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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

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    $\begingroup$ 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 $(1, 1, 1, 1)$. $\endgroup$
    – Theo Bendit
    Sep 27, 2018 at 2:37
  • $\begingroup$ 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. $\endgroup$
    – MAS
    Sep 27, 2018 at 2:44
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    $\begingroup$ 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. $\endgroup$
    – Theo Bendit
    Sep 27, 2018 at 2:47

2 Answers 2

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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

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

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    $\begingroup$ That's great! =) $\endgroup$
    – Siong Thye Goh
    Sep 27, 2018 at 3:02
  • $\begingroup$ Excellent method $\endgroup$
    – MAS
    Sep 27, 2018 at 11:13
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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.

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  • $\begingroup$ 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 . $\endgroup$
    – MAS
    Sep 27, 2018 at 2:42
  • $\begingroup$ So if we use Gram-Schmidt method then we need a basis $\endgroup$
    – MAS
    Sep 27, 2018 at 2:43
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    $\begingroup$ 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. $\endgroup$
    – Siong Thye Goh
    Sep 27, 2018 at 2:48
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    $\begingroup$ There is a pattern that easily adapts to any dimension... $\endgroup$
    – Will Jagy
    Sep 27, 2018 at 3:01

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