Soccer & Linear Algebra0417

When a soccer game starts, the ball lies in the center of the field and is then moved around the field for 45 minutes by shifting and turning. At the beginning of the second half the ball is again on the center of the field. We show with simple means of linear algebra that either an infinite number of points on the surface are always in exactly the same position as in the original state or exactly 2.

First of all, the displacements of the ball, which are carried out during the 1st half, add up trivially to the zero vector. They can therefore be neglected. This leaves a finite number of rotations $$A_1, ..., A_n \in \mathbb{R}^{3 \times 3}$$ with $$A$$ orthogonal and $$\det(A_k) = 1 \,\, \forall \,\, k \in \{1,...,n\}$$ . For every two rotations $$A_i, A_j$$ applies:

$$(A_i A_j)^T (A_i A_j) = A_j^T A_i^T A_i A_j = A_j^T (A_i^T A_i) A_j = A_j^T E_3 A_j = A_j^T A_j = E_3$$

and

$$\det(A_i A_j) = \det(A_i) \cdot \det(A_j)=1 \cdot 1 = 1.$$

This means that $$A_i A_j$$ again a rotation, which is why $$A_1 ... A_n$$ also a single rotation.

If now $$A_1 ... A_n = E_n$$, then obviously all points of the surface of the ball are exactly at the starting point - in the other (more probable) case the eigenvector of $$A_1 ... A_n$$ is equal to its axis of rotation with the eigenvalue $$1$$. Thus exactly these two points, which lie on the axis of rotation, are mapped onto themselves.

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