Linear Algebra Second Exam Review Questions 2016

MATH 240, Second Exam REVIEW QUESTIONS 1. The inverse is computed by augmenting A with I, then performing row reduction until A becomes I (and I becomes A−1 ).   −1 0 −1 3  A−1 =  2 1 0 1 2 2. Suppose A, B and X are n × n matrices, with A, X, and A − AX invertible, and suppose (A − AX)−1 = X −1 B First, explain why B is invertible, then solve the equation for X. If you need to invert a matrix, explain why it is invertible. SOLUTION: B is invertible because it is the product of invertible matrices- That is, B = X(A − AX)−1 We want to isolate X, so we should factor the right side: B = X(I − X)−1 A−1 BA = X(I − X)−1 ⇒ Now, (I − X) is invertible because A is invertible and A − AX = A(I − X) is invertible. Therefore, BA(I − X) = X ⇒ BA − BAX = X ⇒ BA = X + BAX = (I + BA)X Now, is I + BA invertible? Yes, because from the last equation, BA = (I + BA)X and BA and X are invertible. Finally, invert: (I + BA)−1 BA = X  3. Let A =  1 2 . Find A−1 using the formula, then solve Ax = [3, 5]T . 5 12 SOLUTION: First, we see that −1 A 1 = 12 − 10  and −1 x=A b= 1 12 −2 −5 1  13 −5   4. Show that, if AB is invertible, then so is A (assume A, B are n × n). Hint: If AB is invertible, then there is a matrix W so that ABW = I. SOLUTION: Using the hint, ABW = I, or more precisely, there is a matrix D = BW so that AD = I. Then by the Invertible Matrix Theorem (p 129, part (k)), A is invertible. 5. Let S be the parallelogram whose vertices are (−1, 1), (0, 4), (1, 2) and (2, 5). Use determinants to find the area of S. SOLUTION: If you plot the points, you should see which vectors we can take for the determinant:             0 −1 1 1 −1 2 u= − = v= − = 4 1 3 2 1 1 Using these as the columns of the matrix A, the area is: |1 − 6| = 5.       a b c a + 2g b + 2h c + 2i g h i 6. Let A =  d e f , B =  d + 3g e + 3h f + 3i , and C =  2d 2e 2f . g h i g h i a b c If det(A) = 5, find det(B), det(C), det(BC). SOLUTION: To get B from A, we take 2r3 + r1 → r1 and 3r3 + r2 → r2 . These row operations do not change the determinant, so det(A) = det(B) = 5. To get the determinant of C, multiply the determinant of A by −2 (negative because of the row swap); we get −10. The determinant of BC is then 5 · −10 = −50 7. Assume that A and B are row equivalent, where:     1 2 −2 0 7 1 0 4 0 −3  −2 −3  0 1 −3 0 1 −1 −5  5  ,   A= B =  −3 −4  0 0 0 −2 −3  0 1 −4  3 6 −6 5 1 0 0 0 0 0 (a) State which vector space contains each of the four subspaces, and state the dimension of each of the four subspaces: SOLUTION: Here are the spaces and dimensions: i. The row space of A is a subspace of IR5 and has dimension 3 (this is the number of pivot rows). ii. The column space of A is a subspace of IR4 and has dimension 3 (this is also known as the rank of A and is the number of pivot columns). 2 iii. The null space of A is a subspace of IR5 and has dimension 2 (the number of free variables). iv. The null space of AT is a subspace of IR4 and has dimension 1 (because the dimension of the column space is 3, and they should add to 4). (b) Find a basis for Col(A): We see from the RREF that the pivot columns are cols 1, 2, and 4. Be sure to use the columns from the original matrix A! The basis is the set       1 2 0        −3   −1  −2  , ,   −3   −4   −2       3 6 5 (c) Find a basis for Row(A): We can use the reduced rows (written as columns) for the basis:       1 0 0               0 1 0        4  ,  −3  ,  0            0   0   1        −3 5 −4 (d) Find a basis for Null(A): Solve Ax = 0, and the vectors are the basis vectors:     x1 = −4x3 + 3x5 −4 3        −5   x2 = 3x3 − 5x5  3        x3 = x3 ⇒   1  ,  0    0   4   x4 = 4x5      x5 = x5 0 1 8. Determine if the following sets are subspaces of V . Justify your answers.      