Summer Session Practice Final Exam

Transcription

1 Math 2F Summer Session 25 Practice Final Exam Time Limit: Hours Name (Print): Teaching Assistant This exam contains pages (including this cover page) and 9 problems. Check to see if any pages are missing. Enter all requested information on the top of this page, and put your initials on the top of every page, in case the pages become separated. You may not use your books, notes, or any calculator on this exam. You are required to show your work on each problem on this exam. The following rules apply: Work on this practice final ONLY AFTER you have reviewed the homework and lecture problems. There could be mistakes and the problems could be more or less difficult than the actual final exam questions. Organize your work, in a reasonably neat and coherent way, in the space provided. Work scattered all over the page without a clear ordering will receive very little credit. Mysterious or unsupported answers will not receive full credit. A correct answer, unsupported by calculations, explanation, or algebraic work will receive no credit; an incorrect answer supported by substantially correct calculations and explanations might still receive partial credit. If you need more space, use the back of the pages; clearly indicate when you have done this. Do not write in the table to the right. Problem Points Score Total: 9

2 Math 2F Practice Final Exam - Page 2 of. ( points) Find a least-squares solution of Ax = b where: 2 A = and b = 2 5 Solution: Step : Find a basis for Col A. Place basis vectors into a matrix W. Notice that A has two pivot columns, so its first two columns form a basis for Col A. 2 rref(a) =, W = 2 Step 2: Compute b = W (W T W ) W T b. W T W = [ ], (W T W ) = 2 72 [ ] (W T W ) W T b =.5 b = W (W T W ) W T b = [ ] [ ] 2, W T 9 b =, 2 Step : Solve Aˆx = b for ˆx, which is your least squares solution. [ ] Form [A b ] and solve to get ˆx =..5 In this example, Ax = b has no solution (try it!). as-close as you can get and still be inside Col A. Aˆx = b = b = 5 Instead, Aˆx doesn t equal b, but it is

3 Math 2F Practice Final Exam - Page of 2. ( points) Show that if x is an eigenvector of the matrix product AB and Bx, then Bx is an eigenvector of BA. Solution: Since x is an eigenvector of the matrix product AB, we know that there must be some number λ where ABx = λx and x just by the definition of eigenvalues and eigenvectors. Furthermore, we are told that Bx. This is crucial because eigenvectors are defined to be non-zero vectors, and we are hoping to show Bx is an eigenvector. By associativity, ABx = A(Bx) = λx. Now, left multiply both sides by B, to get BA(Bx) = B(λx) = λ(bx). Clearly, Bx is an eigenvector of BA.

4 Math 2F Practice Final Exam - Page of. This question concerns diagonalizable matrices. [ ] (a) (5 points) Is the matrix A = diagonalizable? If so, explain why and diagonalize it. If not, explain why not. Solution: By Theorem 6. (p. 269), the eigenvalues of this triangular matrix are the entries on the main diagonal so λ =. [ ] [ ]} rref([a I ]) = x 2 = so basis for Nul(A I) : because for any x in Nul (A I), x = r We can only find one linearly independent eigenvector [ ] for A. Any other eigenvector corresponding to λ = will be a multiple of, so by Theorem 5.5 (p. 282), A is not diagonalizable (even though it is invertible). 2 2 (b) (5 points) Is the matrix A = 2 diagonalizable? If so, explain why and diagonalize it. If not, explain why not. Solution: First, find the eigenvalues of A. 2 λ 2 [ ] det(a λi) = det( 2 λ 2 λ 2 ) = ( λ) det( ) = ( λ)(2 λ)( λ) = λ λ Thus when λ = or λ = 2, det(a λi) = so they are the eigenvalues. Second, form A I and A 2I and check the dimensions of the eigenspaces (aka Nul (A I) and Nul (A 2I)). 2 rref([a I ]) = [ ] x = 2x so basis for Nul(A I) : 2 because for any x in Nul (A I), x = s + t rref([a 2I ]) = x = x 2 x = because for any x in Nul (A 2I), x = s 2, so basis for Nul(A 2I) : A is diagonalizable by Theorem 5.5, since we have found linearly independent eigenvectors (verify this!). Also, A is diagonalizable by Theorem 5.7 (p.285) since it satisfies

