Algebra 437247.First solution: Assume first that all numbers are integers. Whenever we choose a
number, the sum of the remaining ones is even; hence the parity of each number is the
same as the parity of the sum of all. And so all numbers have the same parity.
By subtracting one of the numbers from all we can assume that one of them is zero.
Hence the numbers have the same parity as zero. After dividing by 2, we obtain 2n+ 1
numbers with the same property. So we can keep dividing by 2 forever, which is possible
only if all numbers are zero. It follows that initially all numbers were equal.
The case of rational numbers is resolved by multiplying by the least common multiple
of the denominators. Now let us assume that the numbers are real. The reals form an
infinite-dimensional vector space over the rationals. Using the axiom of choice we can
find a basis of this vector space (sometimes called a Hammel basis). The coordinates of
the 2n+1 numbers are rational, and must also satisfy the property from the statement
(this follows from the fact that the elements of the basis are linearly independent over
the rationals). So for each basis element, the corresponding coordinates of the 2n+ 1
numbers are the same. We conclude that the numbers are all equal, and the problem is
solved.
However, this solution works only if we assume theaxiom of choiceto be true. The
axiom states that given a family of sets, one can choose an element from each. Obvious
as this statement looks, it cannot be deduced from the other axioms of set theory and
has to be taken as a fundamental truth. A corollary of the axiom is Zorn’s lemma, which
is the actual result used for constructing the Hammel basis. Zorn’s lemma states that if
every totally ordered subset of a partially ordered set has an upper bound, then the set
has a maximal element. In our situation this lemma is applied to families of linearly
independent vectors with the ordering given by the inclusion to yield a basis.
Second solution: The above solution can be improved to avoid the use of the axiom of
choice. As before, we prove the result for rational numbers. Arguing by contradiction
we assume that there exist 2n+1 real numbers, not all equal, such that whenever one
is removed the others can be separated into two sets withnelements having the sum of
their elements equal. If in each of these equalities we move all numbers to one side, we
obtain a homogeneous system of 2n+1 equations with 2n+1 unknowns. In each row of
the coefficient matrix, 1 and−1 each occurntimes, and 0 appears once. The solution to
the system obviously contains the one-dimensional vector spaceVspanned by the vector
( 1 , 1 ,..., 1 ). By hypothesis, it contains another vector that does not lie inV. Solving
the system using Gaussian elimination, we conclude that there must also exist a vector
with rational coordinates outside ofV. But we already know that this is impossible. The
contradiction proves that the numbers must be all equal.
248.Letλ 1 ,λ 2 be the eigenvalues ofA. Then−λ 1 I 2 and−λ 2 I 2 both belong to
C(A),so
0 =|det(A−λiI 2 )|≥|λi|^2 , fori= 1 , 2.