriptsize\begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}}, \quad
{\scriptsize\begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}}, \quad
{\scriptsize\begin{bmatrix}  0 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}}
\right\}.
$$
