Mathematics 244 – Linear Algebra Solutions – Review True-False Questions for Exam 2 March 24, 2004

A) If T : V → W is linear, dim(V) = 5, dim(W) = 7, then T is surjective.

This is FALSE. For instance, letting T be the zero mapping gives a counterexample since Im(T) = {0_{W} } ≠ W.

B) Every linear T : V → W, dim(V) = 5, dim(W) = 7, is T is injective.

This is FALSE. For instance, letting T be the zero mapping gives a counterexample since Ker(T) = V ≠ {0_{V} }.

C) Some linear T : V → W is linear, dim(V) = 5, dim(W) = 7, then T is injective.

This is TRUE. For instance, let β = {v_{1}, …, v_{5}} be a basis for V, and let γ = {w_{1}, …, w_{5}} be any linearly independent subset of W. Then the unique linear T : V → W with T (v_{i}) = w_{i}, i = 1, …, 5 is injective. Proof: We show this by showing T satisfies Ker(T) = {0_{V} }. Suppose T (v) = 0_{W}. Then since β is a basis for V,

v = a_{1}v_{1} + ⋯ + a_{5}v_{5}

for some a_{i} ∈ R. But then, since T is linear,

0_{W} = T (v) = a_{1}T (v_{1}) + ⋯ + a_{5}T (v_{5})

Since γ is linearly independent, this implies a_{i} = 0 for all i, which says v = 0_{V}. Hence Ker(T) = {0_{V} }.

D) If {v_{1}, …, v_{n}} in V is linearly dependent, and T : V → W is linear then {T (v_{1}), …, T (v_{n})} is linearly dependent.

This is TRUE. If {v_{1}, …, v_{n}} is linearly dependent, then there exist scalars a_{i}, not all equal to zero, such that

a_{1}v_{1} + ⋯ + a_{n}v_{n} = 0_{V}

But then T linear shows

T (a_{1}v_{1} + ⋯ + a_{n}v_{n}) = a_{1}T (v_{1}) + ⋯ + a_{n}T (v_{n}) = T (0_{V}) = 0_{W}

This implies that {T (v_{1}), …, T (v_{n})} is linearly dependent since some a_{i} ≠ 0.

E) If T : R^{n} → R^{n} is linear then Ker(T) ∩ Im(T) = {0}.

This is FALSE. A counterexample is given by the mapping T : R^{2} → R^{2} with standard matrix

[T] = ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} )

(as in a problem from the last problem set). For this T, Ker(T) = Im(T) = Span((1, 0)), so Ker(T) ∩ Im(T) = Span((1, 0)).

F) If T : V → V is linear and β = {v_{i}} and β′ = {cv_{i}} for some scalar c ≠ 0, then [T]_{β}^{β} = [T]_{β′}^{β′}.

This is TRUE. The scalars in the jth column of [T]_{β}^{β} are the a_{ij} in

T (v_{j}) = a_{1j}v_{1} + ⋯ + a_{nj}v_{n}

By linearity and one of the “distributive laws” for scalar multiplication:

T (cv_{j}) = cT (v_{j})
= c(a_{1j}v_{1} + ⋯ + a_{nj}v_{n})
= a_{1j}(cv_{1}) + ⋯ + a_{nj}(cv_{n})

Hence the jth column of [T]_{β′}^{β′} is the same as the jth column of [T]_{β}^{β} for all j, so the matrices are equal.

G) If A is an n×n matrix and c is a fixed real number then the set W = {v ∈ R^{n} : Av = cv} is a vector subspace of R^{n}.

This is TRUE. Let v, w ∈ W and let d be an arbitrary scalar. Then by distributivity of matrix multiplication over matrix sums,

A(dv + w) = dAv + Aw
= dcv + cw
= c(dv + w)

Hence dv + w ∈ W, and W is a vector subspace.

H) If T = I_{R^{2}} is the identity mapping, and β, β′ are two bases for R^{2}, then

[T]_{β}^{β′} = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) .

This is FALSE. A counterexample is: Let β = {(3, 4), (1, −1)}, and let β′ be the standard basis. Then

[T]_{β}^{β′} = ( \begin{matrix}3 & 1 \\ 4 & −1 \end{matrix} )

(look at the columns).

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 {v_{1}, ..., v_{n}} is a basis for U, then we know {T (v_{1}), ..., T (v_{n})} spans T (U). Furthermore, since T is an isomorphism, it is injective. Hence, if c_{1}T (v_{1}) + ⋯ + c_{n}T (v_{n}) = 0, then T (c_{1}v_{1} + ⋯ + c_{n}v_{n}) = 0 which implies c_{1}v_{1} + ... + c_{n}v_{n} = 0, which implies c_{i} = 0 for i = 1, ...n. Thus {T (v_{1}), ..., T (v_{n})} is linearly independent, hence a basis for Im(U). Hence the dimensions of U and T (U) are equal.

J) If v_{1}, …, v_{n} are vectors in a vector space V, and w_{1}, …, w_{n} are vectors in a vector space W, then there always exists some linear T : V → W satisfying T (v_{i}) = w_{i} for all i = 1, ..., n.

This is FALSE because we could consider a case where the v_{i} are linearly dependent, but the w_{i} are linearly independent. There is no linear mapping that takes a linearly dependent set to a linearly independent set (see question D above).

K) If T : R^{n} → R^{n} is linear, and Ker(T) + Im(T) = R^{n}, then Ker(T) ∩ Im(T) = {0}.

This is TRUE. By the Dimension Theorem, dim Ker(T)+dim Im(T) = n. The equality Ker(T) + Im(T) = R^{n} is given. Hence if β is a basis for Ker(T) and γ is a basis for Im(T), then β ∪ γ is a basis for R^{n}, because it spans R^{n} and contains exactly n vectors. If Ker(T) ∩ Im(T) contained some nonzero vector, v, then there would be an equation v = w for some v ∈ Ker(T) and some w ∈ Im(T) but then from v − w = 0, expanding v in terms of β and w in terms of γ, there would be a linear dependence on the vectors in β ∪ γ. This is impossible, so Ker(T) ∩ Im(T) = {0}.