a V = IR3 • H =  b  , a ≥ 0, b ≥ 0, c ≥ 0 ,   c SOLUTION: This set is not closed under scalar multiplication (for example, if c = −1).    a + 3b        a − b   • H=  , a, b in IR , V = IR4  2a + b       4a SOLUTION: This is the span of [1, 1, 2, 4]T and [3, −1, 1, 0]T , so H is a subspace. • H = {f : f 0 (x) = f (x)} , V = C 1 [IR] (C 1 is the space of differentiable functions where the derivative is continuous). SOLUTION: We could show this directly: 3 – The zero function is in H since the derivative of 0 is 0. – If u, v are in H, then u0 = u and v 0 = v. Therefore, (u + v)0 = u0 + v 0 = u + v, so u + v is in H. – If u ∈ H, then u0 = u. Therefore, (cu)0 = cu0 = cu, so cu is in H. • H is the set of vectors in IR3 whose first entry is the sum of the second and third entries, V = IR3 . SOLUTION: Rewriting H algebraically, it is the set of vectors in IR3 so that       a+b 1 1  a  = a 1  + b 0  b 0 1 Since H is the span of two vectors, it is a subspace. 9. Prove that, if T : V 7→ W is a linear transformation between vector spaces V and W , then the range of T , which we denote as T (V ), is a subspace of W . SOLUTION: Show that the three parts to the definition hold true• Since 0 ∈ V (V is a subspace) and T (0) = 0, then 0 ∈ T (V ). • Let w1 and w2 be in T (V ). Show that the sum is in T (V ). Since w1 , w2 ∈ T (V ), there are vectors, v1 , v2 in V such that T (v1 ) = w1 and T (v2 ) = w2 . Now by linearity, we have: w1 + w2 = T (v1 ) + T (v2 ) = T (v1 + v2 ) Therefore, w1 +w2 is the image of v1 +v2 which is in V (because V is a subspace). Therefore, w1 + w2 is in T (V ). • Let w ∈ T (V ), and show cw ∈ T (V ). Since w ∈ T (V ), there exists v ∈ V such that T (v) = w. We also know that cv ∈ V because V is a subspace. By the linearity of T , cw = cT (v) = T (cv) and so cw ∈ T (V ). 10. Let H, K be subspaces of vector space V . Define H + K as the set below: H + K = {w | w = u + v, for some u ∈ H, v ∈ K} SOLUTION: • Since 0 ∈ H and K, and 0 + 0 = 0, then 0 ∈ H + K 4 • Let w1 , w2 ∈ H + K. Show that the sum is in H + K: Since w1 , w2 ∈ H + K, there are vectors, u1 , u2 in H and vectors, v1 , v2 in K such that w1 = u1 + v1 ⇒ w1 + w2 = (u1 + u2 ) + (v1 + v2 ) w2 = u2 + v2 with u1 + u2 ∈ H and v1 + v2 ∈ K because H, K are subspaces. • Let w ∈ H + K. Show that cw ∈ H + K: If w ∈ H + K, then there are vectors u ∈ H and v ∈ K so that w = u + v. Therefore, cw = cu + cv and since H and K are subspaces, cu ∈ H and cv ∈ K. Therefore, cw is a sum of something in H and something in K, so it is in H + K. 11. Let A be an n × n matrix. Write statements from the Invertible Matrix Theorem that are each equivalent to the statement “A is invertible”. Use the following concepts, one in each statement: (a) Null(A) (b) Basis (c) Rank SOLUTION: (a) Null(A) = {0} (b) The columns of A form a basis for IRn (c) The rank of A is n. 12. Is it possible that all solutions of a homogeneous system of ten linear equations in twelve variables are multiples of one fixed nonzero solution? Discuss. SOLUTION: No. If we have 10 equations in 12 variables, we must have at least two free variables, and so the null space of the corresponding matrix is the span of at least two linearly independent (nonzero) vectors. 13. Show that {1, 2t, −2 + 4t2 } is a basis for P2 . SOLUTION: Using the isomorphism theorem to connect P2 to IR3 , we look to see if the images of the candidate vectors are linearly independent. If they are, since P2 is three dimensional and we have three linearly independent vectors, then they must form a basis for P2 . To convert the vectors into their corresponding coordinates (using the standard basis for P2 ):       1 0 −2 1 7→  0  2t → 7  2  − 2 + 4t2 7→  0  0 0 4 These are clearly linearly independent (upper triangular). If it wasn’t clear, form a matrix using these vectors as the columns, and do row reduction. If they are linearly independent, there should be a pivot in every column. 