Given a square matrix , prove that the sum of its eigenvalues is equal to the trace of , and the product of its eigenvalues is equal to the determinant of .
This proof requires the investigation of the characteristic polynomial of , which is found by taking the determinant of .
Observe that .
Let be the roots of an n-order polynomial.
Since the eigenvalues are the roots of a matrix polynomial, we can match to . Therefore it is clear that
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