Let A, B and C be nxn matrices such that A=BC-CB. Show that the trace of A (denoted Tr(A)) is 0, where the trace of an nxn matrix is defined as the sum of the entries along the leading diagonal.

This is the type of question which requires some out of the box thinking. A sketch is presented below: We start by defining A = (aij), B = (bij) and C = (cij). We then use the definition of matrix multiplication to write A in ''sigma notation''. The trace of A, we have been told, is defined as sum(aii) from i=1 to i=n (or j could equally be used). Taking the sum from i=1 to i=n to our expression for BC-CB, but with i's replaced with j's, and setting equal to sum(aij) from i=1 to i=n we end up with the difference of two ''double sums''. The two sums are equal (obviously, or by spotting a pattern after writing out) and so their difference is 0. Hence, the trace of A is 0. i.e Tr(A)=0

DG
Answered by Daniel G. Further Mathematics tutor

2859 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

Prove that 1+4+9+...+n^2 = n(n+1)(2n+1)/6.


A=[5k,3k-1;-3,k+1] where k is a real constant. Given that A is singular, find all the possible values of k.


How far is the point (7,4,1) from the line that passes through the points (6,4,1) and (6,3,-1)?


Prove by induction that n! > n^2 for all n greater than or equal to 4.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences