Review problems for Math 511

Transcription

1 Review problems for Math 511 A Eremenko Spring Evaluate the determinants: , No electronic devices and no partial credit! Hint: you should not multiply numbers with three or more digits But you can easily add them Solution For the second determinant, subtract third column from the second to obtain = Expanding on the second column we obtain = where we added twice the second column to the first one = , Which of the following matrices A have the property that A is similar to A 017? a, b, c

2 Solution All three The eigenvalues are ±1 for a and e ±πi/3 for b, so they are diagonalizable As ±1 017 = ±1, A 017 = A Also e ±017 πi/3 = e ±πi/3, so the same holds for b Matrix c has repeated eigenvalue 1, and it is not diagonalizable, so has the Jordan form It is easy to see that J = J 017 = , has the same eigenvalues, and also is not diagonalizable, so it must have the same Jordan form, thus it is similar to the original matrix 3 Find the lengths and directions of the principal semi-axes of the ellipse 5x 1 +1x 1 x +10x = 13 and sketch the ellipse Solution Matrix A = has eigenvalues λ 1 = 1 and λ = 14, with eigenvectors v 1 = 3, T and v =,3 T The ellipsoid has equation x T Ax = 13, so the directions of the principal axes are v 1,v, and the semiaxes are a = and b = 13/ Solve the following recurrences: a n+1 = a n a n 1, a 0 = 0, a 1 = 1 a n+1 = 3a n +a n 1, a 0 = 0, a 1 = 1 Solution For any such equation, the general solution is a n = c 1 λ n 1 +c λ n, if λ 1 λ For example, for the first equation λ 1, are solutions of the characteristic equation λ λ+1 = 0

3 We write these solutions as λ 1 = 1 i 3 Then the initial conditions give = e πi/3, λ = 1+i 3 = e iπ/3 c 1 + c = 0, c 1 λ 1 + c λ = 1 Solving this by Cramer s rule, we obtain Thus a n = c 1 = 1 λ λ 1, c = 1 λ λ 1 1 λ λ 1 λ n λ n 1 = 3 sin πn 3 For the second equation, eigenvalues are and 1/, and the solution is easier 5 a Is there a matrix A such that A 0 but A = 0? b Is there a matrix A such that A 0 but A 3 = 0? In each case either give an example of such a matrix or explain why it does not exist Solution a Yes The Jordan cell with zero eigenvalue b No If A 3 = 0 the only eigenvalue must be 0 If A = 0 then A = 0 If A 0, it must be similar to a Jordan cell J with zero eigenvalues But this Jordan cell satisfies J = 0 thus A = 0 6 Find out which of the following quadratic forms can be transformed to each other by a linear change of the independent variables: P = xy+z, Q = x +y +z xy yz xz, R = xy+xz+yz Solution The signatures are P : +,+,, Q : +,+,, and R : +,,, so P and Q are congruent, but R is not congruent to either P or Q 3

4 7 Find out which of the following three matrices are similar to each other and which are not: A = 1 3 1, B = 0 4 0, C = The answer must be justified Solution Similar;ar matrices must have the same characteristic polynomial The easiest characteristic polynomial to write is that of C, and it is 4 λ 1 λ Now it is easy to see that eigenvalues of A and B are the same 4 and 1, and 4 is of multiplicity for both A and B are also symmetric thus they are both diagonalizable, have the same eigenvalues, including multiplicity, so they are similar For matrix C the characteristic polynomial is the same but there is only one-dimensional eigenspace for eigenvalue 4, therefore it must have a Jordan cell of size, and thus it is not similar to A or B 8 Is there a 3 3 matrix such that A 4 = 0 but A 3 0? Either give an example or explain why it does not exist Solution No See problem 5b 9 a Let A and B be two square matrices of the same size, one of them non-singular Show that AB is similar to BA b Under the same assumptions show that deti AB = deti BA 1 c More difficult Show that 1 is true even for rectangular matrices, A of size m n and B of size n m, and the units I are of appropriate size in each side of the formula Solution a A 1 ABA = BA, BABB 1 = BA b Similar matrices have the same determinant A 1 I ABA = I BA, BI ABB 1 = I = BA c We have I A det = det B I I A B I I A 0 I 4 = det I 0 B I BA = deti BA,

5 and similarly det I A B I = det I A 0 I I A B I I AB 0 = det B I = deti AB 10 a Show that every square matrix with the property A 017 = 1 is diagonalizable b Is it true that very matrix with the property A = 0 is diagonalizable? If true, explain why If not give an example Solution a All eigenvalues must be 017-th roots of 1 In particular there are no zero eigenvalues If there is a Jordan block of size at least with non-zero eigenvalue, no power of this Jordan block can be diagonalizable Thus A must be diagonalizable bno, itisnottrue: anexampleisa Jordancellwithzeroeigenvalue 11 In a railroad station, a worker walks along the train and kicks the car axles with a hummer to check whether there is a crack in an axle How does she detect a crack? How does the sound from an axle with a crack differs from the sound of an undamaged axle? Solution There are different ways of thinking what a crack is One can think of a crack as a decreased stiffness removing some elastic spring from the mathematical model of the axle This will decrease all eigenvalues of the stiffness matrix, thus decrease all frequencies in the sound it produce One can also think of a crack as removing a restriction increasing the number of degrees of freedom So the smallest eigenvalue of the matrix describing the axle with a crack will be smaller than the smallest eigenvalue of the matrix describing the system without a crack As the pitch of the tone is defined by the smallest eigenvalue, the pitch of a cracked axle will be lower, no matter what model of the crack one uses 1 Suppose that a square matrix A has the following property: whenever all coordinates of a vector x are positive, the coordinates of Ax are negative Does it follow that all eigenvalues of A are negative? If it does, explain why, if not, give a counterexample 5

6 Solution No Counterexample: Suppose that a system Ax = b of m linear equations with 017 unknowns with b = 1,0,,0 T has a unique solution Does it follow that a he null-space of A is trivial? b the rank of A equals 017? c m = 017? d A is invertible? e A T A is invertible? f AA T is invertible? g The rows of A are linearly independent? h the columns of A are linearly independent? i the rows of A are linearly independent? j Ax = b has a solution with every b R m? 14 Let us consider the linear operator on the space of 5 5 matrices, LA = A T Find the eigenvalues and describe the eigenspaces What are the dimensions of eigenspaces? Is this operator diagonalizable =does there exist a basis made of eigenvectors? Solution As L = I the eigenvalues can be only ±1 The eigenvectors corresponding to λ = 1 are symmetric matrices They make an eigenspace of dimension nn + 1/ Eigenvectors corresponding to λ = 1 are skew symmetric matrices They make an eigenspace of dimension nn 1/ As the sum of these dimensions is n, there is a basis consisting of eigenvectors 15 a On the space of all polynomials, consider three operators: the operator of differentiation D, the operator of multiplication on the independent variable X, and the identity operator I Show that DX XD = I b Can we find two 5 5 matrices which satisfy AB BA = I? Hint: what is the trace of the LHS? 6

