# 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.

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

• Apply Gram-Schmidt to the basis $$((1, 1, 1, 1), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1)),$$ or any other basis beginning with $(1, 1, 1, 1)$. – Theo Bendit Sep 27 '18 at 2:37
• Is not in that case the solution will be particular , I mean there can other vectors also which can not be conclude using Gram-schmidt method. – M. A. SARKAR Sep 27 '18 at 2:44
• Yes, it will be particular. If you range over all such bases, then you will obtain all orthonormal bases beginning with $\left(\frac12,\frac12,\frac12,\frac12\right)$, though not uniquely. – Theo Bendit Sep 27 '18 at 2:47

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

$$\left( \begin{array}{rrrrrrrrrr} 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 2 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 3 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 4 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 5 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 6 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 7 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 9 \end{array} \right).$$

• That's great! =) – Siong Thye Goh Sep 27 '18 at 3:02
• Excellent method – M. A. SARKAR Sep 27 '18 at 11:13

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.

• But if we consider the Hadamard matrix , then the solution will be particular. There may be other vectors which are perpendicular to $(1,1,1,1)$ except the last 3 column vectors in the Hadamard vectors and its multiples . – M. A. SARKAR Sep 27 '18 at 2:42
• So if we use Gram-Schmidt method then we need a basis – M. A. SARKAR Sep 27 '18 at 2:43
• Theo has given you a basis right? Note that answer is not unique. If you want to describe the set, you have already done so in your post. – Siong Thye Goh Sep 27 '18 at 2:48
• There is a pattern that easily adapts to any dimension... – Will Jagy Sep 27 '18 at 3:01