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I have these set of vectors;

$$(1,-1,-1,1)$$ $$(1,-1,1,-1)$$ $$(1,1,-1,-1) $$ $$(A,B,C,D)$$

The conditions on A,B,C,D are that they have to be either be 1 or -1, these elements have to add up to 0 and they have to be orthogonal to all the above vectors. Can someone show me a method of finding A,B,C and D?

Jnyeboah93
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  • Maybe start by considering the first three equations to be the vectors $(1, -1, -1, 1), (1, -1, 1, -1), (1, 1, -1, -1)$ and calculate a vector that is perpendicular to all of them? Then see if any point on this vector satisfies the conditions? – user141592 Oct 22 '14 at 13:43
  • @Stefanos sorry I meant all of the above!! – Jnyeboah93 Oct 22 '14 at 13:44
  • @Stefanos Yes it does. It comes from four contrasts being orthogonal to each other. – Jnyeboah93 Oct 22 '14 at 13:49
  • @Stefanos Your solution is orthogonal to the other vectors but the elements in your solution do not add up to 0. – Jnyeboah93 Oct 22 '14 at 13:55
  • @Jnyeboah93 Yes, sorry. There is no solution with entries $\pm1$. See my post below – Jimmy R. Oct 22 '14 at 14:06

1 Answers1

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Two contrasts (i.e. vectors with $\pm1$ that sum up to zero) are orthogonal if their inner product (sumproduct of their coordinates) is equal to zero. Thus the vector $(A,B,C,D)$ is orthogonal to a vector of the form $(c_1, c_2, c_3,c_4)$ where $c_i=\pm1$ if $$c_1A+c_2B+c_3C+c_4D=0$$ In the case above this yields $$\begin{cases}A-B-C+D=0\\A-B+C-D=0\\A+B-C-D=0\\A+B+C+D=0\end{cases}$$ where the last equation ensures that the vector $(A,B,C,D)$ is also a contrast, i.e. it's entries sum up to zero.

If you sum up all equations, you have that $$4A=0 \implies A=0$$ which means that there is no solution with the given criteria (entries equal to $\pm1$ and orthogonal to the above vectors).

Jimmy R.
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