Non-definite n-by-n matrix

21 Views Asked by At

How does one prove that, if $n$ is even and the symmetric $n \times n$ matrix has a negative determinant, then this matrix is non-definite?

1

There are 1 best solutions below

0
On

A symmetric matrix has real eigenvalues, and you're given that the product of an even number of these is negative.

Is it possible for all of the eigenvalues to be positive?

Negative?