Find the nonzero vectors $u,v,w$ that are perpendicular to the vector $(1,1,1,1)$ and to each other Find the nonzero vectors $u,v,w$ that are perpendicular to the vector $(1,1,1,1)$ and to each other.
Answer:
If I follow algebra, then I get complicated results to solve it as follows:
Let $u=(u_1,u_2,u_3,u_4), \ v=(v_1,v_2,v_3,v_4) , \ w=(w_1,w_2,w_3,w_4)$
Then, $u \cdot (1,1,1,1)=v \cdot (1,1,1,1)=w \cdot (1,1,1,1)=0$
Also $u \cdot v=w \cdot u=v \cdot w=0$.
These gives us
$u_1+u_2+u_3+u_4=0, \\ v_1+v_2+v_3+v_4=0 , \\ w_1+w_2+w_3+w_4=0, \\ u_1v_1+u_2v_2+u_3v_3+u_4v_4=0, \\ u_1w_1+u_2w_2+u_3v_3+u_4v_4=0, \\ v_1w_1+v_2w_2+v_3w_3+v_4w_4=0. $
But how to solve for $u_i, v_i,w_i, \ i=1,2,3,4$ from here?
Does there exit any other easy method?
Help me out 
 A: as columns
$$
\left(
\begin{array}{rrrr}
1&-1&-1&-1 \\
1& 1&-1&-1 \\
1&0 &2&-1 \\
1&0&0&3
\end{array}
\right)
$$
Pattern, done correctly,  works in any dimension
$$    
 \left(  \begin{array}{rrrrrrrrrr}
  1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1   \\
  1  &  1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1   \\
  1  &  0  &  2  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1   \\
  1  &  0  &  0  &  3  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1   \\
  1  &  0  &  0  &  0  &  4  &  -1  &  -1  &  -1  &  -1  &  -1   \\
  1  &  0  &  0  &  0  &  0  &  5  &  -1  &  -1  &  -1  &  -1   \\
  1  &  0  &  0  &  0  &  0  &  0  &  6  &  -1  &  -1  &  -1   \\
  1  &  0  &  0  &  0  &  0  &  0  &  0  &  7  &  -1  &  -1   \\
  1  &  0  &  0  &  0  &  0  &  0  &  0  &  0  &  8  &  -1   \\
  1  &  0  &  0  &  0  &  0  &  0  &  0  &  0  &  0  &  9   
\end{array}
  \right).
  $$
A: Consider the Hadamard matrix of which we know that the columns form an orthogonal basis of $\mathbb{R}^4$.
$$\begin{bmatrix} 1 & 1 & 1 & 1 \\  1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1\end{bmatrix}$$
The other columns would give you a solution. 
Alternatively, use Gram-Schmidt process. 
