0
answers
0
watching
217
views
10 Nov 2019
Prove Theorem i.e. prove that trace(AB) = trace(BA). Let A and B be matrices of size m times n and n times m respectively (so the both products AB and BA are well defined). Then trace(AB) = trace(BA) We leave the proof of this theorem as an exercise, see Problem below. There are essentially two ways of proving this theorem. One is to compute the diagonal entries of AB and of DA and compare their sums. This method requires some proficiency in manipulating sums in notation. If you sue not comfortable with algebraic manipulations, there is another way. We can consider two linear transformations, T and T1, acting from Mnxm to R = R1 defined by T(X) = trace(AX) , T1(X) = trace(XA) To prove the theorem it is sufficient to show t hat T = T1 the equality for X = B gives the theorem. Since a linear transformation is completely defined by its values on a generating system, we need Just to check the equality on some simple matrices. for example on matrices Xj, k, which has all entries 0 except the entry 1 in the intersection of jth column and kth row.
Prove Theorem i.e. prove that trace(AB) = trace(BA). Let A and B be matrices of size m times n and n times m respectively (so the both products AB and BA are well defined). Then trace(AB) = trace(BA) We leave the proof of this theorem as an exercise, see Problem below. There are essentially two ways of proving this theorem. One is to compute the diagonal entries of AB and of DA and compare their sums. This method requires some proficiency in manipulating sums in notation. If you sue not comfortable with algebraic manipulations, there is another way. We can consider two linear transformations, T and T1, acting from Mnxm to R = R1 defined by T(X) = trace(AX) , T1(X) = trace(XA) To prove the theorem it is sufficient to show t hat T = T1 the equality for X = B gives the theorem. Since a linear transformation is completely defined by its values on a generating system, we need Just to check the equality on some simple matrices. for example on matrices Xj, k, which has all entries 0 except the entry 1 in the intersection of jth column and kth row.