DEPARTMENT OF MATHEMATICS

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1 DEPARTMENT OF MATHEMATICS Ma322 - EXAM #2 Spring 22 March 29, 22 Answer: Included answers DO NOT TURN THIS PAGE UNTIL YOU ARE INSTRUCTED TO DO SO. Be sure to show all work and justify all your answers. There are 8 problems and a total of 9 pages including this one. No other sheets, books, papers are allowed. Maximum Actual Problem Score Score Bonus 4 Total NAME:

2 . (a) Let A = 2 3. Write down the augmented matrix (A I) and find its REF. Answer: (b) Use your work to find the consistency matrix G for the matrix A. Answer: ( ). Use your consistency matrix to find a value of t for which the vector 2 + t is in Col(A). 5 Answer: Solve ( ) 2 + t 5 (c) You are given that certain row operations give: (M I) = =. This gives t = (d) ( Use this to determine ) a consistency matrix for M. Answer: G = (e) Use the consistency matrix to show why the vector v = 9 27 ( ) is in Col(M). Answer: Check Gv =. (f) Using your above calculations or otherwise, solve the system of 2 equations M X = v. Answer: Let H = Then we may solve HMX = Hv. These have ( last) two rows zero 9 and the first two rows give the answer X =. 2

3 2. Let A = 3 2. (a) Calculate A. Calculator answer is accepted. Answer:. 2 (b) Calculate det(a). This must be calculated by a suitable expansion. Work must be shown. Answer: We carry out C 2 + C to get det 2 = ()(+)((2)() (3)()) =. 3 (c) Calculate adj(a). You may use above work, but write down the correct formula that you use. Answer: Using the formula adj(a) = det(a)a we get the negative of the inverse above. (d) Calculate the determinant 2t 2 t 2 3 Your final answer must be in factored form. Hint: It is best to use expansion by convenient rows or columns to bring out the factors. Answer: Expanding by column, ( 2t)(2 t ( 2)(3)) = ( 2t)(8 t).. 2

4 3. Consider an abstract vector space V with a given basis B = ( b b 2 b 3 ). You should answer various questions concerning this vector space. (a) You are given vectors c = b 2b 2 + b 3, c 2 = 3b 5b 2, c 3 = b 2b 2. Calculate the coordinate vectors [c ] B, [c 2 ] B, [c 3 ] B. 3 Answer: [c ] B = 2, [c 2 ] B = 5, [c 3 ] B = 2 (b) Let W be the subspace of V spanned by c 2, c 3. What is the dimension of W and why? Answer: The rank of the matrix of coordinate vectors of c 2, c 3 is 3 exactly 2, since it has a non zero 2 2 determinant 5 2 and no bigger determinant. Hence the two vectors are independent. Since they span W, they form a basis of W. Hence dim(w ) = 2. (c) Find a vector u V such that ( c 2 c 3 u ) is a basis for V. You must explain why it is a basis. The explanation is what will be graded. Answer: We take u = b 3 with [u] B =.. The matrix of the coordinate vectors of c 2, c 3, b is easily seen to be rank 3 and hence they span a 3 dimensional space. Since this is a subspace of the 3 dimensional space V, they span V and hence form its basis. (d) Do the three vectors ( ) b c 2 c 3 form a basis of V? You must prove your claim. Answer: No. All three vectors are inside the 2 dimensional space Span{b, b 2 } and so can span a space of dimension at most 2. Hence, they don t span V. Indeed the span of c 2, c 3 is already seen to be the same as span of b, b 2 and we could also argue that b, c 2, c 3 are dependent, since b would be in the span of c 2, c 3. Explicitly, b = 2c 2 5c 3. 3

5 4. Complete the following definition. Let V be a real vector space. (a) A set of vectors {v, v 2,, v r } in V is said to be linearly dependent if: Answer: there exist scalars c,, c r such that c v + +c r v r =. (b) A set of vectors {v, v 2,, v r } in V is said to be a basis of V if: Answer: they span V and are linearly independent. (c) The dimension of V is defined to be: Answer: the number of elements in any basis of V. (d) Recall that P is the vector space of all polynomials in x with real coefficients. Given any integer m, show that there are more than m polynomials in x which are linearly independent. Explain why this means dim(p ) is infinite. Answer: Consider the m + polynomials, x,, x m. By the definition of polynomials, they are independent. To prove dim(p ) = we argue by contradiction. Suppose, if possible P has finite dimension, m, say. Then it cannot have more than m independent polynomials. But we exhibited m + independent polynomials in P above. Contradiction! Thus, we have proved that dim(p ) =. 4