5 14. Let T : V → W be a 1 − 1 and linear transformation on vector space V to vector space W . Show that if {T (v1 ), T (v2 ), T (v3 )} are linearly dependent vectors in W , then {v1 , v2 , v3 } are linearly dependent vectors in V . SOLUTION: If {T (v1 ), T (v2 ), T (v3 )} are linearly dependent, then there is a nontrivial solution to c1 T (v1 ) + c2 T (v2 ) + c3 T (v3 ) = 0 Since T is linear, we can write this equation as: T (c1 v1 + c2 v2 + c3 v3 ) = T (0) If T is 1 − 1, we can say that T (a) = T (b) implies that a = b. In this case, we then can say that c1 v 1 + c2 v 2 + c3 v 3 = 0 So that we again get a non-trivial solution, and these vectors are linearly dependent. 15. Use Cramer’s Rule to solve the system: 2x1 + x2 = 7 −3x1 + x3 = −8 x2 + 2x3 = −3 SOLUTION: We’ll need some determinants: 2 1 0 −3 0 1 0 1 2 =4 7 1 0 −8 0 1 −3 1 2 2 7 0 −3 −8 1 0 −3 2 =6 = 16 2 1 7 −3 0 −8 0 1 −3 = −14 So the solution is: x1 = 6 3 = 4 2 x2 = 16 =4 4 x3 = −14 −7 = 4 2   −6 12 16. Let A = , and w = [2, 1]T . Is w in the column space of A? Is it in the null −3 6 space of A? SOLUTION: The first question is asking if w is in the span of the columns of A. By inspection, the weights are −1/3 and 0, so yes. The second question is asking if Aw = 0. In this case, it is true as well. Side Note: It was only a coincidence that this vector was in both the column space as well as the null space. If A was m × n, then the column space would be in IRm and the null space would be in IRn , so these would definitely be different. 6 17. Prove that the column space is a vector space using a very short proof, then prove it directly by showing the three conditions hold. SOLUTION: For the short proof, we know that any spanning set is automatically a subspace. The column space is the span of the columns of A. Now for the longer proof using the definition: Recall that a vector b is in the column space of A if and only if b = Ax for some x. (a) Since A0 = 0, 0 ∈ Col(A). (b) Let b1 , b2 ∈ Col(A). Then there exists x1 , x2 so that A(x1 ) = b1 A(x2 ) = b2 therefore b1 + b2 = A(x1 ) + A(x2 ) = A(x1 + x2 ) This shows that b1 + b2 is in the column space of A. (c) Let b ∈ Col(A). Then Ax = b for some x in the domain, and cb = cAx = A(cx) Therefore, cb is in the column space of A for all constants c. 18. If A, B are 4 × 4 matrices with det(A) = 2 and det(B) = −3, what is the determinant of the following (if you can compute it): SOLUTIONS: (a) det(AB) = −6, (b) det(A−1 ) = 1/2, (c) det(5B) = 53 (−3), (d) det(3A − 2B) is unknown with what is given, (e) det(B T ) = −3 19. True or False, and give a short reason: (a) If det(A) = 2 and det(B) = 3, then det(A + B) = 5. SOLUTION: False. det(A + B) 6= det(A) + det(B). (b) Let A be n × n. Then det(AT A) ≥ 0. SOLUTION: True: det(AT A) = det(AT )det(A) = (det(A))2 ≥ 0. (c) If A3 is the zero matrix, then det(A) = 0. SOLUTION: True- If A3 = 0, then 0 = det(A3 ) = (det(A))3 , so det(A) = 0. (d) IR2 is a two dimensional subspace of IR3 . SOLUTION: False. IR2 is not a subspace of IR3 , they are completely different vector spaces. However, a two dimensional subspace in IR3 is isomorphic to IR2 . (e) Row operations preserve the linear dependence relations among the rows of A. SOLUTION: False. Swapping rows will especially change the relations- However, row operations do NOT change the linear dependence relations among the columns of A. 7 (f) The sum of the dimensions of the row space and the null space of A equals the number of rows of A. SOLUTION: False. The sum is the number of columns of A. 