Linear Algebra- Final Exam Review 2010

Linear Algebra- Final Exam Review 1. Let A be invertible. Show that, if v1 , v2 , v3 are linearly independent vectors, so are Av1 , Av2 , Av3 . NOTE: It should be clear from your answer that you know the definition. SOLUTION: We need to show that the only solution to: c1 Av1 + c2 Av2 + c3 Av3 = 0 is the trivial solution. Factoring out the matrix A, A(c1 v1 + c2 v2 + c3 v3 ) = 0 Think of the form Ax̂ = 0. Since A is invertible, the only solution to this is x̂ = 0, which implies that the only solution to the equation above is the solution to c1 v1 + c2 v2 + c3 v3 = 0 Which is (only) the trivial solution, since the vectors are linearly independent. (NOTE: Notice that if the original vectors had been linearly dependent, this last equation would have non-trivial solutions). 2. Find the line of best first for the data: x 0 1 2 3 y 1 1 2 2 Let A be the matrix formed by a column of ones and a column containing the data in x. If we write the original system as We form the normal equations AT Aβ = AT y    1 14 6 4 −6 T T −1 A A= ⇒ (A A) = 6 4 20 −6 14 1 The solution is x̂ = (AT A)−1 AT y = 10 [4, 9]T   −3 0 3. Let A = . (a) Is A orthogonally diagonalizable? If so, orthogonally diagonalize 0 0 it! (b) Find the SVD of A. SOLUTION: Yes it is (it is symmetric). The eigenvalues are the diagonal elements, λ = −3, 0. The eigenvectors are [1, 0]T and [0, 1]T , respectively, so P = I. Because one eigenvalue is negative, the SVD is slightly different. The singular values are √ sigma = 9 = 3 and 0. We can put the negative sign with the matrix V to compensate (so that the factorization works out). NOTE: The SVD of a matrix, like the eigenvector decomposition of a matrix, need not be unique. Any representatives from the eigenspaces will make the factorization work for the diagonalization, and in the SVD, ± signs are chosen to be sure the diagonal entries of Σ are positive. 1 4. Let V be the vector space spanned by the functions: f1 (x) = x sin(x) f2 (x) = x cos(x) f3 (x) = sin(x) f4 (x) = cos(x) Define the operator D : V → V as the derivative. (a) Find the matrix A of the operator D relative to the basis f1 , f2 , f3 , f4 SOLUTION: Df1 = sin(x) + x cos(x) = f3 + f2 This implies that in coordinates,    1 0  0   1   D  0 = 1 0 0 Similarly, we should find that:      0 −1 0  1   0   0     D  0  =  0 , D 1 0 1 0        0   0   =  ,   0  1    0 0  0   0   D  0  =  −1 1 0     The matrix for D is the matrix formed from the columns above (in order). (b) Find the eigenvalues of A. SOLUTION: Taking the determinant of A − λI, we find that the characteristic polynomial is (I used the third column): (λ2 + 1)2 = 0 so λ = ±i, ±i. (c) Is the matrix A diagonalizable? SOLUTION: No, not using real numbers. 5. Short answer: (a) Let H be the subset of vectors in IR3 consisting of those vectors whose first element is the sum of the second and third elements. Is H a subspace? SOLUTION: One way of showing that a subset is a subspace is to show that the subspace can be represented by the span of some set of vectors. In this case,       a+b 1 1      a =a 1 +b 0  b 0 1 Because H is the span of the given vectors, it is a subspace. 2 (b) Explain why the image of a linear transformation T : V → W is a subspace of W SOLUTION: Maybe “Prove” would have been better than “Explain”, since we want to go through the three parts: i. 0 ∈ T (V ) since 0 ∈ V and T (0) = 0. ii. Let u, v be in T (V ). Then there is an x, y in V so that T (x) = u and T (y) = v. Since V is a subspace, x + y ∈ V , and therefore T (x + y) = T (x) + T (y) = u + v so that u + v ∈ T (V ). iii. Let u ∈ T (V ). Show that cu ∈ T (V ) for all scalars c. If u ∈ T (V ), there is an x in V so that T (x) = u. Since V is a subspace, cu ∈ V , and T (cu) ∈ T (V ). By linearity, this means cT (u) ∈ T (V ). (OK, that probably should not have been in the short answer section)   1 2 3 (c) Is the following matrix diagonalizable? Explain. A =  0 5 8  0 0 13 SOLUTION: Yes. The eigenvalues are all distinct, so the corresponding eigenvectors are linearly independent. (d) If the column space of an 8 × 4 matrix A is 3 dimensional, give the dimensions of the other three fundamental subspaces. Given these numbers, is it possible that the mapping x → Ax is one to one? onto? SOLUTION: If the column space is 3-d, so is the row space. Therefore the null space (as a subspace of IR4 ) is 1 dimensional and the null space of AT is 5 dimensional. Since the null space has more than the zero vector, Ax = 0 has non-trivial solutions, so the matrix mapping will not be 1-1. Since the column space is a three dimensional subspace of IR8 , the mapping cannot be onto.   1 1 2 2 6. Find a basis for the null space, row space and column space of A, if A =  2 2 5 5  0 0 3 3 The basis for the column space is the set containing the first and third columns of A. A basis for the row space is the set of vectors [1, 1, 0, 0]T , [0, 0, 1, 1]T . A basis for the null space of A is [−1, 1, 0, 0]T , [0, 0, −1, 1]T . 7. Find an orthonormal basis for W = Span {x1 , x2 , x3 } using wait until the very end to normalize all vectors at once):      0 0     0   , x2 =  1  , x3 =  x1 =    1   1  1 2 Gram-Schmidt (you might  1 1  , 1  1 Using Gram Schmidt (before normalization), we get      0 0 3  0   2   1     v1 =   1  , v2 =  −1  , v3 =  1 1 1 −1 3     8. Find the QR factorization of the matrix A[x1 x2 x3 ], defined in the previous problem. SOLUTION: Gets a bit messy by hand- We can write it symbolically as R = QT A, where the columns of Q are the normalized vectors from the previous problem. 9. Given the QR factorization of a matrix A, show that the least squares solution to Ax = b is x̂ = R−1 QT b. (Hint: Start with Ax̂, and see where you go). SOLUTION: Using the hint, and substitute the expression: Ax̂ = (QR)R−1 QT b = QQT b which is the orthogonal projection of b into the column space of Q, which was the column space of A. Therefore, we obtain b̂; the element of the column space closest to b. 10. Using the QR factorization in the previous problem for matrix A, find the least squares solution to Ax = b, where b = [1, 0, 0, 0]T . SOLUTION: As in the text (6.4), we see that even small problems can be messy- Try finding the solution in Matlab (sorry- I thought it would work out a little cleaner by hand). 11. Let Pn be the vector space of polynomials of degree n or less. Let W1 be the subset of Pn consisting of p(t) so that p(0)p(1) = 0. Let W2 be the subset of Pn consisting of p(t) so that p(2) = 0. Which of the two is a subspace of Pn ? SOLUTION: First consider W1 . The zero polynomial is in W1 . The condition states that either p(0) or p(1) must be zero, but not both- That could cause a problem. In fact, what happens with a sum? Let f (t) = p(t) + q(t). Then f (0)f (1) = (p(0) + q(0))(p(1) + q(1)) = p(0)p(1) + p(0)q(1) + p(1)q(0) + q(0)q(1) The middle terms will cause a problem. For example, how about p(0) = 4, p(1) = 0 and q(0) = 0, but q(1) = 3? They each satisfy the condition for W1 , but the sum will not. Therefore, W1 is not a subspace of Pn . For W2 , we will have a subspace (check the three conditions). 12. For each of the following matrices, find the characteristic equation, a basis for each eigenspace:      1 0 7 2 3 −1  A= B= C= 0 2 −4 1 1 3 1 0 the eigenvalues and  1 0  1 SOLUTION: For matrix A, λ = 3, 5. Eigenvectors are [1, 1]T and [2, 1]T , respectively. For matrix B, for λ = 3+i, an eigenvector is [1, i]T . The other eigenvalue and eigenvector are the complex conjugates. For matrix C, expand along the 2d row. λ = 2 is a double eigenvalue with eigenvectors [0, 1, 0]T and [1, 0, 1]T . The third eigenvalue is λ = 0 with eigenvector [−1, 0, 1]T . 