49th Putnam 1988

Problem A6

V is an n-dimensional vector space. Can we find a linear map A : V → V with n+1 eigenvectors, any n of which are linearly independent, which is not a scalar multiple of the identity?



Answer: no.

Let the n+1 corresponding eigenvalues have sum k. Given any eigenvector with eigenvalue λ, the remaining n eigenvectors form a basis (since they are linearly independent) which diagonalises the matrix of the linear map. The resulting matrix has trace k - λ. But trace is independent of basis, so all the λ are the same and the diagonalised matrix is just a multiple λ of the identity. In other words, the linear map is just a scalar multiple of the identity.



49th Putnam 1988

© John Scholes
1 Jan 2001