Algebra 439250.First solution: The eigenvalues are the zeros of the polynomial det(λIn−aA−bAt).
The matrixλIn−aA−bAtis a circulant matrix, and the determinant of a circulant matrix
was the subject of problem 211 in Section 2.3.2. According to that formula,
det(λIn−aA−bAt)=(− 1 )n−^1n∏− 1j= 0(λζj−aζ^2 j−b),whereζ=e^2 πi/nis a primitiventh root of unity. We find that the eigenvalues ofaA+bAt
areaζj+bζ−j,j= 0 , 1 ,...,n−1.
Second solution: Simply note that forζ=e^2 πi/nandj= 0 , 1 ,...,n−1,( 1 ,ζj,ζ^2 j,
...,ζ(n−^1 )j)is an eigenvector with eigenvalueaζj+bζ−j.
251.Letφbe the linear transformation of the spaceRnwhose matrix in a certain basis
e 1 ,e 2 ,...,enisA. Consider the orthogonal decompositions of the spaceRn=kerφ⊕T,
Rn=Imφ⊕S. Setφ′=φ|T. Thenφ′:T →Imφis an isomorphism. Letγ′be
its inverse, which we extend to a linear transformationγof the whole ofRnby setting
γ|S =0. Thenφγ φ=φ′γ′φ′ =φ′onT andφγ φ=0onT⊥=kerφ. Hence
φγ φ=φ, and we can chooseBto be the matrix ofγin the basise 1 ,e 2 ,...,en.
(Soviet Union University Student Mathematical Olympiad, 1976)
252.The map that associates to the angle the measure of its projection onto a plane is
linear in the angle. The process of taking the average is also linear. Therefore, it suffices
to check the statement for a particular angle. We do this for the angle of measureπ,
where it trivially works.
Remark.This lemma allows another proof of Fenchel’s theorem, which is the subject of
problem 644 in Section 4.1.4. If we defined the total curvature of a polygonal line to be
the sum of the “exterior’’ angles, then the projection of any closed polygonal line in three-
dimensional space onto a one-dimensional line has total curvature at leastπ+π= 2 π
(two complete turns). Hence the total curvature of the curve itself is at least 2π.
(communicated by J. Sullivan)
253.The first involutionAthat comes to mind is the symmetry with respect to a hyper-
plane. For that particular involution, the operatorB=^12 (A+I)is the projection onto
the hyperplane. Let us show that in general for any involutionA, the operatorBdefined
as such is a projection. We have
B^2 =
1
4
(A+I)^2 =
1
4
(A^2 + 2 AI+I^2 )=
1
4
(I+ 2 A+I)=B.
There exists a basis ofVconsisting of eigenvectors ofB. Just consider the decomposition
ofVinto the direct sum of the image ofBand the kernel ofB. The eigenvectors that
form the basis are either in the image ofB, in which case their eigenvalue is 1, or in