Math 115 Midterm Solutions

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1 Math 115 Midterm Solutions 1. (25 points) (a) (10 points) Let S = {0, 1, 2}. Find polynomials with real coefficients of degree 2 P 0, P 1, and P 2 such that P i (i) = 1 and P i (j) = 0 when i j, for i, j S. (b) (10 points) Show that P 0, P 1, and P 2 are linearly independent in the vector space of all polynomials of degree at most 2. (c) (5 points) Show that P 0, P 1, and P 2 span the space of all polynomials of degree at most 2. State clearly any theorems you might use. Solution to Problem 1. (a) The P i can be found using the Lagrange interpolation formula. So P 0 = (X 1)(X 2) (0 1)(0 2) = 1/2X 2 3/2X + 1, P 1 = (X 0)(X 2) (1 0)(1 2) = X 2 + 2X and P 2 = (X 0)(X 1) (2 0)(2 1) = 1/2X 2 1/2X. (b) Suppose ap 1 + bp 2 + cp 3 = 0. Evaluating at 0, 1, 2 respectively, we find a = 0, b = 0, and c = 0. For example, at X = 0 the relation becomes ap 1 (0) + bp 2 (0) + cp 3 (0) = 0. Since P 2 (0) = P 3 (0) = 0 and P 0 (0) = 1, this says a = 0. Thus, the P i are linearly independent. (c) By what we have just seen, the P i span a vector space W of dimension three. However, each P i has degree two, so this W is a subspace of the space of all polynomials of degree at most two. We know that this space has dimension three too. Hence, by Theorem 1.11, the two spaces are the same, and so every polynomial of degree at most two is a linear combination of the P i. 1

2 2. (25 points). (a) (15 points) Let V be a finite dimensional vector space and let L be a linearly independent subset. Show that there exists a basis of V which contains L. (b) (10 points) Illustrate this theorem by finding a basis of F 4 which contains (1, 1, 1, 3) and (1, 1, 1, 1). Solution(s) to Problem 2. Let B be a finite spanning set of V. Since V is finite dimensional, such a B exists. Let T be a subset of the union of L and B which contains L, is linearly independent, and has the largest number of elements for any subset with these two properties. Thus any subset of the union of B and L with T + 1 elements or more must be linearly dependent. Suppose T does not span V. Then there exists a v in V that is not in the span of T. Then the set gotten by adding v to T is linearly independent. (You can quote this. It was done in class and was on the last quiz, but here is a proof again: Consider a linear combination of the form av+(linear comb. of vectors in T ) = 0. Then, if a = 0, this really a linear comb. of elements of T. Since T is a linearly independent set, the coefficients must be all 0. On the other hand, suppose a 0. Then the equation av+(linear comb. of vectors in T ) = 0 can be solved for v, showing that v would in this case be in the span of T. Since we are assuming this is not true, the only case allowed is that where a = 0, and hence, as we saw first, all coefficients are zero. Thus the set consisting of T enlarged by adding v is linearly independent.) Now we can conclude the proof: Since the enlarged set consisting of T with v added is larger than T and linearly independent, T can t be both maximal and fail to span V. Since we assumed it IS maximal, T is therefore a linearly independent spanning set of V, i.e. a basis of V. If you don t wish to argue with a maximal set as above, you can construct the basis inductively by adding suitable elements v j of B to L one at a time. Thus, if you have already constructed a linearly independent set T j 1 = T {v 1,, v j 1 } you are done if T j 1 actually spans V and otherwise you choose any v j not in the span of T j 1 and form T j = T {v 1,, v j } which is linearly independent (again quoting this fact, or proving it; take each 2

