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?

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) (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
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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