You can
put this solution on YOUR website! Given two n x n matrices A and B where AB=BA how does one show that the
determinant of (A^2 + B^2) >=0?
What level of linear algebra you are studying?
It seems we have to use eigenvectors to prove it.
AB=BA (commute) implies there is a basis of non-zero eigenvector say
{vi | i=1,2,..n} in
or
such that Avi = civi,Bvi = divi for some scalar (eigenvalues)
ci,di for each i.
Since for each i, we have (
)(vi) =
=
= (
) vi.
Also, note that det(
) equals to the product of eigenvalues
of the matrix
.
Hence, det(
) =
(means product)
Try to read carefully and understand the above proof.
Good luck!
Kenny
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