18.06 Problem Set 8 Solution Due Wednesday, 22 April 2009 at 4 pm in Total: 160 points.

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1 86 Problem Set 8 Solution Due Wednesday, April 9 at 4 pm in -6 Total: 6 points Problem : If A is real-symmetric, it has real eigenvalues What can you say about the eigenvalues if A is real and anti-symmetric A A T? Give both a general explanation for any n n A similar to what we did in class and in the book and check by finding the eigenvalues a anti-symmetric example matrix Solution 5 points proof + 5example If λ is an eigenvalue of A with a nonzero eigenvector v, that is Av λv Then, on one hand, we have v H Av v H λv λ v, and on the other hand, v H Av v H A T v Av H v λv H v λ v Since v is nonzero, v > We conclude that λ λ This implies that λ is purely imaginary, that is the real part of λ is zero For example, we take A Since A is anti-symmetric, its diagonal entries must be zero We then solve deta λi λ + to get λ i, λ i They are purely imaginary numbers Problem : Find an orthogonal matrix Q that diagonalizes A that Q T AQ Λ where Λ is diagonal What is Λ? 6, ie so 6 7 Solution points Since A is real-symmetric, we should be able to get orthonormal eigenvectors, and then Q is just the matrix whose columns are the eigenvectors as in class and the textbook, and Λ is the diagonal matrix of eigenvalues So, we just solve for the eigenvalues and eigenvectors of A To get the eigenvalues, we solve deta λi λ 5λ 5, obtaining λ and λ 5 Since the eigenvalues are distinct, the eigenvectors are automatically orthogonal, and we just need to normalize them to have length : 6 / 5 λ, A λ I, v, q λ 5, A λ I , v 6, q / 5 / 5 / 5

2 Hence, we have Q , Λ 5 Problem 3: Even if the real matrix A is rectangular, the block matrix B A A T is symmetric An eigenvector x of B satisfies B x λ x with: y A y y x, z A T λ, z z and thus A z λ y and A T y λ z a Show that λ is also an eigenvalue of B, with the eigenvector y, z T b Show that A T A z λ z, so that λ is an eigenvalue of A T A c Show that λ is also an eigenvalue of AA T by finding a corresponding eigenvector d If A I, find all four eigenvalues and eigenvectors of B Solution 5 points a We check this by direct computation y A z λ y B z A T y λ z y λ z Hence λ is also an eigenvalue of B, with the eigenvector b Again, we check by direct computation A T A z A T λ y λa T y λλ z λ z Hence, λ is an eigenvalue of A T A with eigenvector z y z c By symmetry, it is not hard to guess that y may be an eigenvector of AA T Indeed, AA T y Aλ z λa z λλ y λ y

3 Hence, λ is an eigenvalue of AA T with eigenvector y d We can use the results from part c to solve this quickly If A I, then A T A I has an eigenvalue λ But the eigenvalues of I are, so λ, so λ must be or But from a, if is an eigenvalue then must also be an eigenvalue and vice versa, so the eigenvalues are and Furthermore, from c, y is given by the eigenvectors of A T A I, which are just, T and, T Furthermore, since we were given that A T y λ z, it follows that z ± y for λ ± Hence, the eigenvectors of λ are: and the eigenvectors of λ are: x, x x 3, x 4 There are also more laborious ways to get the same result For example, once we know that λ ±, we could compute the nullspace of B ± I Problem 4: True or false give a reason if true, or a counter-example if false a A matrix with real eigenvalues and real eigenvectors is symmetric b A matrix with real eigenvalues and orthogonal eigenvectors is symmetric c The inverse of a symmetric matrix is symmetric d The eigenvector matrix S of a symmetrix matrix is symmetric e A complex symmetric matrix has real eigenvalues f If A is symmetric, then e ia is symmetric g If A is Hermitian, then e ia is Hermitian 3

4 Solution 35 points a False For example, A has eigenvalues λ 3 and λ 3 with eigenvectors v and v, respectively REMARK: For an upper triangular matrix A, the eigenvalues are exactly the elements on the diagonal b False For example, A REMARK: True is acceptable if the students assume that A is diagonalizable, since we had handled only diagonalizable matrices up to that point in the course In this case, we may normalize the eigenvectors so that they form an orthonormal basis Now, we write A S ΛS Since the eigenvectors form an orthonormal basis, S S T Hence, is symmetric A T S T ΛS T S T Λ T S T T S T ΛS A c True If A is a symmetric matrix, A T A A T A T AA T I T I; this says A T is also the inverse of A, which is of course the same as A Hence A is symmetric d False For example, A Then deta λi λ λ 3 λ, λ 3 We solve the eigenvectors to be v, v respectively So, S Another way to see this is false is that one can freely multiply any column of S by a nonzero constant, but this will not preserve the symmetry e False For example, A i, the matrix The eigenvalue is i; it is not a real number f True Indeed, e ia T e iat e ia g False e ia H e iah e iah e ia It is typically not the same as e ia Taking A, the matrix, would be a good enough example because e ia e i, which is not a real number 4

