Let $\textbf{v} = (1, 1, 1, 1)$. Find a basis for... Let $\textbf{v}=(1,1,1,1)$. Find a basis for $\{\textbf{u}\in\Bbb{R}^4\ |\ \textbf{u}\cdot\textbf{v}=0\}$
How can I do this? In particular, I do not understand $\textbf{u}\cdot\textbf{v}=0.$
 A: Let
$$u=\begin{pmatrix}x\\y\\z\\w\end{pmatrix}$$
Then,
$$u\cdot v=0\Rightarrow x+y+z+w=0$$
$$w=-x-y-z$$
$$u=\begin{pmatrix}x\\y\\z\\-x-y-z\end{pmatrix}=x\begin{pmatrix}1\\0\\0\\-1\end{pmatrix}+y\begin{pmatrix}0\\1\\0\\-1\end{pmatrix}+z\begin{pmatrix}0\\0\\1\\-1\end{pmatrix}$$
The set
$$\left\{\begin{pmatrix}1\\0\\0\\-1\end{pmatrix},\begin{pmatrix}0\\1\\0\\-1\end{pmatrix},\begin{pmatrix}0\\0\\1\\-1\end{pmatrix}\right\}$$ span the space and are linearly independent. They form a basis.
A: Denote $W=\{u\in\mathbb{R}^4\mid u\cdot v=0\}$, we claim that the set
$\beta=\left\{
\begin{pmatrix}-1\\1\\0\\0\end{pmatrix},
\begin{pmatrix}-1\\0\\1\\0\end{pmatrix},\begin{pmatrix}-1\\0\\0\\1\end{pmatrix}
\right\}$ is a basis for $W$. First, given 
$u=(u_1,u_2,u_3,u_4)\in W$, then we have
$$0=u\cdot v=
\begin{pmatrix}u_1\\u_2\\u_3\\u_4\end{pmatrix}\cdot
\begin{pmatrix}1\\1\\1\\1\end{pmatrix}=
u_1+u_2+u_3+u_4,$$ that is, $u_1=-u_2-u_3-u_4$. So we can write
$$u=\begin{pmatrix}-u_2-u_3-u_4\\u_2\\u_3\\u_4\end{pmatrix}
=u_2\begin{pmatrix}-1\\1\\0\\0\end{pmatrix}
+u_3\begin{pmatrix}-1\\0\\1\\0\end{pmatrix}
+u_4\begin{pmatrix}-1\\0\\0\\1\end{pmatrix},$$
which follows that $\beta$ spans $W$. Next, we show that $\beta$ is linearly independent, but it is easy to prove so I leave it to you. Hence we conclude that
$\beta$ is a basis for $W$, and the claim is proved.
A: You want to find the set of all vectors that are perpendicular to $v$ which is in $\mathbb{R}^4$. The cardinality for such a basis will be 3 (why?).
Hint
If you define the dot product as the standard method, then you wish to find the set of solutions to:
$$(1)x_1 + (1)x_2 + (1)x_3 + (1)x_4 = 0$$
