Show that if A is Diagonalizable, then Tr(A) is the sum of the eigenvalues.
Thank you for the opportunity to help you with your question!
Tr(A) - Tr(B)=0
Tr(A - B)=0
but since A=B
showing that Tr(A) =Tr(B)
and for A Diagonalizable, Tr(A) is the sum of the eigenvalues as shown from the above solution
Content will be erased after question is completed.
Enter the email address associated with your account, and we will email you a link to reset your password.
Forgot your password?