
Adam D. answered 08/16/22
Applied Mathematics Graduate and Experienced Tutor
a. False. Counterexample:
Let A = {{1, 1}, {0, 0}}, B = {{-1, 1}, {0, 0}}, and x = {{-1}, {1}}.
Then, x is in the null space of A, since Ax = 0.
However, Bx = {{2}, {0}}, so Mx = ABx = {{1, 1}, {0, 0}} * {{2}, {0}} = {{2}, {0}} != 0
Therefore x is not in the null space of M = AB.
b. True. Proof:
x is in the null space of B, so Bx = 0
So, (AB)x = A(Bx) = A*0 = 0.
c. False. Counterexample:
Let A = {{1, 1}, {0, 0}}, B = {{-1, 1}, {0, 0}}, and x = {{1}, {1}}.
Then, x is in the null space of AB, but only because x is in the null space of B, as Bx = 0, meaning ABx = A*0 = 0.
x is not in the null space of A, since Ax = {{2}, {0}}.
d. False. Counterexample:
Let A = {{-1, 1}, {0, 0}}, B = {{1, 0}, {1, 0}}, and x = {{1, 0}.
Then, ABx = 0, but Bx = {{1}, {1}} != 0.