(A) We know if T: U → V be a L.T and dim U =m,dim V=n .Then if m>n then then T can not be one-one.
here U = R
4 so its dim is 4
and V = R
3 so its dim is 3
Thus T can not be one-one
(B) We know if V be a V .S of all n x n, skew symmetric matrices, then
So here n = 4, so dim V = 4.3/2 = 6
(C) Clearly characteristic polynomial of A will be of degree n. while characteristic polynomial of T will be of degree n
2.
So they can not be equal.
Thus option A.B,C can not be true,
hence option (D) is correct.
That vector space can have basis as {(1,0),(0,1)} or {(1,1),(0,1)} or {(1,1),(1,0)}