7 Solution a is a direct calculation The answer to b is no, because trab BA = trab trba = 0 16 a Show that for all square matrices of the same size trab = trba b Show that the function A,B = trab is not a dot product on the space of n n matrices, but tra T B is c Show by example that in general trabc trbac d Show that trabc = trbca and in general trace of a product does not change when the multiples are interchanged cyclically 17 Show that for every square matrix A we have NA T A = NA and conclude that ranka T A = rankaa T = ranka Does this last equality hold for rectangular matrices as well? Solution If Ax = 0 then A T Ax = 0 Now suppose that A T Ax = 0, and denote Ax = y Then y is in the column space of A and also in the null space of A T But we know that these two spaces are orthogonal, thus we must have y = 0, which means that x NA Let A be of size m n and of rank r Then dimna = n r and dimna T A = n ranka T A We just proved that dimensions null spaces are equal, therefore the ranks are also equal Replacing A by A T which is of the same rank, we obtain that the rank of AA T is the same as r 18 Show that every orthogonal projector is positive semi-definite Solution Eigenvalues of an orthogonal projector can be only 0 and 1 Indeed, for all vectors in the column space of P we have Px = x, and for all vectors in the null space Px = 0 As the whole space is the direct sum of the column and null space they are orthocomplements to each other, because the orthogonal projector is symmetric, we obtain the result 19 a Which of the following matrices gives a rotation of R 3? 1/3 /3 /3 /3 1/3 /3 /3 /3 1/3, 7 1/3 /3 /3 /3 1/3 /3 /3 /3 1/3

8 b For this matrix, determine the axis and cosine of the angle of rotation Solution We need the characteristic polynomial of the first matrix To simplify the calculation, multiply it by 3, and let A = The characteristic polynomial of A is λ 3 +λ 3λ+7 = x 3x +3x+9 We hope that the first matrix represents a rotation, therefore it is expected to have 1 as an eigenvalue, so A is expected to have 3 as an eigenvalue, this simplifies factorization of the characteristic polynomial We also notice that the second matrix is obtained from the first by multiplying the last column by 1 Therefore its determinant will be negative, because the determinant of the first matrix is 1 So the second matrix cannot be a rotation Now the first matrix is orthogonal with determinant 1 so it is a rotation The eigenvector with eigenvalue 1 is the same as the eigenvector A with eigenvalue 3 This vector can be obtained by solving 4 x = 0, so we can take 1,1,0 T for the direction of the axis The other eigenvalues of A are 3±3i 3/, so eigenvalues of the first matrix are 1±i 3/ = e ±π/3, so cosine of the rotation angle is 1/ 0* Find the positive definite square root of the matrix 10 5 A = 5 5 Solution Eigenvalues are λ 1 = , 5 3 5

9 Corresponding matrix of eigenvectors is B = Now we have A = B ΛB 1, Λ = diag λ 1, λ, where λ j are positive These square roots can be found by trial and error: λ 1 = 5+ 5, Performing the multiplication we obtain A = λ = 5 5 It is easy to check that this answer is positive definite, and that its square equals A 1 Write this matrix as a product of a positive definite symmetric matrix left and an orthogonal matrix right: A = Solution We want to find an orthogonal matrix Q so that AQ 1 = S is symmetric and positive definite Notice that the most general form of a orthogonal matrix is Q 1 = p q q p, p +q = 1 So the condition that AQ 1 is symmetric means q p = 3p q which implies that p = 0,q = ±1 Taking q = 1 we obtain a symmetric S which is not positive definite, so take q = 1 and obtain =

10 Comments To 10: Why no power of Jordan block with non-zero eigenvalue of size k is diagonalizable? n-th power of this Jordan block is λi +N n = λ n I +nλ n 1 N +linear combination of higher powers ofn, where N = Powers of N higher than first have zeros on the main diagonal and on the next diagonal above the main Therefore, the element a,1 of the matrix λi +N n is nλ n 1 0 The only eigenvalue of λi +N n is λ n To find the eigenvectorswehavetosolvethesystembx = 0, whereb = λi+n n λ n I in the matrix B of this system, all elements on the main diagonal and below it are zero So the element b,1 = a,1 0 is the pivot, and therefore the rank of this matrix is at least 1 So it cannot have ks linearly independent eigenvectors, and thus is not diagonalizable To 11: Here is a story of computer modeling of a famous bell with a crack: To 1: The general form of a orthogonal matrix Suppose it is a x b y Orthogonality implies: ax+by = 0, a +b, bx+ay = ±1 3 The last equation says that the determinant is ±1 Considering, 3 as a system of equations with respect to x,y we obtain This proves the statement x = b ±a = b, y = a +b a +b = ±a 10