# Finding 4 perpendicular unit vectors

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.

• In which dimension? – Matias Heikkilä Sep 6 '15 at 19:48
• It isn't specified. – PiscesGamer Sep 6 '15 at 19:53
• Hint: start with $(1/2, 1/2, 1/2, 1/2)$ and change suitable pairs of $+$ to $-$... – Peter Franek Sep 6 '15 at 19:54
• it has to be at least four since there can be no solution in a dimension lower than that. – Matias Heikkilä Sep 6 '15 at 19:54
• Yes I did that, but I don't believe that's all I have to do is it? – PiscesGamer Sep 6 '15 at 19:57

One vector could be $$\left(\begin{matrix}\frac 12\\\frac 12\\\frac 12\\\frac 12\end {matrix}\right)$$ Then the others can be $$\left(\begin{matrix}\frac 12\\\frac 12\\-\frac 12\\-\frac 12\end {matrix}\right)$$$$\left(\begin{matrix}\frac 12\\-\frac 12\\\frac 12\\-\frac 12\end {matrix}\right)$$$$\left(\begin{matrix}\frac 12\\-\frac 12\\-\frac 12\\\frac 12\end {matrix}\right)$$

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.

• Wait, do I really only need to write these vectors, no required equations once I know what they are. – PiscesGamer Sep 6 '15 at 20:40
• How can I know for sure they are perpendicular. – PiscesGamer Sep 6 '15 at 20:56
• Check the dot product of every pair – Ben Grossmann Sep 7 '15 at 2:33

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.