3 T j 1 for T in the proof in the last paragraph). Since B does span, there must be a j for which T j spans V. (Otherwise you could continue, adding all of B, and still not get a spanning set, which is impossible.) Part b. of Problem: Conceptual method: Using the unique linear combination ae 3 + be 4 of the standard basis vectors e 3 and e 4, we can make any vector of the form (0, 0, a, b). If we put v = (1, 1, 1, 3) and w = (1, 1, 1, 1), then y = 1/2(w + v) = (1, 0, 1, 2) and z = 1/2(v w) = (0, 1, 0, 1). Thus, there is a unique linear combination of y and z (and hence of w and v) which gives a vector with any given entries in the first two positions. Adding in a suitable unique linear combination of e 3 and e 4 (which will not change the first two entries), we can adjust to get any entries we like in the last two positions also. Therefore v, w, e 3 and e 4 span F 4 and only one linear combination will give any given vector, i.e. v, w, e 3 and e 4 are linearly independent. Alternatively, you can conclude linear independence as follows: Suppose that {v, w, e 3, e 4 } were not a basis. Then some vector in the set could be eliminated without changing the span. Continuing in this way, we would eventually get a basis for F 4 with fewer than 4 elements. But this would contradict the fact that all bases have the same number of elements, and {e 1,, e 4 } is a basis. You could actually use any pair of standard basis vectors of F 4 in this argument (suitably altered). Part b. of Problem: Direct Method. (also using e 3 and e 4 but other choices will work). Solve (a, b, c, d) = αv + βw + γe 3 + δe 4. We get four equations: a = α + β, b = α β, c = α + β + γ, d = 3α β + δ Solving these we get α = 1/2(a + b), β = 1/2(a b), and γ = c a, and δ = d + 3α + β = d + 2a + b. Thus v, w, e 3 and e 4 span F 4, and since these solutions are unique, they are linearly independent (i.e. if a = b = c = d = 0 then we must have α = β = γ = δ. 3

4 3. (25 points) (a) (10 points) State the dimension theorem for a linear transformation T : V W of finite dimensional vector spaces. (b) (15 points)consider the space P 4 of polynomials of degree 4 (with real coefficients). Consider the map T from P 4 to R 2 defined by T (P ) = (P (1), P (1)). Compute (proving your answer) the dimension of the kernel of T. Solution to Problem 3. (a) Let T : V W be a linear transformation, where V and W are vector spaces of finite dimension. Let R(T ) be the range of T of dimension r, and let N(T ) be the null space (kernel ) of T, of dimension n. Then dim(v ) = r + n. (b) T (1) = (1, 0), and T (X 1) = (0, 1). T is a linear transformation and R(T ) contains a basis of R 2. Hence R(T ) has dimension two. Since P 4 has dimension 5, N(T ) has dimension 3 by the dimension theorem. 4

5 4. (25 points) Let P 2 be the space of polynomials of degree 2 with real coefficients. Let T : P 2 P 2 be the map defined by T (P )(X) = P (X + 1). (a) (15 points) Show that T is a linear map from P 2 to P 2. (b) (10 points) Compute the matrix [T ] α of T relative to the standard basis α = {1, X, X 2 } of P 2. Solution to Problem 4. (a) Let α F. Let P (X) = ax 2 +bx +c and Q(X) = dx 2 +ex +f be polynomials. Then T (αp ) = (αa)(x + 1) 2 + (αb)(x + 1) + (αc) = α(a(x + 1) 2 + b(x + 1) + c)t = αt (P ) and T (P + Q) = (a + d)((x + 1) 2 + (b + e)(x + 1) + (e + f) = (a(x + 1) 2 + b(x + 1) + e) + (d(x + 1) 2 + e(x + 1) + f) = T (P ) + T (Q). Hence T is linear. (b) T (1) = 1, T (X) = X + 1, T (X 2 ) = (X + 1) 2 = X 2 + 2X + 1. Hence [T ] α = (a ij ) where a 11 = 1, a 21 = 0, a 31 = 0, a 12 = 1, a 22 = 1, a 32 = 0, a 13 = 1, a 23 = 2, a 33 = 1. 5