A = 1 6 (y 1 8y 2 5y 3 ) Therefore, a general solution to this system is given by

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1 Mid-Term Solutions 1. Let A = For which triples (y 1, y 2, y 3 ) does AX = Y have a solution? Solution. The following sequence of elementary row operations: R 1 R 1 /3, R 1 2R 1 + R 2, R 3 R 1 + R 3, R 2 (3/5)R 2, R 1 = (1/3)R 2 + R 1, R 3 = (8/3)R 2 + R 3, R 3 ( 5/6)R 3, R 1 = ( 3/5)R 3 + R 1, R 2 = (1/5)R 3 + R 2 reduces the augmented matrix to its row-reduced echelon form given below Let (y 1 2y 2 y 3 ) (y 1 2y 2 + y 3 ) (y 1 8y 2 5y 3 ). Therefore, a general solution to this system is given by (x 1, x 2, x 3 ) = ( 1 2 (y 1 2y 2 y 3 ), 1 6 (y 1 2y 2 + y 3 ), 1 6 (y 1 8y 2 5y 3 )). 2. In each of following, verify whether the S forms a basis for the vector space V. (a) S = {(2, 5, 1), (4, 1, 2), (6, 4, 1)}, V = R 3. (b) S = {(1, 0, i), (1 + i, 1 i, 1), (i, i, i)}, V = C 3. Solution. (a) Since (2, 5, 1) + (4, 1, 2) = (6, 4, 1), S is a linearly dependent set, and hence cannot form a basis for R 3. (b) Suppose that there exists a, b, c C such that a(1, 0, i) + b(1 + i, 1 i, 1) + c(i, i, i) = (0, 0, 0). Then we get the follwing system of linear equations: a + (1 + i)b + ic = 0 (1 i)b + ic = 0 ia + b + ic = 0 (1) 1

2 Multiplying the first equation in system (1) with i and adding to the third, we obtain: ib + (i 1)c = 0 (2) Using equation (2) and the second equation of (1), we can conclude that b = c = 0, which upon substitution in the first equation in system (1) yields a = 0. Thus S is a linearly independent set of three vectors in C 3, and consequently S forms a basis for C Let V, W be finite dimensional vector spaces over a field K. Let T : V W be a linear map. (a) Define Ker T. (b) Prove that T is injective if and only if Ker T = {0}. Solution. (a) Let V, W be finite dimensional vector spaces over a field K. Let T : V W be a linear map. Then the kernel of T, denoted by Ker T, is defined by Ker T = {v V : T (v) = 0}. (b) Suppose that T is injective and T (v) = 0. Then v = 0, which would imply that Ker T = {0}. Conversely, let us assume that Ker T = {0}. If for v, w V we have that T (v) = T (w), then T (v w) = 0, that is v w Ker T. Since Ker T = {0}, this would imply that v = w, thereby showing that T is injective. 4. Let V, W be finite dimensional vector spaces over a field K. Let T : V W be a linear map. (a) State the Rank-Nullity Theorem for T. (b) Let dim V = dim W. Show that T 1 exists if and only if T is either injective or surjective. Solution. (a) Rank-Nullity Theorem: Let V, W be finite dimensional vector spaces over a field K. Let T : V W be a linear map. Then dim V = dim(ker T ) + dim(im T ). (b) If T 1 exists, then T has to be a bijective map. Conversely, let us assume that T is injective. Then Ker T = {0} and by the Rank-Nullity Theorem dim V = dim (Im T ). Since dim V = dim W, we have that 2

3 dim (Im T ) = dim W, which would imply that W = Im T, that is, T is surjective. Alternatively, if we assume that T is surjective, then W = Im T, and since dim V = dim W, the Rank-Nullity Theorem would imply that Ker T = {0}, which shows that T is injective. 5. Let M = {A : A = (A ij ) n n and A ij R}, and let tr : M R be defined as tr(a) = A ii, for A M. (a) Show that tr is a linear map. (b) Show that tr(ab) = tr(ba), for A, B M. (c) If B is invertible, then show that tr(b 1 AB) = tr(a). Solution. (a) By definition, tr(a+b) = (A+B) ii = (A ii +B ii ) = A ii + B ii = tr A+tr B, A, B M. Also, tr(ca) = (ca) ii = c A ii = ctr A, c R and A M. Therefore, T is a linear map. (b) For all A, B M, we have that tr(ab) = (AB) ii = A ik B ki = k=1 B ki A ik = k=1 (BA) kk = tr(ba), k=1 which proves the result. (c) Since B is invertible, we have that B 1 exists. Since tr(ab) = tr(ba), we have that tr(b 1 AB) = tr(b 1 (AB)) = tr((ab) B 1 ) = tr(a(bb 1 )) = tr(ai n ) = tr(a), thereby proving the result. 3