4   p(−1) 13. Define T : P2 → IR3 by: T (p) =  p(0)  p(1) (a) Find the image under T of p(t) = 5 + 3t. SOLUTION: [2, 5, 8]T (b) Show that T is a linear transformation. SOLUTION: We show it using the definition. i. Show that T (p + q) = T (p) + T (q):       p(−1) + q(−1) p(−1) q(−1) T (p + q) =  p(0) + q(0)  =  p(0)  +  q(0)  = T (p) + T (q) p(1) + q(1) p(1) q(1) ii. Show that T (cp) = cT (p) for all scalars c.     cp(−1) p(−1) T (cp) =  cp(0)  = c  p(0)  = cT (p) cp(1) p(1) (c) Find the kernel of T . Does your answer imply that T is 1 − 1? Onto? (Review the meaning of these words: kernel, one-to-one, onto) SOLUTION: Since the kernel is the set of elements in the domain that map to zero, let’s see what what the action of T is on an arbitrary polynomial. An arbitrary vector in P2 is: p(t) = at2 + bt + c, and:   a−b+c  c T (at2 + bt + c) =  a+b+c For this to be the zero vector, c = 0. Then a − b = 0 and a + b = 0, so a = 0, b = 0. Therefore, the only vector mapped to zero is the zero vector. Side Remark: Recall that for any linear function T , if we are solving T (x) = y, then the solution can be written as x = xp + xh , where xp is the particular solution (it solves T (xp ) = y), and T (xh ) = 0 (we said xh is the homogeneous part of the solution). So the equation T (x) = y has at most one solution iff the kernel is only the zero vector (if T was realized as a matrix, we get our familiar setting). Therefore, T is 1 − 1. The mapping T will also be onto (see the next part). (d) Find the matrix for T relative to the basis {1, t, t2 } for P2 . (This means that the matrix will act on the coordinates of p). You can verify that the matrix representation for T is:   1 −1 1  0 0 1  1 1 1 5 From this we see that the matrix is invertible, so the mapping is both 1-1 and onto (as a mapping from IR3 to IR3 ). 14. Let v be a vector in IRn so that kvk = 1, and let Q = I − 2vvT . Show (by direct computation) that Q2 = I. SOLUTION: This problem is to practice matrix algebra: Q2 = (I −2vvT )(I −2vvT ) = I 2 −2IvvT −2vvT I +4vvT vvT = I −4vvT +4v(1)vT = I 15. Let A be m × n and suppose there is a matrix C so that AC = Im . Show that the equation Ax = b is consistent for every b. Hint: Consider ACb. SOLUTION: Using the hint, we see that ACb = b. Therefore, given an arbitrary vector b, the solution to Ax = b is x = Cb. 16. If B has linearly dependent columns, show that AB has linearly dependent columns. Hint: Consider the null space. SOLUTION: If B has linearly dependent columns, then the equation Bx = 0 has nontrivial solutions. Therefore, the equation ABx = 0 has (the same) non-trivial solutions, and the columns of AB must be linearly dependent. 17. If λ is an eigenvalue of A, then show that it is an eigenvalue of AT . SOLUTION: Use the properties of determinants. Given |A − λI| = |(A − λI)T | = |AT − λI T | = |AT − λI| the solutions to |A − λI| = 0 and |AT − λI| = 0 are exactly the same.     2 −2 18. Let u = and v = , Let S be the parallelogram with vertices at 0, u, v, and 1 1 u + v. Compute the area of S. SOLUTION: The area of the parallelogram formed by two vectors in IR2 is the determinant of the matrix whose columns are those vectors. In this case, that would be 4.       a b c a + 2g b + 2h c + 2i g h i 19. 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: This question reviews the relationship between the determinant and row operations. The determinant of B is 5. The determinant of C is 10. The determinant of BC is 50.        6 3 4 4 20. Let B = , and C = . Write down the matrices that −5 −6 5 7 take [x]C to [x]B and from [x]B to [x]C . 6 SOLUTION: Since x = PB [x]B and x = PC [x]C , then [x]C = PC−1 PB [x]B [x]B = PB−1 PC [x]C Doing the calculations, PC−1 PB  = −1/2 5 −1 5  PB−1 PC  = 2 2 2/5 1/5  21. Define an isomorphism: A one-to-one linear transformation between vector spaces (see p. 251) 22. Let  B= 1 −3   , 2 −8    −3 , 7 Find at least two B−coordinate vectors for x = [1, 1]T . SOLUTION: The null space of the matrix is spanned by [5, −1, 1]T (found by row reduction. The particular part of the solution is [5, −2, 0]T . So we can find an infinite number of solutions. 23. Let B = {b1 , . . . , bn } be a basis for vector space V . Explain why the B−coordinate vectors of {b1 , . . . , bn } are the columns of the n × n identity matrix: SOLUTION: [b1 ]B = [1, 0, 0, . . . , 0]T because b1 = 1b1 + 0b2 + · · · + 0bn . A similar argument shows that [bi ]B = ~ei . 24. Find the volume of the parallelepiped formed by 0, a,b, c, a + b, c + b, c + a, and the sum of all three.       0 −2 −1 a= 3  b= 0  c= 0  −1 1 1 The volume is the determinant. In this case, expand along the second row, and get 3. 7