Math 272 Practice Problems Involving Linear Transformations

Math 272 Practice Problems Involving Linear Transformations 1. Suppose that T : V → W is a linear transformation. Prove that T is one-to-one if and only if the only solution to T (v) = 0 is v = 0. 2. For each of the following transformations, determine the kernel and the range and whether the transformation is one-to-one and/or onto. (a) T : R2 → R2 , T (x, y) = (2x − 3y, 5x + y). (b) T : R2 → R2 , T (x, y) = (8x + 4y, 2x + y). (c) T : R3 → R2 , T (x, y, z) = (x − y, y − z). (d) T : R2 → R3 , T (x, y) = (2x − 3y, 5x + y, y). 3. Suppose that a linear transformation T : Rn → Rn is defined by T v = Av, where A is an n × n matrix with detA=0. Can T be one-to-one? Can T be onto? Explain. 4. If A is an m × n matrix and the linear system Ax = b has at least one solution for every vector b in Rm , what is the range of TA : Rn → Rm ? 5. Let T be a transformation from the set of polynomials of degree 2 or less to R2 defined by T (p) = (p(1), p(−1)). What are the range and kernel of T ? Is T one-to-one? Is it onto? 6. If T : V → W is a linear transformation that is both one-to-one and onto, then for each vector w in W there is a unique vector v in V such that T (v) = w. Prove that the inverse transformation T −1 : W → V defined by T −1 (w) = v is linear. 7. Let V be a vector space and let v1 , v2 ∈ V . This gives subspaces Span{v1 } and Span{v2 } of V . Prove that Span{v1 } + Span{v2 } = Span{v1 , v2 }, where Span{v1 } + Span{v2 } = {u + v| u ∈ Span{v1 }, v ∈ Span{v2 }}. 8. Let V be a vector space and assume that v1 , v2 ∈ V are linearly independent. Given v3 ∈ V , prove that v1 , v2 , v3 are linearly dependent if and only if v3 ∈ Span{v1 , v2 }. Prove this directly from the definitions. 9. Let V be a vector space and assume that v, w ∈ V . Prove that Span{v, w} = Span{v+ w, v − w}. 10. Prove that the kernel of a linear transformation T : V → W is a subspace of V . You should prove this directly from the appropriate definitions. 11. Prove that the range of a linear transformation T : V → W is a subspace of W . You should prove this directly from the appropriate definitions. 12. Let T : V → W be linear and one-to-one. Suppose v1 , · · · , vn ∈ V are linearly independent. Prove that T (v1 ), · · · , T (vn ) are linearly independent. You should prove this directly from the appropriate definitions. 13. Let T : V → W be linear and let v1 , · · · , vn be a basis of V . Also assume that T (v1 ), · · · , T (vn ) are linearly independent. Prove that T is one-to-one. You should prove this directly from the appropriate definitions. 14. Let T : V → W and S : W → Z be linear maps between vector spaces. Prove that the composition S ◦ T : V → Z is linear. You should prove this directly from the appropriate definitions. 15. Let T : V → W be linear and onto. Also assume that v1 , · · · , vn span V . Prove that T (v1 ), · · · , T (vn ) span W . You should prove this directly from the appropriate definitions. 16. Let T : V → W be linear. Also assume that v1 , · · · , vn ∈ V have the property that T (v1 ), · · · , T (vn ) span W . Prove that T is onto. You should prove this directly from the appropriate definitions. Selected Solutions 1. First suppose T is one-to-one, so T (u) = T (v) implies u = v. For any linear transformation, T (0) = 0. If T (u) = 0, then T (v) = T (0), so we must have v = 0. Now suppose the only solution to T (v) = 0 is v = 0. If T (u) = T (v), then T (u − v) = 0 (using linearity of T ) and so u − v = 0 (using the assumption), that is, u = v. This implies that T is one-to-one. 2. (a) range(T )=R2 , ker(T )={0}. T is both one-to-one and onto (so invertible). (b) range(T )=Span{[4, 1]}, ker(T )=Span{[−1, 2]}. T is neither one-to-one nor onto. (c) range(T )=R2 , ker(T )=Span{[1, 1, 1]}. T is onto but not one-to-one. (d) range(T )=Span{[2, 5, 0], [−3, 1, 1]}, ker(T )={0}. T is one-to-one but not onto. 3. T v = Av where |A| = 0 implies that T is singular so neither one-to-one nor onto. In particular, the kernel of T and the null space of A contain more than the zero vector, so T is not one-to-one. Also, the columns of A must be linearly dependent since |A| = 0, so the n columns do not span Rn , implying that range(T )=col(A) is not all of Rn and T cannot be onto. 4. The range of TA must be Rm since for every b ∈ Rm we can find x ∈ Rn such that Ax = b. 5. T (a2 x2 + a1 x + a0 ) = (a2 + a1 + a0 , a2 − a1 + a0 ). The range of T is all of R2 , and the kernel of T is Span{x2 − 1}. T is not one-to-one, but it is onto. 6. Take any w1 , w2 ∈ W and scalar c. Let v1 = T −1 (w1 ) and v2 = T −1 (w2 ). Using the linearity of T and that T −1 (T (v)) = v for all v ∈ V , T −1 (cw1 + w2 ) = T −1 (cT (v1 ) + T (v2 )) = T −1 (T (cv1 + v2 )) = cv1 + v2 = c T −1 (T (v1 )) + T −1 (T (v2 )) = c T −1 (w1 ) + T −1 (w2 ). Therefore T −1 is linear. 7. First take any u ∈ Span{v1 } + Span{v2 }, so we can write u = u1 + u2 for some vectors u1 ∈ Span{v1 } and u2 ∈ Span{v2 }. We can write u1 = c1 v1 and u2 = c2 v2 for some scalars c1 and c2 , so u = c1 v1 +c1 v2 ∈ Span{v1 , v2 }. Therefore Span{v1 }+Span{v2 } ⊆ Span{v1 , v2 }. Now take any u ∈ Span{v1 , v2 }, so u = c1 v1 + c1 v2 for some scalars c1 and c2 . Let u1 = c1 v1 ∈ Span{v1 } and u2 = c2 v2 ∈ Span{v2 }. Then u = u1 +u2 ∈ Span{v1 }+Span{v2 }. Therefore Span{v1 , v2 } ⊆ Span{v1 } + Span{v2 }. We conclude that Span{v1 } + Span{v2 } = Span{v1 , v2 }.