18.06 Linear Algebra, Quiz 3 Solutions

18.06 Quiz 3 Solutions Your PRINTED name is: 1. Your recitation number is 2. 3. 1. (40 points) Suppose u is a unit vector in Rn , so uT u = 1. This problem is about the n by n symmetric matrix H = I − 2u uT . (a) Show directly that H 2 = I. Since H = H T , we now know that H is not only symmetric but also . Solution Explicitly, we find H 2 = (I − 2uuT )2 = I 2 − 4uuT + 4uuuT uuT (2 points): since uT u = 1, H 2 = I (3 points). Since H = H T , we also have H T H = 1, implying that H is an orthogonal (or unitary) matrix. (b) One eigenvector of H is u itself. Find the corresponding eigenvalue. Solution Since Hu = (I − 2uuT )u = u − 2uuT u = u − 2u = −u, λ = −1. (c) If v is any vector perpendicular to u, show that v is an eigenvector of H and find the eigenvalue. With all these eigenvectors v, that eigenvalue must be repeated how many times? Is H diagonalizable? Why or why not? Solution For any vector v orthogonal to u (i.e. uT v = 0), we have Hv = (I−2uuT )v = v − 2uuT v = v, so the associated λ is 1. The orthogonal complement to the space spanned by u has dimension n−1, so there is a basis of (n−1) orthonormal eigenvectors with this eigenvalue. Adding in the eigenvector u, we find that H is diagonalizable. (d) Find the diagonal entries H11 and Hii in terms of u1 , . . . , un . Add up H11 + . . . + Hnn and separately add up the eigenvalues of H. Solution Since ith diagonal entry of uuT is u2i , the i diagonal entry of H is Hii = 1−2u2i � � (3 points). Summing these together gives ni=1 Hii = n − 2 ni=1 u2i = n − 2 (3 points). � Adding up the eigenvalues of H also gives λi = (1) − 1 + (n − 1)(1) = n − 2 (4 points). 2. (30 points) Suppose A is a positive definite symmetric n by n matrix. (a) How do you know that A−1 is also positive definite? (We know A−1 is symmetric. I just had an e-mail from the International Monetary Fund with this question.) Solution Since a matrix is positive-definite if and only if all its eigenvalues are positive (5 points), and since the eigenvalues of A−1 are simply the inverses of the eigenvalues of A, A−1 is also positive definite (the inverse of a positive number is positive) (5 points). (b) Suppose Q is any orthogonal n by n matrix. How do you know that Q A QT = Q A Q−1 is positive definite? Write down which test you are using. Solution Using the energy text (xT Ax > 0 for nonzero x), we find that xT QAQT x = (QT x)T A(QT x) > 0 for all nonzero x as well (since Q is invertible). Using the positive eigenvalue test, since A is similar to QAQ−1 and similar matrices have the same eigenvalues, QAQ−1 also has all positive eigenvalues. (5 points for test, 5 points for application) (c) Show that the block matrix ⎡ ⎤ A A ⎢ ⎥ B=⎣ ⎦ A A is positive semidefinite. How do you know B is not positive definite? Solution First, since B is singular, it cannot be positive definite (it has eigenvalues of 0). However, the pivots of B are the pivots of A in the first n rows followed by 0s in the remaining rows, so by the pivot test, B is still semi-definite. Similarly, the first n upper-left determinants of B are the same as those of A, while the remaining ones are 0s, giving another proof. Finally, given a nonzero vector ⎡ ⎤ ⎢ x ⎥ u=⎣ ⎦ y where x and y are vectors in Rn , one has uT Bu = (x+y)T A(x+y) which is nonnegative (and zero when x + y = 0). 3. (30 points) This question is about the matrix ⎡ ⎤ 0 −1 ⎢ ⎥ A=⎣ ⎦ 4 0 . (a) Find its eigenvalues and eigenvectors. ⎡ ⎤ ⎢ 2 ⎥ Write the vector u(0) = ⎣ ⎦ as a combination of those eigenvectors. 0 Solution Since det(A − λI) = λ2 + 4, the eigenvalues are 2i, −2i (4 points). Two associated eigenvectors are [1 − 2i]T , [1 2i]T , though there are many other choices (4 points). u(0) is just the sum of these two vectors (2 points). (b) Solve the equation du = Au starting with the same vector u(0) at time t = 0. dt In other words: the solution u(t) is what combination of the eigenvectors of A? Solution One simply adds in factors of eλt to each term, giving ⎡ ⎤ ⎡ ⎤ 1 ⎥ ⎢ 1 ⎥ −2it ⎢ u(t) = e2it ⎣ ⎦+e ⎣ ⎦ −2i 2i (c) Find the 3 matrices in the Singular Value Decomposition A = U Σ V T in two steps. –First, compute V and Σ using the matrix AT A. –Second, find the (orthonormal) columns of U. Solution Note that AT A = V ΣT U T U ΣV T = V Σ2 V T , so the diagonal entries of Σ are simply the positive roots of the eigenvalues of ⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎢ 0 4 ⎥ ⎢ 0 −1 ⎥ ⎢ 16 0 ⎥ AT A = ⎣ ⎦⎣ ⎦=⎣ ⎦ −1 0 4 0 0 1 i.e. σ1 = 4, σ2 = 1. Since AT A is already diagonal, V is the identity matrix. The columns of U should satisfy Au1 = σ1 v1 , Au2 = σ2 v2 : by inspection, one obtains ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ 0 ⎥ ⎢ −1 ⎥ ⎢ 0 −1 ⎥ u1 = ⎣ ⎦ , u2 = ⎣ ⎦,U = ⎣ ⎦ 1 0 1 0