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MA 237-102 Linear Algebra I Homework 5 Solutions 3/3/10 1. Which of the following transformations are linear? Justify your answers. (i) T : P2 → P3 defined by T (p(t)) = t3 p′ (0) + t2 p(0); For any polynomials p(t) and q(t) we have T ((p + q)(t)) = t3 ((p + q)′ (0))+t2 (p+q)(0) = t3 p′ (0) + t2 p(0) + t3 q ′ (0) + t2 q(0) . Similarly for any real number a we have T (ap(t)) = t3 ((ap)′ (0)) + t2 (ap)(0) = a t3 p′ (0) + t2 p(0) . It follows that T is linear. (ii) T : P1 → P2 defined by T (p(t)) = tp(t) + p(0); For any polynomials p(t) and q(t) we have T ((p + q)(t)) = t ((p + q)(t)) + (p + q)(0) = (tp(t) + p(0)) + (tq(t) + q(0)) . Similarly for any real real number a we have T (ap(t)) = t ((ap)(t)) + (ap)(0) = a (tp(t) + p(0)) . Thus T is linear. (iii) T : P1 → P2 defined by T (p(t)) = tp(t) + 1. If T (p(t)) = tp(t) + 1, then T (3p(t)) = t(3p(t)) + 1 = 3tp(t) + 1, which is not the same as 3T (p(t)) = 3tp(t) + 3. Thus T is not linear. 2. Let V and W be vector spaces and suppose T : V → W is a linear transformation. (i) Show that T (0V ) = 0W ; For any vector X in V we have T (0V ) = T (0 · X) = 0 · T (X) = 0W . (ii) Show that the set of vectors X in V with the property that T (X) = 0 is a subspace of V (this subspace is called the kernel of T ); If T (X) = 0 and T (Y ) = 0, then for any real numbers s and t we have T (sX + tY ) = sT (X) + tT (Y ) = s0 + t0 = 0. Thus the kernel of T is a subspace of V. (iii) Show that the set of vectors Y in W such that the equation T (X) = Y is solvable is a subspace of W (this subspace is called the image of T ). Assume that Y1 and Y2 are in the image of T . Then there are (not necessarily unique) vectors X1 and X2 so that T (X1 ) = Y1 and T (X2 ) = Y2 . Then for any real numbers s and t we have T (sX1 + tX2 ) = sT (X1 ) + tT (X2 ) = sY1 + tY2 , so that sY1 +tY2 is also in the image of T . a subspace of W. Thus the image of T is 3. (i) Find the matrix representation for the linear map T : R3 → R2 given by x 2y − z T y = . 3x − y z We apply T to the standard basis vectors in R3 : 1 0 T , = 0 3 0 0 2 , T = 1 −1 0 0 −1 . T = 0 0 1 It follws that the matrix representation of T is 0 2 −1 . AT = 3 −1 0 (ii) Find a basis for the kernel of T . The kernel of T is exactly the nullspace of AT , so we row-reduce AT as follows: 3 −1 0 0 2 −1 . 0 2 −1 3 −1 0 Solving the corresponding homogeneous system gives z = t, 2y − z = 0 ⇒ y = t/2 3x − y = 0 ⇒ x = t/6, so the solution is t/6 1/6 t/2 = t 1/2 . t 1 It follows that this vector is a basis for the nullspace of AT , and hence for the kernel of T . (iii) Find a basis for the image of T . The image of T is the column space of AT . From the row-reduction above, we see that the first two columns of AT are its pivot columns. Thus a basis for the column space of AT , and hence also for the image of T , is 0 2 , . 3 −1