5 Math 2F Practice Final Exam - Page 5 of conditions 7a., 7b., and 7c. (check!). Thus, we can write A = P DP where: 2 2 P = and D =

6 Math 2F Practice Final Exam - Page 6 of. Consider the matrix A = 2 (a) (5 points) Find the eigenvalues and eigenvectors of A. Solution: A is a triangular matrix, so its eigenvalues are on the main diagonal:,, 2 (Theorem 5. on p. 269). By the way, by Theorem 5.6 (p. 28), A is diagonalizable since it has distinct eigenvalues. To find the eigenvectors, look at the nontrivial solutions of (A I)x =, (A + I)x =, and (A 2I)x = : rref([a I ]) = x 2 = x = so basis for Nul(A I) : so an eigenvector corresponding to the eigenvalue is /2 rref([a + I ]) = 2x = x 2 so basis for Nul(A + I) : 2 x = so an eigenvector corresponding to the eigenvalue - is 2 / rref([a 2I ]) = / x = x so basis for Nul(A 2I) : x 2 = x so an eigenvector corresponding to the eigenvalue 2 is (b) (5 points) Find the solution to the system of differential equations x (t) = Ax(t) where x() = 6 Solution: A general solution is of the form x(t) = c e t + c 2 e t 2 + c e 2t We have x() =, and notice that P = 2. 6 c Then c 2 = P = 6 = c

7 Math 2F Practice Final Exam - Page 7 of Consider the matrix A = 2 and assume that it is row equivalent to the matrix B = (a) (5 points) List rank A and dim Nul A. Solution: Looking at B, the reduced echelon form of A, reveals that A has 2 pivot columns, so rank A = dim Col A = 2. Since A has total columns, by the Rank Theorem (p. 56), dim Nul A = 2. (b) (5 points) Find bases for Col A and Nul A. Find a simple example of a vector that belongs to Col A, as well as an example of a vector that belongs to Nul A. Solution: Again, we saw by the reduced echelon form of A that the first column and second column of A are pivot columns. Thus, we take the first two columns of A as a basis for Col A. Our basis is:, For the Nul space of A, we are looking for a set of linearly independent vectors that will span the subspace of all the solutions of Ax =. x = x 5x rref([a ]) = rref([b ]) x 2 = 5/2x x 5 Thus x = r 5/2 + s, so any vector x in Nul A is a linear combination of 5/2, 5 and, with coefficients/weights r and s. A basis for Nul A is: 5 5/2, Recall that for any matrix A, the zero vector is a solution to Ax =. So the zero vector belongs to the Nul space of A. An example of a vector that belongs to Col A is any column of A since it is a linear combination of the columns of A with a coefficient of in front of that column and zeros as the weights on the other columns.

