Math 2040: Matrix Theory and Linear Algebra II Solutions to Assignment 2

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1 ection 4.5 Dimension Math 2040: Matrix Theory and Linear Algebra II olutions to Assignment Problem Restatement: Find (a) a basis, and (b) state the dimension of the subspace a + 6b c { 6a 2b 2c 9a + 5b + c ; a, b, c R}. a + b + c 6 Final Answer: (a) { 6 9, 2 5 } is a basis of the R subspace. (b) The dimension of the subspace is 2 since a basis for it has two vectors in it. Work: (a) { a + 6b c 6a 2b 2c 9a + 5b + c a + b + c ; a, b, c R} = pan{ 6 9, 6 2 5, 2 }. The first and third vectors are multiples of each other, and the first and second vectors are linearly independent of each other so we can remove the third to get a basis (that is, a linearly independent set of spanning vectors) of the subspace. (b) No work required Problem Restatement: Determine the dimension of W = pan{ 2, 4, 8 6, 0 } Final Answer: dim(w ) =. Work: dim(w ) = dim(ol ince there are pivot columns, dim(w ) =. ) = dim(ol Problem Restatement: Determine the dimensions of N ul(a) and ol(a) for ).

2 A = Final Answer: dim(ol(a)) = Rank(A) = and dim(nul(a)) = 6 Rank(A) = 6 =. Work: A is in echelon form and we count pivot columns in A Problem Restatement: Let B be the basis of P 2 consisting of the first three Laguerre polynomials, {, t, 2 4t + t 2 }, and let p(t) = 7 8t + t 2. Find the coordinate vector of p relative to B. Final Answer: [p(t)] B = 5 4. Work: everal equivalent methods to solve this problem are possible. One of thesefollows. Let = {, t, t 2 } be the standard basis of P 2. [p(t)] B = P B (p(t)) = PB (P 7 8 ) = (P B P ) 7 8 = (P P B) 7 8. Therefore, if b = 7 8, then [p(t)] B is the solu- tion of (P P B)x = b. Now P P B is easy to construct; it is the matrix Passing to the augmented matrix and row reducing to reduced echelon form gives us [P P B b] = Therefore, [p(t)] B = Problem Restatement: Let H be an n-dimensional subspace of an n-dimensional vector space V. how that H = V. Final Answer: hoose a basis B = {b,..., b n } of H, so that H = pan(b). By the Basis Theorem, page 257, B is also a basis of V, so V = pan(b). Therefore, H = V, since they are each equal pan(b).. 2

3 ection 4.6 Rank Problem Restatement: Assume that A is row equivalent to B where A = and B = Without calculation, list Rank(A) and dim(nul(a)). Then find a basis for ol(a), Row(A) and Nul(A). Final Answer: Rank(A) = and dim(nul(a)) = 2. A basis of ol(a) is given by the first, third, and fifth columns of A. A basis of Row(A) is given by the first three rows of A (or of B). 5 A basis of Nul(A) is { 0 0, 0 /2 } (see work below). 0 0 Work: A B / Ax = 0, we have x 2 and x 4 free, giving us x = x 2 5x 4, x 2 = x + 0x 4, x = 0x 2 + (/2)x 4, x 4 = 0x 2 + x 4, x 5 = 0x 2 + 0x 4., so if x is a solution of the homogeneous system Problem Restatement: If a 6 matrix A has rank, find dim(nul(a)), dim(row(a)), and Rank(A T ). Final Answer: A : R R 6, A having 6 rows and columns. dim(nul(a)) = Rank(A) = = 0, by the Rank Theorem. dim(row(a)) = Rank(A), so dim(row(a)) =, and Rank(A T ) = Rank(A) so Rank(A T ) =.

4 Problem Restatement: If the null space of a 7 6 matrix A is 5-dimensional, what is the dimension of the column space of A? Final Answer: The dimension of the column space of A is Rank(A) by definition. Then, by the Rank Theorem, we have dim(ol(a)) = 6 dim(nul(a)) = 6 5 = Problem Restatement: If A is a 6 4 matrix, what is the smallest possible dimension of Nul(A)? Final Answer: Zero is the smallest possible dimension of the null space of a 6 4 matrix A. Work: A : R 4 R 6, A having 6 rows and 4 columns. 0 dim(nul(a)), but if {e, e 2, e, e 4, e 5, e 6 } is the standard basis of R 6 and A = [e e 2 e e 4 ], say, then A is one to one and, therefore, Nul(A) = {0}, so dim(nul(a)) = 0. Therefore, if A is a 6 4 matrix, the smallest possible dimension of Nul(A) is Problem Restatement: uppose A is m n and b is in R m. What has to be true about the two numbers Rank([A b]) and Rank(A) in order for the equation Ax = b to be consistent? Final Answer: It must be that Rank([A b]) = Rank(A). Work: In order for Ax = b to be consistent, the last column of the augmented matrix [A b] cannot be a pivot column. Therefore, in order for Ax = b to be consistent, Rank([A b]) must equal the number of pivot columns of A. ection 4.7 hange of Basis Problem Restatement: Let B = {b, b 2 } and = {c, c 2 } be bases for a vector space V, and suppose b = c + 4c 2 and b 2 = 5c c 2. a. Find the change-of-coordinate matrix from B to. b. Find [x] for x = 5b + b 2. Final Answer: a. P B = [ 5 4 ] [ 0. b. [x] = Work: a. P B = P P B = [P P [ Be P P Be ] 2 ] = [P b P b 2] = [[b ] [b 2 ] ] 5 = [[ c + 4c 2 ] [5c c 2 ] c ] = b. [x] = [5b + b 2 ] = 5[b ] + [b 2 ] = P B = 5 + = 4 ]. [ 0 ]. 4