5 Problem 5: For which s is A positive definite? s 4 4 A 4 s s Solution 5 points Method : We do elimination and check the pivots are all positive numbers s 4 4 s 4 4 A 4 s 4 s s s 4 4 s 4 6 s 6 s s s 4 4 s 4 4 s s s s 4 6 s 6 s s s s s s s /s 6 s We need all pivots to be bigger than The first pivot is s, which gives s > The second pivot is s 6, which gives s s 6 > Since we already have s >, s > 6 and hence s > 4 Now, the third s pivot > gives Hence s > 8 s 6/s > 4 6/s s 6 > 4s + 6 s 4 3s + 56 > 6s + 8s + 56 s 4 48s 8s > s 3 48s 8 > s + 4 s 8 > Method : Since A is symmetric, it is diagonalizable We just need to make all the eigenvalues of A positive Consider s λ 4 4 deta λi det 4 s λ s λ s λ s λ s λ 3 48s λ 8 s λ + 4 s λ 8 5

6 Hence the eigenvalues are λ λ s + 4 and λ 3 s 8 If A is positive definite, s + 4 >, s 8 > s > 8 REMARK: The factorization does not come out from nowhere One way to see it is that we know beforehand, when s λ 4, A λi is of rank and hence there should be a -dimensional nullspace So, we are expecting two eigenvalues s + 4 The s λ 8 term might be harder to guess But knowing two roots of the equation, one can get the third by high-school synthetic division One could also look up the formula for cubic equations on Wikipedia, but that s a lot more messy! REMARK: There is another criterion by taking the determinants of the main diagonal submatrices, which we did not cover in the class It says that a symmetric matrix A is positive definite if and only if the determinant of upper left r r block is > for any r including r n, which corresponds to the determinant of A Problem 6: If A has full column rank, and C is positive-definite, show that A T CA is positive definite Recall that A T CA is an important matrix; for example, it arose in lecture 3 on graphs and networks, section 8 of the text Solution points Since C is positive-definite, y T Cy > for any y in R n Now, we need to show that z T A T CAz > for any z in R n We can rewrite it as z T A T CAz Az T CAz Since A has full column rank, NA {} and in particular, Az in R n Therefore, we have Az T CAz > This implies that A T CA is positive definite Problem 7: For f x, y x 4 /4 + x + x y + y and f x, y x 3 + xy x, find the second-derivative matrices H and H, where: H x y x x y y Find the minimum point of f and check that H is positive-definite there Find the saddle point of f look only where the first derivatives are zero, and check that H has two eigenvalues with opposite signs 6

7 Solution points + For f x, y, we first solve for the stationary point x y x f x x3 + x + xy, f y x + y From, we have y x / Plug this into, we have x and hence the only critical point is x y At this point, 3x H y x x x y y It is positive definite and hence, is a minimal point of f x, y REMARK: The problem for f x 4 /4 + x y + y as originally stated, you get a curve of minima x + y, and H is only positive semidefinite For f x, y, we first solve for the stationary point f x 3x + y, f y x This implies that y At this point,, H x y x x y y 6x The eigenvalues of H at, is the solution to deth λi λ, which are λ and λ They are with opposite signs and hence, is a saddle point of f x, y Problem 8: a Give an explicit formula for u k A k u, where A T and u 7

8 b Although you should find that A s eigenvalues and eigenvectors are not real, give explicit values for u, u, u, u 3, showing that your formula gives real results c u k+n u k for what values of n? Solution points +5+5 a Solving deta λi λ + gives λ i and λ i i λ i, A λ I, v i i i λ i, A λ I, v i i Hence, Then, we solve c i i c u k c λ k v + c λ k v i i k i c c + + i i +i i k i REMARK: As said in class, when A is a real matrix, having a complex eigenvalue λ with eigenvector v, then λ is also an eigenvalue and v is one of the eigenvectors Moreover, when we try to solve for c and c, we have c c One can further simplify the result we have to be i Re ii u k Re i k k i Re ii k+ b We have u i + + i i i u i i + + i i i i u i + + i i i u 3 i i + + i i i i 8

9 A faster way might be to use u k+ A u k to compute u, u, u 3 after we computed u c Since i k+4 i k and i k+4 i k, We have u k+4 u k From this, we can of course say u k+4m u k for any m Another way to see this is to check first several terms and find out that u 5 u Then we know u k+4 A k u 5 A k u u k Problem 9: For what real values of s does d u/dt A u have exponentially growing solutions, where s A? 3 Solution points We need only to find the eigenvalues of A Solving deta λi λ +4λ+3 s would give the two eigenvalues of A, λ 4± 4+8s The system has exponentially growing solution if and only if one of the solution has a solution with positive real part If 4 + 8s <, ie, the eigenvalues are complex, the real part is equal to, which will not give exponentially growing solution Thus, 4 + 8s, ie, we have two real eigenvalues The bigger one should be greater than in order to get the exponentially growing solution 4 ± 4 + 8s > 4 + 8s > 4 s > 3/ 9

18.06 Problem Set 8 - Solutions Due Wednesday, 14 November 2007 at 4 pm in

18.06 Problem Set 8 - Solutions Due Wednesday, 14 November 2007 at 4 pm in 806 Problem Set 8 - Solutions Due Wednesday, 4 November 2007 at 4 pm in 2-06 08 03 Problem : 205+5+5+5 Consider the matrix A 02 07 a Check that A is a positive Markov matrix, and find its steady state