20. Let the matrix A and its RREF, RA , be given as below:    1 1 7 2 2 1  3 0   0 9 3 4   A=  −3 1 −5 −2 3  RA =  0 2 2 14 4 2 0 1 1 0 0  0 0   1  0 0 1 0 0 3 4 0 0 0 1 0 0  0 −4 0 7   1 −5  0 0 so that the columns of A are a1 , · · · , a5 . Similarly, define Z and its RREF, RZ , as:   4 5 3 4  5 6 5 −3   Z=  10 −3 9 −106  4 10 2 44  1  0 Rz =   0 0 Label the columns of Z as z1 , · · · , z5 . (a) Find the rank of A and a basis for the column space of A (use the notation a1 , etc.). Similarly, do the same for Z: SOLUTION: The rank is the dimension of the column space (the number of pivot columns in the RREF), so in this case it is 3. The corresponding columns of A are a1 , a2 , a5 and these form a basis for Col(A). Similarly, the rank of Z is 3 and a basis for Col(Z) is z1 , z2 , z3 . (b) You’ll notice that the rank of A is the rank of Z. some columns of A and Z:    5 3 1 0 1 1 2 4    3 0 4 5 6 5   0 1   −3 1 3 10 −3 9  →  0 0 2 2 2 4 10 2 0 0 Here is a row reduction using  0 −1 2 −1 0 1 3 0   1 2 0 2  0 0 0 0 Are the subspaces spanned by the columns of A and Z equal? SOLUTION: The RREF tells us that z1 , z2 and z3 can be written in terms of a1 , a2 and a5 . The subspaces are both three dimensional, so if we have any set of 3 linearly independent vectors (in the subspace) we must have a basis (therefore, the basis for Col(Z) could be used for Col(A) and vice-versa). (c) Let B and be the set of basis vectors used for the column spaces of A found in (a). Find the change of coordinates matrix PB that changes the coordinates from 8 B to the standard basis, then find the coordinates of z1 with respect to B (Hint: The second part does not rely on the first). SOLUTION: The matrix is formed from the columns of A: PB = [a1 a2 a5 ] The coordinates of z1 are shown in the row reduction in part (b): z1 = −a1 + a2 + 2a5 so the coordinate vector is [−1, 1, 2]T . (d) Find the coordinates of z4 using the basis vectors in z1 , z2 , z3 . SOLUTION: From the initial setup, we see that   −4 [z4 ]C =  7  −5 21. Short Answer: (a) Define the kernel of a transformation T : SOLUTION: The kernel is the set of all x such that T (x) = 0. (b) Define the dimension of a vector space: SOLUTION: The dimension is the number of basis vectors for the vector space, with the special cases of dimension 0 for the vector space containing only the zero vector, and if the vector space cannot be spanned by a finite set, it is infinite dimensional. (c) We said that Pn is isomorphic to IRn+1 . What is the isomorphism? SOLUTION: The isomorphism is the coordinate mapping. That is: p(t) = a0 + a1 t + · · · + an tn ∈ Pn 7→ [a0 , a1 , · · · , an ]T ∈ IRn+1 (d) If C is 4 × 5, what is the largest possible rank of C? SOLUTION: The largest possible rank is 4. (If C has its largest possible rank, it is said to be full rank). What is the smallest possible dimension of the null space of C? SOLUTION: The dimension of the null space is the number of free variables in the RREF of the matrix (or the number of non-pivot columns). If the rank is 4, the dimension of the null space must be at 1. (Note: If the rank were smaller than 4, the extra dimensions are added to the null space). 9 (e) If A is a 4 × 7 matrix with rank 3, find the dimensions of the four fundamental subspaces of A. SOLUTION: If A has rank 3, then the dimension of the row space is 3, the dimension of the null space is 4, the dimension of the column space is 3, and the dimension of Null(AT ) is 1. (f) Show that the coordinate mapping (from n−dimensional vector space V to IRn ) is onto. SOLUTION: Let v1 , . . . , vn be a basis for V . Then for any vector c in IRn , the associated vector in V is given by: c1 v 1 + · · · + cn v n which is in V because V is a vector space. Therefore, the coordinate mapping is onto. 22. Let A be m × n and let B be n × p. Show that the rank(AB) ≤ rank(A). (Hint: Explain why every vector in the column space of AB is in the column space of A). SOLUTION: Using the hint, if y ∈ Col(AB), then y = ABx for some x. Therefore, y = A(Bx) = Aw, so y must be in the column space of A. Therefore, the column space of AB is a subspace of the column space of A, so dim(Col(AB)) ≤ dim(Col(A)) which gives the result, since the rank is the dimension of the column space. 