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? 

 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^n }}} or {{{(R^n)}}} such that Avi = civi,Bvi = divi for some scalar (eigenvalues)
 ci,di for each i.

 Since for each i, we have ({{{A^2 + B^2}}})(vi) = {{{A^2(vi) + B^2(vi)}}} 
 = {{{ci^2* vi + di^2* vi}}} = ({{{ci^2 + di^2 }}}) vi.

  Also, note that det({{{A^2 + B^2}}}) equals to the product of eigenvalues
 of the matrix {{{A^2 + B^2}}}.
 Hence, det({{{A^2 + B^2}}}) = {{{PI(ci^2 + di^2 ) >=0}}} (means product)

 Try to read carefully and understand the above proof.

 Good luck!

 Kenny