5 Problem Restatement: Let D = {d, d 2, d } and F = {f, f 2, f } be bases for a vector space V, and suppose f = 2d d 2 + d, f 2 = d 2 + d, and f = d + 2d. a. Find the change-of-coordinate matrix from F to D. b. Find [x] D for x = f 2f 2 + 2f. Final Answer: a. P D F = b. [x] D = 4 7 Work: ee above for methodology, replacing B by F and by D throughout and adjusting for dimension Problem Restatement: Let B = {, } and = {, } be bases for R 2. Find the change-of-coordinated matrix from B to and the change-of-coordinated matrix from to B. 2 2 Final Answer: P B = and P 4 B =. 4 Work: P B = P P B = = =. P 4 B = P B =. Other calculations will lead to the same 4 answers Problem Restatement: In P 2, find the change-of-coordinates matrix from the basis B = { t 2, 2 + t 5t 2, + 2t} to the standard basis. Then write t 2 as a linear combination of the polynomials in B. Final Answer: P B = t 2 = ( t 2 ) 2(2 + t 5t 2 ) + ( + 2t). 5 0 Work: The standard basis is = {, t, t 2 }. P B = P P B = [P P Be P P Be 2 P P Be ] = [P ( t2 ) P (2+t 5t2 ) P (+2t)] = [t 2 ] B = P B = (P P B) (t2 ) = P P P (t2 ) = P 0 0 B = P B e. B P

6 Therefore, [t 2 ] B is the solution of the equation P B x = e. Passing to the augmented matrix and row reducing, gives us [P B e ] = x. Problem Restatement: alculate the change of coordinate matrix P B where B = { 2t + t 2, 5t + 4t 2, 2t + t 2 } and = { t 2, 2 + t 5t 2, + 2t}. You may assume B and are each basis of P 2. Final Answer: P B = Work: P B = P P B = P P P P B = (P P ) (P P B) = P P B. Therefore, P B is the solution of the equation P X = P B. The matrices P and P B are easy to formulate (see above). They are P B = and 4 P = Passing to the augmented matrix and row reducing allows us to 5 0 solve for X = P B. [P P B ] = = [I P B ]

7 ection 5. Eigenvectors and Eigenvalues Problem Restatement: Is λ = an eigenvalue of A = corresponding eigenvector ? If so, find one Final Answer: λ = is an eigenvalue of A. A corresponding eigenvector is Work: A λi = Problem Restatement: Let A = A corresponding to the eigenvalue λ = 4. [ /2 Final Answer: { } or, equivalently, { /2 Work: A λi = Problem Restatement: Let A = Final Answer: { Find a basis of the eigenspace of 4 2 A corresponding to the eigenvalue λ = , 0 0 }. 0 4 Work: A λi = ] } Find a basis of the eigenspace of Problem Restatement: onstruct an example of a 2 2 matrix with only one distinct eigenvalue.. 7

8 Final Answer: By Theorem of section 5., we can take any 2 2 triangular matrix with two equal diagonal elements. One example, therefore, is the 2 2 zero matrix. Another example is the 2 2 identity matrix I 2. Many other examples are possible Problem Restatement: how that if A 2 is the zero matrix, then the only eigenvalue of A is 0. Final Answer: uppose A 2 = 0. We must show 0 is an eigenvalue of A, and we must show that if λ is an eigenvalue of A then λ = 0. First, we show 0 is an eigenvalue of A. To do this, in R n, choose any vector v 0 (tacitly, we are assuming A is an n n matrix with n > 0). Let Av = b. There are two cases to consider; b = 0 and b 0. In the first case, if b = 0 then b = 0v, and this gives us Av = 0v with v 0, allowing us to conclude 0 is an eigenvalue of A. In the other case, we assume b 0. Then because A 2 = 0, we get 0 = Ab upon multiplying on the left both sides of Av = b by A. But 0b = 0, so we get 0b = Ab with b 0, allowing us to conclude again that 0 is an eigenvalue of A. Therefore, in both cases (b = 0 and b 0), 0 is shown to be an eigenvalue of A. This concludes the first part of the proof. Next, we show that if λ is an eigenvalue of A then λ = 0. To do this, assume λ is an eigenvalue of A. Therefore, there is a vector v 0 such that Av = λv. Then A 2 v = A(λv) = λav = λλv = λ 2 v. But then 0 = λ 2 v because A 2 v = 0v = 0. Thus, 0 = λ 2 because v 0. Finally, because λ is a scalar, 0 = λ 2 implies 0 = λ as required, and this completes the second part of the proof. 8