Suppose that all eigenvalues of are positive real numbers. Show that

**Solution**

Let the eigenvalues of be , . Consider the Jordan normal form of ; this Jordan form is an upper triangular matrix that has the eigenvalues of in the main diagonal. Let this matrix be called . Furthermore , as a matrix and its Jordan normal form are similar. As is upper triangular, its inverse is given by an upper triangular matrix whose diagonal entries are the inverses of the diagonal entries of . That is,

is an upper-triangular matrix, and its determinant can be computed simply as the product of the elements of the diagonal. Hence,

due to the inequality for all .

**Note:**The main point here is that if is an eigenvalue of then is an eigenvalue of . Taking into account that

yields a shorter solution to the problem. No need for upper triangular matrices.