NAME MATH 304 Examination 2 Page 1

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1 NAME MATH 4 Examination 2 Page. [8 points (a) Find the following determinant. However, use only properties of determinants, without calculating directly (that is without expanding along a column or row or otherwise). Explain your answer. 2 det Solution: The determinant is, which can be verified in advance using your calculator. To show that it s without expanding along columns or rows, you can do two things. First, if you row reduce the matrix into row echelon form, you obtain the matrix 2 / 2/ 8/25. This matrix has determinant by multiplying along the diagonal (since it s upper triangular). Two matrices that are row equivalent do not necessarily have the same determinant, but the determinant of one is a non-zero multiple of the determinant of the other. Thus, our original determinant must also be. Or, you can solve the problem this way. Use your calculator to find that the determinant is in advance. Once you know that the determinant should be, that means we should look for linear dependencies among the columns and among the rows. So then we look for one, and we find that Row 2 2 (Row ) = Row 4. Having a linear dependency among the rows then implies that the matrix has determinant, which is what we wanted to show. (b) Let v = 5 2, v 2 = 4 9 2, v = 2 9 9, v 4 =. Are v, v 2, v, v 4 linearly independent or linearly dependent? Explain. Solution: They must be linearly dependent. The matrix has determinant (so is singular). The rows and columns of such a matrix must be linearly dependent.

2 NAME MATH 4 Examination 2 Page 2 (c) Using v, v 2, v, v 4 from (b), is it the case that Span(v, v 2, v, v 4 ) = R 4? Solution: No, they cannot span all of R 4. Any spanning set of R 4 must contain at least 4 linearly independent vectors. Our set contains only 4 vectors, which are not linearly independent. 2. [4 points Consider the following vectors in R : v =, v 2 =, v =, v 4 =, v 5 = (a) Are v, v 2,..., v 5 linearly independent or linearly dependent? How do you know? If they are linearly dependent, find a non-trivial linear dependency among them. Solution: They must be linearly dependent. The dimension of R is, so any set of 4 or more vectors must be linearly dependent. There are of course several dependencies to choose from, but here is one: v + v + v 4 + v 5 =. (b) Find a subset of {v, v 2,..., v 5 } that is a basis of R. Justify your answer. You need to find of the vectors that are linearly independent. It turns out that any of v,..., v 5 will do. We will pick v, v 2, v. They will form a basis if we can show that they are linearly independent and that they span R. They are linearly independent because the matrix has determinant 56, which is. Any three linearly independent vectors in R must also span R, so v, v 2, v must also span R.. [8 points Consider the following subset of 2 2 matrices, {[ } a b V = R 2 2 a + d =. c d (a) Is V a vector space? Explain.

3 NAME MATH 4 Examination 2 Page Solution: Yes. It is closed under addition and scalar multiplication, and so it is a subspace of R 2 2. (b) What is the dimension of V? Solution: The dimension is. 4. [4 points Let A be the matrix 2 A = 2. 5 Find a basis for the nullspace N(A). Justify your answer. Solution: To find a basis for N(A), we first put A in reduced row echelon form: The homogenous systems of linear equations defined by both matrices have the same set of solutions, so 7α + 2β N(A) = 2α β α α, β R. β Therefore, This gives us our basis: 7 N(A) = α 2 [ 7 [ 2 2, + β 2 α, β R.. These two vectors span N(A) by the equation above. They are linearly independent because neither is a non-zero multiple of the other. 5. [2 points Let A = are bases for R. {[ (a) Why is A a basis for R? [ [ } 2,, and B = {[ 2 [ [ },,. Both A and B Solution: The vectors in A are linearly independent because the matrix formed from them has determinant (by multiplying down the diagonal). Any linearly independent vectors in R must form a basis since R is -dimensional.

4 NAME MATH 4 Examination 2 Page 4 (b) What are the change of basis matrices U A and U B for A and B? Solution: The change of basis matrices are 2 2 U A = U B =. Some people put the inverses of these matrices instead, which was fine for full credit on this part. Of course it was still important to use the right matrices in parts (c) (d). (c) Let v = [ 62. Express v in terms of the basis A. That is, find [v A. Solution: [v A = U A = 6 2 = 8 A. (d) Let u R have representation [u B = terms of the standard basis for R? [ 2 B. What is the representation of u in Solution: u = U B 2 B 2 4 = 2 = 2. 2 (e) Using u from part (d), what is [u A? Solution: [u A = U A = [4 points Determine whether each of the following statements is either True or False. (a) Suppose S is a subset of R n so that Span(S) = R n. Then every vector v R n can be expressed as a linear combination of elements of S in only one way. Solution: False. S must be a basis for this to be true. (b) For vectors v, v 2,..., v k R n, the dimension of Span(v, v 2,..., v k ) is at least k.

5 NAME MATH 4 Examination 2 Page 5 Solution: False. The dimension is at most k. (c) Suppose v and v 2 are linearly independent vectors in R. Then there is a vector v R so that {v, v 2, v } is a basis for R. Solution: True. Any linearly independent set in a vector space can be enlarged to form a basis. (d) Suppose S is a subset of a vector space V. If Span(S) V, then there must be a vector v V such that v / Span(S). Solution: True. If S is not a spanning set, then that means that there is definitely something in V that is not in the span of S. (e) If v, v 2, v, and v 4 R n are linearly dependent, then Span(v, v 2, v, v 4 ) = Span(v, v 2, v ). Solution: False. It would only be true if the non-trivial dependency among v,..., v 4 involves v with a non-zero coefficient. (f) Let v, v 2, v, and v 4 be vectors in R n. Suppose {v, v 2, v } are linearly indpendent and {v 2, v, v 4 } are linearly independent. Then {v, v 2, v, v 4 } are linearly independent. Solution: False. It is easy to construct counterexamples in any vector space of dimension at least. For example, if n happens to be, then the four vectors must be linearly dependent. (g) Suppose W is a subspace of a vector space V. If dim W = dim V, then W = V. Solution: True. If both spaces have the same dimension, then there is a basis of W with the same number of elements as dim W = dim V. But any linearly independent subset of V containing the same number of elements as the dimension of V must also be a basis for V. So any basis for W is also a basis for V i.e. W and V must be the same (they re both the spans of the same set). 7. [2 points Suppose {u, v, w} is a basis for a vector space V. Show that is also a basis for V. {u + v + w, v + w, w}

6 NAME MATH 4 Examination 2 Page 6 Solution: There are several ways to do this problem. Here is one way. First we show that u + v + w, v + w, and w are linearly independent. Suppose with a, b, c R. Then gathering terms: a(u + v + w) + b(v + w) + cw =, au + (a + b)v + (a + b + c)w =. Since {u, v, w} is a basis for V, we know that u, v, w are linearly independent. Therefore the last equation implies that a = a + b = a + b + c =. We quickly find from this that the only solution is a = b = c =. Thus, u + v + w, v + w, and w are linearly independent, which is what we wanted to show. To conclude that they form a basis, we need only observe that we have a set of linearly independent vectors in a vector space of dimension. It must also span, so it is a basis.