4 6. Prove or disprove the following statements. (a) Let V and W be finite dimensional vector spaces over a field K such that dim V dim W. Then any linear map T : V W has a nonzero kernel. (b) Any subset of a linearly independent set is linearly independent. (c) Any subset of a linearly dependent set is linearly dependent. Solution. (a) This statement is false. For n < m, consider the injective linear map T : K n K m defined by T (x 1,..., x n ) = (x 1,... x n, x n+1,... x m ). Since ker T = {0}, T is clearly a counterexample to the statement. (b) This statement is true. Let A = {v 1,... v n } be a linearly independent set in a vector space V over a field K. Without loss of generality, for k < n, let B = {v 1,... v k } be a subset of A. Suppose we assume on the contrary that B is linearly dependent. Then there exists c 1,..., c k K, with not all c i = 0, such that c 1 v c k v k = 0. Let c i = 0, for k+1 i n. Then c 1 v c k v k +c k+1 v k c n v n = 0, with not all c i = 0, which implies that A is a linearly dependent set. But this contradicts our earlier assumption that A is a linearly independent set. (c) This statement is false. Let V be a vector space of dimension n over a field K and let B = {v 1,..., v n } be a basis for V. If we take a nonzero vector v V \ B, then the maximal linear independence of B would imply that A = {v, v 1,..., v n } is a linearly dependent set. Since B A, this is an apparent counterexample to the statement. (Bonus). Answer any one of the following questions. (a) Let V be a vector space over a field K and let T be a linear operator on V such that T 2 = T. Prove that V = Ker T Im T. Solution. We are given a linear map T : V V such that T 2 (v) = (T T )(v) = T (T (v)) = T (v), v V. In other words, T (v T (v)) = 0, that is, v T (v) Ker T, v V. Now every v V can be expressed v = (v T (v)) + T (v), where v T (v) Ker T and T (v) Im T. Since both Im T and Ker T are subspaces of T, we have that V = Im T + Ker T. It remains to show that V = Im T Ker T = {0}. Suppose we assume on the contrary that there is a nonzero v V. Since v 4

5 Ker T, we have that T (v) = 0. Also, since v Im T, there exists nonzero u V such that T (u) = v, that is, T (u) v = 0. Wde can see that 0 = T (0) = T (T (u) v) = T 2 (u) T (v) = T (u) = v. Hence v = 0, which is true for any v Im T Ker T. Therefore, Im T Ker T = {0} and V = Ker T Im T. (b) Give a counterexample to show that 4(b) does not hold good for infinite dimensional vector spaces. Solution. Example 1. Let V = R, or the vector space of all sequences in R. (Note that proving R is a vector space over R is analogous to the proof for R n. Therefore, this is left as an exercise). Consider the map T : V V defined by T ((x 1, x 2, x 3,...)) = (x 2, x 3, x 4,...), or more formally, if a = {a n } n=1 V, then T (a) = {a n+1 } n=1. For a, b V and c R, we have that T (a + b) = {a n + b n } n=1 = {a n } n=1 + {b n } n=1 = T (a) + T (b), and T (ca) = {ca n } n=1 = c{a n } n=1 = ct (a). Therefore, T is a linear map. For any x = {x n } n=1 V, T ((a, x 1, x 2, x 3,...)) = (x 1, x 2, x 3,...), a R, which shows that T is surjective but not injective. Therefore, T is a counterexample to 4(b) for the infinite dimensional case. Example 2. Alternatively, in the same setting as above, the map T ((x 1, x 2, x 3,...)) = (0, x 1, x 2, x 3,...) is an injective map but not a surjective map. Example 3. Let V be denote vector space of all real-valued differentiable functions on [a, b], where a, b R and a < b. The derivative operator D on V is clearly not injective as the derivative of any constant is 0. However, D is surjective as any f V has an antiderivative F by the Fundamental Theorem of Calculus. 5

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