Proof of the rank-trace theorem

The previous post discussed the motivation for and application of the rank-trace theorem. This post will give a proof. Suppose A is a real symmetric matrix. The rank-trace inequality says where tr is the trace operator, the sum of the elements along the diagonal of the matrix. Terse proof Here’s the proof in a nutshell: diagonalize A and use the Cauchy-Schwarz inequality. Detailed proof Now let’s unpack that. Any real symmetric matrix A is similar to a matrix D with the eigenvalues of A along the diagonal. The trace of a matrix stays the same under a similarity transformation, i.e. multiplying by P on one side and its inverse on the other side. So without loss of generality we may as well assume A is diagonal. The rank of a matrix equals the number of non-zero eigenvalues, so a vector containing the non-zero eigenvalues of A has length r where r is the rank of A. Define w to be the vector of dimension r consisting of all 1’s. Then by the Cauchy-Schwarz inequality we have Cyclic trace…