SOLUTION: 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?

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Question 22654: 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?
Answer by khwang(438) About Me  (Show Source):
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
F%5En+ or %28R%5En%29 such that Avi = civi,Bvi = divi for some scalar (eigenvalues)
ci,di for each i.
Since for each i, we have (A%5E2+%2B+B%5E2)(vi) = A%5E2%28vi%29+%2B+B%5E2%28vi%29
= ci%5E2%2A+vi+%2B+di%5E2%2A+vi = (ci%5E2+%2B+di%5E2+) vi.
Also, note that det(A%5E2+%2B+B%5E2) equals to the product of eigenvalues
of the matrix A%5E2+%2B+B%5E2.
Hence, det(A%5E2+%2B+B%5E2) = PI%28ci%5E2+%2B+di%5E2+%29+%3E=0 (means product)
Try to read carefully and understand the above proof.
Good luck!
Kenny