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
4375 Views
See similar Further Mathematics A Level tutorsRelated Further Mathematics A Level answers
All answers ▸A spring with a spring constant k is connected to the ceiling. First a weight of mass m is connected to the spring. Deduce the new equilibrium position of the spring, find its equation of motion and hence deduce its frequency f.
Prove De Moivre's by induction for the positive integers
A particle is undergoing circular motion in a horizontal circle, that lies within the smooth surface of a hemispherical bowl of radius 4r. Find the distance OC (explained in diagram) if the angular acceleration of the particle is equal to root (3g/8r).
It is given that f(x)=(x^2 +9x)/((x-1)(x^2 +9)). (i) Express f(x) in partial fractions. (ii) Hence find the integral of f(x) with respect to x.