Putnam 48/B6 solution

8th Putnam 1948

Problem B6

Do either (1) or (2):

(1) Take the origin O of the complex plane to be the vertex of a cube, so that OA, OB, OC are edges of the cube. Let the feet of the perpendiculars from A, B, C to the complex plane be the complex numbers u, v, w. Show that u² + v² + w² = 0.

(2) Let (a_ij) be an n x n matrix. Suppose that for each i, 2 |a_ii| > ∑₁ⁿ |a_ij|. By considering the corresponding system of linear equations or otherwise, show that det a_ij ≠ 0.

Solution

(1)

(2) If the det is non-zero, then we can find x_inot all zero so that ∑ x_i a_ij = 0 for each j. Take k so that |x_k| ≥ |x_i| for all i. Then |∑_{i not k} x_ia_ik| ≤ ∑_{i not k} |x_i a_ik| ≤ |x_k| ∑_{i not k} |a_ik| < |x_k| |a_kk|, so we cannot have ∑ x_ia_ik = 0. Contradiction.

8th Putnam 1948