23. Let T : IRn → IRm be a linear transformation. (a) If T is one-to-one, what is the dimension of the range of T ? SOLUTION: If T is 1 − 1, then it is invertible on its range. Therefore, it will map the basis vectors of IRn to linearly independent vectors in IRm . These will form the basis for the range, so the range is n dimensional. Alternative: Since T : IRn → IRm , there is an m × n matrix A so that Ax = T (x). If T is 1 − 1, the solution to Ax = 0 is only the trivial solution, so the dimension of the null space is 0, and therefore, the dimension of the column space is n (and that is the range of T ). (b) What is the dimension of the kernel of T if T maps IRn onto IRm ? Explain. SOLUTION: Let A be the m × n matrix for the linear map. If the column space of A has dimension m, then so does the row space, and so the dimension of the null space would be n − m, which is valid only if n ≥ m (wide or square matrix). 24. Find the determinant of the matrix A below:  4 8 8  0 1 0  A=  6 8 8  0 8 8 0 8 2 10 8 0 8 3 0 5 0 7 0 0       SOLUTION: Expand in terms of the second row, so the determinant is the determinant of the smaller and smaller matrices: 4 0 6 0 0 8 1 8 8 8 8 0 8 8 2 8 0 8 3 0 5 0 7 0 0 = 4 6 0 0 8 8 8 2 8 8 3 0 5 7 0 0 4 8 5 =2 6 8 7 0 3 0 = 2 · (−3) 4 5 6 7 = 12 25. Let P be the stochastic matrix below. Find an equilibrium solution to the Markov Chain xi+1 = P x:   0.8 0.5 P = 0.2 0.5 SOLUTION: An equilibrium solution (or fixed point) is a vector x so that x = P x: P x − x = ~0 ⇒ (P − I)x = ~0 Therefore, we look for the null space of P − I. Subtract 1 from each of the diagonal elements of P and perform row reduction. You should get at least one row of zeros:           −0.2 0.5 −2 5 5/2 5 5/7 P −I = ∼ ⇒ x = x2 = x̂2 = x̃2 0.2 −0.5 0 0 1 2 2/7 Any of the three vector representations for x would be acceptable. The last one (as a probability vector) may be easiest to interpret. 26. On any given day, a student is either healthy or ill. Of the students that are healthy today, 95% will be healthy tomorrow. Of the students that are ill today, 55% will still be ill tomorrow. • What is the stochastic matrix for this situation? SOLUTION: We will assume that healthy is first, ill is second (be sure you state your ordering!)   0.95 0.45 P = 0.05 0.55 • Suppose 20% of the students are ill on Monday. What fraction are likely to be ill on Tuesday? On Wednesday? SOLUTION: For this computation, it is easiest to use a calculator (so I would make the numbers work out nicely on any exam). With that, the computations would be to take x = [0.8, 0.2]T , then on Tuesday we have P x and Wednesday we would have P 2 x:        0.85 0.925 0.675 0.8 0.875 2 Px = P x= = 0.15 0.075 0.325 0.2 0.125 From these computations, 15% of the students will be ill on Tuesday, and 12.5% will be ill on Wednesday. 11 • If a student is well today, what is the probability that he or she will be well two days from now? SOLUTION: From the matrix for P 2 , we could set x0 = [1, 0] and we would get:      0.925 0.675 1 0.925 = 0.075 0.325 0 0.075 We interpret this as: There is a probability of 92.5% that the student will be well two days from now (and also a probability of 7.5% that the student will be ill two days from now). 12