8 Math 2F Practice Final Exam - Page 8 of 6. (a) (5 points) Consider an n n matrix A with the property that the row sums all equal the same number s. Show that s is an eigenvalue of A. [Hint: find an eigenvector] Solution: What does it mean for the row sums to all equal the same number s. In order to answer any question, it is important to figure out what it is asking. Let us try a small 2 2 case to make sure we understand the problem. [ ] a b Let A = c d If the row sums equal, that means a + b = s and c + d = s (just by summing along the rows). Aha! An equivalent way to write this is: [ a c ] + [ ] b = d [ ] s s So we may represent this as a matrix equation for any square matrix A. s A. =. = s. s It is clear that s is an eigenvalue of A. (b) (5 points) Explain or demonstrate that the eigenspace of a matrix A corresponding to some eigenvalue λ is a subspace. Solution: Remember, λ is an eigenvalue of a square matrix A when Ax = λx and x is a non-zero vector. Alternatively, this means (A λi)x = has non-trivial solutions (eigenvectors corresponding to λ). The eigenspace consists of the zero vector and all the eigenvectors corresponding to λ. But this is equivalent to the null space of A λi which includes the trivial (zero vector) solution of (A λi)x = as well as the non-trivial (non-zero) solutions. As the null space is definitely a subspace, and the eigenspace is essentially the same, then the eigenspace is a subspace too. If you want to argue that the eigenspace is a subspace without showing that it is equivalent to another subspace, Nul (A λi), you need to check the properties of subspaces. First of all, the eigenspace is defined to have the zero vector (even though zero cannot be an eigenvector) so it satisfies the zero vector requirement. Secondly, we need to check that the eigenspace is closed under addition. Suppose that v and v 2 are eigenvectors corresponding to λ. Is (v + v 2 ) an eigenvector corresponding to λ?. Well, by definition Av = λv and Av 2 = λv 2, and then A(v + v 2 ) = Av + Av 2 = λv + λv 2 = λ(v + v 2 ), so (v + v 2 ) is in the eigenspace of λ. Finally, we need to check that eigenspaces are closed under scalar multiplication. Let v be an eigenvector corresponding to λ. Is cv in the eigenspace as well where c is any scalar value? Well, Av = λv so then c(av) = A(cv) = c(λv) = λ(cv), so yes, cv is also in the eigenspace. After all, if c = then cv is the zero vector which we know is already in the eigenspace, and if c, then cv is an eigenvector corresponding to λ as we have shown.

9 Math 2F Practice Final Exam - Page 9 of 7. (a) (5 points) Suppose a 6 8 matrix A has four pivot columns. What is dim Nul A? Is Col A = R? Why or why not? Solution: Since A has 6 rows, any vector that belongs to Col A is a linear combination of the columns of A which have 6 entries each, for example, a vector like b = b b 2 b b b 5 b 6 assuming that Ax = b for some x. A member of the vector space R is a vector with four entries, for example x = 2. So, there is no way that a subspace with vectors of four entries can equal a subspace with vectors of 6 entries. Nevertheless, as A has pivot columns, a basis for Col A has linearly independent vectors that each have 6 entries. Thus, the dimension of Col A=. Also, if A has pivot columns out of 8 total columns, then it must have non-pivot columns. Alternatively, by the Rank Theorem (p. 56), rank A + dim Nul A = + dim Nul A = 8. Thus, the dimension of the Nul space of A must be. (b) (5 points) If A is a 7 5 matrix, what is the smallest possible dimension of Nul A? Solution: Recall the Rank Theorem (p. 56) which says 5=dim Col A + dim Nul A. Minimizing dim Nul A is equivalent to making rank A = dim Col A as large as possible. Convince yourself by writing a diagram that A could have 5 pivot columns so that dim Col A = 5. This means that dim Nul A =. so the smallest possible dimension of Nul A is.

10 Math 2F Practice Final Exam - Page of 8. ( points) Find the dimension of the subspace spanned by the given vectors. Are the vectors linearly independent? 2 5,,, Solution: Using Theorem.8 (p. 59), it is easy to see that there are more vectors () than there are entries in each vector (), so the vectors are linearly dependent. This means that whatever the dimension of the subspace spanned by these vectors is, it must be less than since one of the vectors is essentially redundant. After all, by definition, a basis that spans a subspace is a set of only linearly independent vectors, so we can t include any vector that makes the set linearly dependent. To find the dimension of the subspace spanned by these vectors, place the vectors into some matrix V and row-reduce: 2 5 V = 2, rref(v) = 2 2 In the reduced echelon form of V, we can see that there are pivot columns, so the dimension of the set of all linear combinations of the columns of V (aka rank V = dim col V ) is. Another way to check if the vectors are linearly independent or not is to investigate whether the homogenous equation V x = has a nontrivial solution. rref([v ]) = Clearly, V has a non-pivot column and thus a free variable which means that it has non-trivial solutions and, as expected, has linearly dependent columns.

11 Math 2F Practice Final Exam - Page of 9. ( points) Choose two statements at random from the list of 2 problems and re-work them...