Proof
The result is trivial if the matrix N is singular, so assume the columns of N are linearly independent. By dividing each column by its length, it can be seen that the result is equivalent to the special case where each column has length 1, in other words if ei are unit vectors and M is the matrix having the ei as columns then
and equality is achieved if and only if the vectors are an orthogonal set, that is when the matrix is unitary. The general result now follows:
For the positive definite case, let P =M*M and let the eigenvalues of P be λ1, λ2, … λn. By assumption, each entry in the diagonal of P is 1, so the trace of P is n. Applying the inequality of arithmetic and geometric means,
so
If there is equality then each of the λi's must all be equal and their sum is n, so they must all be 1. The matrix P is Hermitian, therefore diagonalizable, so it is the identity matrix—in other words the columns of M are an orthonormal set and the columns of N are an orthogonal set.
Many other proofs can be found in the literature.
Read more about this topic: Hadamard's Inequality
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