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The question: Find 4 perpendicular unit vectors whose components are all either 1/2 or -1/2.

I' m just having trouble understanding how to do this.

NOTE: The dimension is not specified.

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  • $\begingroup$ In which dimension? $\endgroup$ – Matias Heikkilä Sep 6 '15 at 19:48
  • $\begingroup$ It isn't specified. $\endgroup$ – PiscesGamer Sep 6 '15 at 19:53
  • $\begingroup$ Hint: start with $(1/2, 1/2, 1/2, 1/2)$ and change suitable pairs of $+$ to $-$... $\endgroup$ – Peter Franek Sep 6 '15 at 19:54
  • $\begingroup$ it has to be at least four since there can be no solution in a dimension lower than that. $\endgroup$ – Matias Heikkilä Sep 6 '15 at 19:54
  • $\begingroup$ Yes I did that, but I don't believe that's all I have to do is it? $\endgroup$ – PiscesGamer Sep 6 '15 at 19:57
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One vector could be $$\left(\begin{matrix}\frac 12\\\frac 12\\\frac 12\\\frac 12\end {matrix}\right)$$ Then the others can be $$\left(\begin{matrix}\frac 12\\\frac 12\\-\frac 12\\-\frac 12\end {matrix}\right)$$$$\left(\begin{matrix}\frac 12\\-\frac 12\\\frac 12\\-\frac 12\end {matrix}\right)$$$$\left(\begin{matrix}\frac 12\\-\frac 12\\-\frac 12\\\frac 12\end {matrix}\right)$$

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First, deduce the number of entries. If a unit-vector has $n$ entries which are all $\pm 1/2$, what is $n$? Use the "Pythagorean theorem".

You should find that $n=4$. There are only $2^4=16$ vectors that you can work with, you just need $4$ that are perpendicular to each other. "Trial and error" is a reasonable way to go here.

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  • $\begingroup$ Wait, do I really only need to write these vectors, no required equations once I know what they are. $\endgroup$ – PiscesGamer Sep 6 '15 at 20:40
  • $\begingroup$ How can I know for sure they are perpendicular. $\endgroup$ – PiscesGamer Sep 6 '15 at 20:56
  • $\begingroup$ Check the dot product of every pair $\endgroup$ – Omnomnomnom Sep 7 '15 at 2:33
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I assume that the problem is in $\mathbb{R}^4$. Vectors $v,w \in \mathbb{R}^n$ are perpendicular if and only if their inner product equals zero. That is what you should use to solve this. You can just work with vectors whose entries are either $1$ or $-1$ since multiplication by a scalar does not affect orthogonality.

If you can't come up with a solution, you can Google something called "a Hadamard matrix" and you'll find one.

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