2 votes 2 votes Let A be a square matrix such that $A^{3}$ = 0, but $A^{2} \neq 0$. Then which of the following statements is not necessarily true? (A) $A \neq A^{2}$ (B) Eigenvalues of $A^{2}$ are all zero (C) rank(A) > rank($A^{2}$ ) (D) rank(A) > trace(A) Linear Algebra isisamplepapers matrix + – neha.i asked Apr 17, 2017 retagged Jun 20, 2017 by Silpa neha.i 1.5k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes Option C can never be true rank (A*B) = min {r(A),r(B)} so rank (A*A) = r(A) Option A is true suppose A2 = A now A3 = A2*A = A*A =A2 and given A3 is 0 this implies A2 is also zero this contradicts statement given in question A2 not eqal to zero so our assumption A2 = A is wrong . this implies A2 is not equal to A For other options i dont have exact answer :) aehkn answered Apr 23, 2017 aehkn comment Share Follow See all 2 Comments See all 2 2 Comments reply Prince Singh 1 commented Oct 30, 2018 reply Follow Share Isn't there a correction because rank (A*B) <= min {r(A),r(B)} 0 votes 0 votes huzii maths commented Sep 9, 2020 reply Follow Share option B will not necessarily true as mentioned that A^2 NOT Zer0 therefore all of its eigen value must be non zero otherwise product of eigen values results zero that leads to A^2 =0 which is not desired 0 votes 0 votes Please log in or register to add a comment.