Determine whether the given set S is a subspace of the vecto

Determine whether the given set S is a subspace of the vector space V. A. V = M_2(R) and S is the set of 2 times 2 matrices A such that A^2 = O (here O is the 2 times 2 zero matrix) B. V = M_3(R) and S is the set of 3 times 3 matrices of rank 1 C. V = M_3(R) and S is the set of 3 times 3 matrices A such that A^2 = A D. V = M_2(R) and 5 is the set of 2 times 2 matrices that commute with the matrix (4 2 7 8) E. V = M_3(R) and S is the set of 3 times 3 matrices A such that the vector 3 is in the (1 3 0) is in the nullspace of A F. V = M_3(IR) and S is the set of 3 x 3 matrices with trace 0 (recall the trace of a matrix is the sum of its diagonal entries)

Solution

Let A and B be two arbitrary elements of S. The A2 = O and B2 = O. However, (A+B)2 = (A +B)(A+B) = A2 +BA +AB +B2 = BA +AB. Since BA +AB need not be equal to O, hence (A+B) does not necessarily belong to S. Hence S is not closed under addition and, therefore S is not a vector space. The rank of a matrix is the number of non-zero rows in its RREF. Since the 3 x 3 zero matrix, which is the additive identity does not have rank 1 (its rank is 0), it does not belong to S. Therefore, S is not a vector space. Let A and B be two arbitrary elements of S. The A2 = A and B2 = B. However, (A+B)2 = (A +B)(A+B) = A2 +BA +AB +B2 = A+B + BA +AB which is not necessarily equal to A+B. Hence (A+B) does not necessarily belong to S. Hence S is not closed under addition and, therefore S is not a vector space. Let the given matrix be denoted by A and let B and C be 2 arbitrary elements of S. Then AB + BA and AC = CA. Also A(B +C) = AB +AC = BA +CA + (B+C)A . Hence B+C commutes with A and , therefore S is closed under addition. Further, if p is a scalar, then A(pB) = pAB = pBA = (pB)A. This means that S is closed under scalar multiplication. Also, the 2 x 2 zero matrix ( denoted by O) belong to S as AO = OA =O. Thus S is a vector space. Let the given vector be denoted by X and let A and B be two arbitrary elements of S. Since X is in null space of both A and B, hence AX = O and BX = O. Further (A+B)X= AX+BX = O+O = O. Hence A+B belongs to S so that S is closed under addition. Further, if p is a scalar, then (pA)X = p(AX) = pO = O. Hence S is closed under scalar multiplication. Also, the 3x3 zero matrix O also belomgs to S as OX = O. Thus, S is a vector space. Let A and B be two arbitrary elements of S. Also, let the elements on the main diagonal of A and B be a1 ,a2 ,a3 and b1 , b2 , b3 respectively. Then a1 + a2 + a3 = 0 and b1 + b2 + b3 = 0. The elements on the main diagonal of A+B are (a1+b1),(a2+b2) and (a3+b3). Now, the trace of A+B is (a1+ b1 ) +      (a2 + b2 ) + (a3+b3 ) = (a1 + a2 + a3) + (b1 + b2 + b3) = 0+0 = 0. Thus A+B belongs to S and hence S is closed under addition.Further,if p is a scalar,then the trace of pA is pa1+pa2+pa3 = p(a1 + a2 + a3) = p*0 = 0. Hence pA belongs to S and therefore S is closed under scalar multiplication. Further, the trace of the 3 x 3 zero matix O is 0 so that O belongs to S. Hence S is a vector space.