6 5. Complete the following definition. Let V, W be real vector spaces. A function T : V W is said to be a linear transformation if the following two conditions hold: (a) For any two vectors v, w V we have: Answer: T (v + w) = T (v) + T (w). (b) For any vector v V and c R we have: Answer: T (cv) = ct (v). (c) Consider a map T : P P given by T (f(x)) = f() + f ()x. Prove that T is a linear transformation using the above definition. It is crucial to check both conditions. Answer: T (f(x) + g(x)) = f() + g() + (f () + g ())x = (f() + f ()x) + (g() + g ()x) = T (f(x)) + T (g(x)). Also T (c(f(x)) = cf() + cf ()x = ct (f(x)). 5

7 6. Consider a linear transformation L : P 3 P 2 defined by L(p(x)) = x 2 p (x) 6p(x). Answer the following questions. All answers must be supported by appropriate arguments and properly described calculations. Unsupported answers will receive no credit. (a) Calculate the images L(), L(x), L(x 2 ), L(x 3 ). Answer:. L() = 6, L(x) = 6x, L(x 2 ) = 4x 2, L(x 3 ) = (b) Choose basis B = ( x x 2 x 3 ) for P 3 and the basis C = ( x x 2 ) for P 2. Use the above calculations and find a matrix A of the transformation L with respect to the bases B, C. Remember: Your matrix would be Answer: A = 6. 4 (c) Using the matrix A or otherwise, calculate a basis for Ker(L). Answer: The matrix has evidently rank 3 (it is in REF), so the null space has dimension 4 3 =. An obvious vector in it and hence a basis for it is:. The corresponding basis for Ker(L) is B times it, so x3. (d) Using the matrix A or otherwise, calculate basis for Im(L). Answer: The Col(A) has dimension 3 = rank(a) and clearly has three independent vectors 6, 6x, 4x 2. So, they form a basis. We could also take the simpler, x, x 2. (e) Prove or disprove the statement that L is injective. Answer: Since Ker(L) is non zero, it is not injective. (f) Prove or disprove the statement that L is surjective. Answer: Since Im(L) has dimension 3 and it is a subspace of the target P 2 of dimension 3, it is equal to P 2. So L is surjective. We could also simply note that the image contains a basis, x, x 2 of P 2, hence is equal to it! 6

8 7. Given a linear transformation T : V W there is a fundamental theorem about dimensions of various vector spaces associated with T : dim(v ) = dim(ker(t )) + dim(im(t )). Answer the following questions related to this theorem. show how the theorem is used. Be sure to (a) Let L : V W be a linear transformation. Suppose that dim(v ) = 5 and dim(w ) 3. Either prove that L is not injective or give an example of such an injective transformation for V and W of indicated dimensions. Answer: Note that dim(im(l)) dim(w ) 3. Let dim(im(l)) = 3 ɛ where ɛ. Then our equation says 5 = dim(ker(l))+3 ɛ. This shows that dim(ker(l)) = ɛ 2. Thus L is not injective. (b) Let L : V W be a linear transformation. Suppose that dim(v ) = 2 and dim(w ) 3. Either prove that L is not surjective or give an example of such an surjective transformation for V and W of indicated dimensions. Answer: Suppose L is surjective, so that dim(im(l)) = dim(w ) 3. So, let dim(im(l)) = 3 + ɛ with ɛ. We will show a contradiction. Our equation becomes 2 = dim(ker(t )) ɛ 3. This is a contradiction. (c) Let L : V W be an injective linear transformation. Suppose V, W are vector spaces such that dim(v ) = 3 = dim(w ). Prove that L is also surjective. Answer: Our equation gives 3 = +dim(im(l)) and thus Im(L) is a subspace of W of the same dimension 3. Hence the two are equal, i.e. the map is surjective. 7

9 8. All questions on this page are about vectors in P 3. (a) Prove that f (x) = + x + x 2 and f 2 (x) = + x + x 3 are linearly independent. Answer: If c f (x) + c 2 f 2 (x) = then the coefficients of x 2, x 3 give equations: c + =, + c 2 =. This shows independence. We could also take their coordinate vectors in the standard basis ( x x 2 x 3 ) and prove them independent. (b) Write down a concrete polynomial f 3 (x) such that f (x), f 2 (x), f 3 (x) are linearly independent. Prove your claim. Answer: f 3 (x) = works. We could work with equations as above or use the matrix of coordinate vectors, namely:. The matrix has rank 3 as evident from the top 3 3 subdeterminant. So the columns, and hence the corresponding polynomials are independent. (c) Prove or disprove the statement Span{f (x), f 2 (x), f 3 (x)} = P 3. Answer: As noted, P 3 has dimension 4, so no set of 3 vectors can be a basis! 8