document.write( "Question 22654: Given two n x n matrices A and B where AB=BA how does one show that the
\n" ); document.write( "determinant of (A^2 + B^2) >=0?
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Algebra.Com's Answer #11250 by khwang(438)\"\" \"About 
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Given two n x n matrices A and B where AB=BA how does one show that the
\n" ); document.write( "determinant of (A^2 + B^2) >=0? \r
\n" ); document.write( "\n" ); document.write( " What level of linear algebra you are studying?\r
\n" ); document.write( "\n" ); document.write( " It seems we have to use eigenvectors to prove it.\r
\n" ); document.write( "\n" ); document.write( " AB=BA (commute) implies there is a basis of non-zero eigenvector say
\n" ); document.write( " {vi | i=1,2,..n} in
\n" ); document.write( " \"F%5En+\" or \"%28R%5En%29\" such that Avi = civi,Bvi = divi for some scalar (eigenvalues)
\n" ); document.write( " ci,di for each i.\r
\n" ); document.write( "\n" ); document.write( " Since for each i, we have (\"A%5E2+%2B+B%5E2\")(vi) = \"A%5E2%28vi%29+%2B+B%5E2%28vi%29\"
\n" ); document.write( " = \"ci%5E2%2A+vi+%2B+di%5E2%2A+vi\" = (\"ci%5E2+%2B+di%5E2+\") vi.\r
\n" ); document.write( "\n" ); document.write( " Also, note that det(\"A%5E2+%2B+B%5E2\") equals to the product of eigenvalues
\n" ); document.write( " of the matrix \"A%5E2+%2B+B%5E2\".
\n" ); document.write( " Hence, det(\"A%5E2+%2B+B%5E2\") = \"PI%28ci%5E2+%2B+di%5E2+%29+%3E=0\" (means product)\r
\n" ); document.write( "\n" ); document.write( " Try to read carefully and understand the above proof.\r
\n" ); document.write( "\n" ); document.write( " Good luck!\r
\n" ); document.write( "\n" ); document.write( " Kenny\r
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