In this video, we will solve a CSIR NET FEB 2022 LINEAR ALGEBRA QUESTION ID 479 problem related to an inner product. Here we will use properties of trace and determinant of a matrix to solve this problem
inner product of A & B =0 does not imply that either A= or B=0. (i.e =0 does not imply A=0 or B=0) where as is greater than or equal to 0 and =0 if and only if A=0. So, if =0 then according to option B, trace(AA)=0 (because =trace(AB)), thus trace(A^2)=0. For simplicity, if we take A=[a b;b d] (real symmetric matrix) then trace(A^2)=a^2+b^2+d^2=0. Now trace(A^2)=0 implies a^2+b^2+d^2=0 which implies a=b=d=0. Therefore, A=[0 0;0 0] which is a zero matrix. Thus trace(A^2)=0 implies A=0.
