1
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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 '18 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. $\endgroup$ – M. A. SARKAR Sep 27 '18 at 2:44
  • 1
    $\begingroup$ 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 '18 at 2:47
3
$\begingroup$

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

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

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ That's great! =) $\endgroup$ – Siong Thye Goh Sep 27 '18 at 3:02
  • $\begingroup$ Excellent method $\endgroup$ – M. A. SARKAR Sep 27 '18 at 11:13
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 . $\endgroup$ – M. A. SARKAR Sep 27 '18 at 2:42
  • $\begingroup$ So if we use Gram-Schmidt method then we need a basis $\endgroup$ – M. A. SARKAR Sep 27 '18 at 2:43
  • 1
    $\begingroup$ 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 '18 at 2:48
  • 1
    $\begingroup$ There is a pattern that easily adapts to any dimension... $\endgroup$ – Will Jagy Sep 27 '18 at 3:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.