a_{i} and b_{i} are reals such that a_{1}b_{2} ≠ a_{2}b_{1}. What is the maximum number of possible 4-tuples (sign x_{1}, sign x_{2}, sign x_{3}, sign x_{4}) for which all x_{i} are non-zero and x_{i} is a simultaneous solution of a_{1}x_{1} + a_{2}x_{2} + a_{3}x_{3} + a_{4}x_{4} = 0 and b_{1}x_{1} + b_{2}x_{2} + b_{3}x_{3} + b_{4}x_{4} = 0. Find necessary and sufficient conditions on a_{i} and b_{i} for this maximum to be achieved.

**Solution**

Solving in terms of x_{3}, x_{4} gives x_{1} = s_{23}/s_{12} x_{3} + s_{24}/s_{12} x_{4}, x_{2} = s_{31}/s_{12} x_{3 }+ s_{41}/s_{12} x_{4}, x_{3} = x_{3}, x_{4} = x_{4}, where s_{ij} = (a_{i}b_{j} - a_{j}b_{i}). Plot the 4 lines s_{23}/s_{12} x_{3} + s_{24}/s_{12} x_{4} = 0, x_{2} = s_{31}/s_{12} x_{3 }+ s_{41}/s_{12} x_{4} = 0, x_{3} = 0, x_{4} = 0 in the x_{3}, x_{4} plane. We get 4 lines through the origin. Evidently x_{1} changes sign if we cross the first, x_{2} changes sign if we cross the second, x_{3} changes sign if we cross the third and x_{4} changes sign if we cross the fourth. So we have a different combination of signs in each sector, but the same combination throughout any given sector.

Thus the maximum is achieved when the four lines are distinct, giving 8 sectors (and hence 8 combinations). This requires that s_{23} and s_{24} are non-zero (otherwise the first line coincides with one of the last two) and that s_{31} and s_{41} are non-zero (otherwise the second line coincides with one of the last two). Finally, the first two lines must not coincide with each other. That requires that s_{31}/s_{41} is not equal to s_{23}/s_{24}. After some slightly tiresome algebra that reduces to s_{34} non-zero. So a necessary and sufficient condition to achieve 8 is that all s_{ij} are non-zero.

© John Scholes

jscholes@kalva.demon.co.uk

14 Jan 2002