Mathematics 44, section 1 -- Linear Algebra

The True-False questions we did not get to in the review session

• I) If T : V -> W is an isomorphism, and U is a vector subspace of V, then dim(U) = dim(T(U)).
  
  This is TRUE. If {A₁, ... , A_b} is a basis for U, then we know {T(A₁), ... , T(A_b)} spans T(U). Furthermore, since T is an isomorphism, it is injective. Hence, if c₁T(A₁) + ... + c_nT(A_b) = 0, then T(c₁A₁ + ... + c_nA_b) = 0 which implies c₁A₁ + ... + c_nA_n = 0, which implies c_i = 0 for i = 1, ... n. Thus {T(A₁), ... , T(A_b)} is linearly independent, hence a basis for Im(U). Hence the dimensions of U and T(U) are equal.

• J) If A₁, ... , A_n are vectors in a vector space V, and B₁,..., B_n are vectors in a vector space W, then there always exists some linear T : V -> W satisfying T(A_i) = B_i for all i = 1,..., n.
  
  This is FALSE because we could consider a case where the A_i are linearly dependent, but the B_i are linearly independent. There is no linear mapping that takes a linearly dependent set to a linearly independent set: As in question I, if c₁A₁ + ... + c_nA_n = 0, with some nonzero coefficient, then T(c₁A₁ + ... + c_nA_n) = c₁T(A₁) + ... + c_nT(A_n) = 0. Thus a linear T always maps linearly dependent sets to linearly dependent sets.

• K) If T: Rⁿ -> Rⁿ is linear, and Ker(T) + Im(T) = Rⁿ, then Ker(T) intersect Im(T)={0}.
  
  This is TRUE. By the Dimension Theorem, dim Ker(T) + dim Im(T) = n. The equality Ker(T) + Im(T) = Rⁿ is given. We know from our formula of dimensions of sums of subspaces (Discussion 4) that n = dim(Ker(T) + Im(T)) = dim Ker(T) + dim Im(T) - dim (Ker(T) intersect Im(T)). Combining these equations, we see dim (Ker(T) intersect Im(T)) = 0, which implies that Ker(T) intersect Im(T) = {0}.