PY 351 Modern Physics Short assignment 4, Nov. 9, 2018, to be returned in class on Nov. 15. You may write your answers on this sheet or on a separate paper. Remember to write your name on top. Please note: the last four pages of this assignment recapitulate the basic facts about vectors and matrices and show how to calculate eigenvalues and eigenvectors of a simple matrix. 1) Find the eigenvalues and the eigenvectors of the following matrix: ( ) 4 6 6 1 (1) 2) Find the eigenvalues and the eigenvectors of the following matrix: 3 2 0 2 5 2 (2) 0 2 3 If you are not familiar with matrices, and/or eigenvalues and eigenvectors of a matrix, please read the following pages. 1
Matrices An N N matrix is a square table of numbers with N rows and N columns. A 2 2 matrix M would have the following form a1,1 a M (3) a 2,1 a 2,2 The numbers a i,j are called the elements of the matrix. Matrices are multiplied by a number c by multiplying all elements by c: ca1,1 ca cm (4) ca 2,1 ca 2,2 Matrices are added element by element. So, if M b1,1 b b 2,1 b 2,2 then M + M a1,1 + b 1,1 a + b a 2,1 + b 2,1 a 2,2 + b 2,2 (5) (6) The multiplication of M times a vector v (v 1, v 2 ) is done row by column as follows a1,1 a M v v1 a1,1 v 1 + a v 2 (7) a 2,1 a 2,2 v 2 a 2,1 v 1 + a 2,2 v 2 Matrices are multiplied by the same row by column rule. M M a1,1 a b1,1 b a1,1 b 1,1 + a b 2,1 a 1,1 b + a b 2,2 a 2,1 a 2,2 b 2,1 b 2,2 a 2,1 b 1,1 + a 2,2 b 2,1 a 2,1 b + a 2,2 b 2,2 (8) The above generalizes to 3 3 matrices etc. (same row by column rule for multiplication of matrix by vector and multiplication of two matrices.) Eigenvalues and eigenvectors A vector v is said to be an eigenvector of a matrix M if the vector one obtains by multiplying v by M is proportional to v: M v λv (9) 2
The constant of proportionality λ is called the eigenvalue of M corresponding to the eigenvector v. Under reasonably general circumstances an N N matrix will have N independent eigenvectors with N corresponding eigenvalues, although these are not necessarily distinct (two different eigenvectors may correspond to the same eigenvalue.) If we write out Eq. 9 explicitly, we get the following system of linear equations for v 1, v 2 : or, bringing the terms with λ to the l.h.s. In terms of M this can be written as a 1,1 v 1 + a v 2 λv 1 a 2,1 v 1 + a 2,2 v 2 λv 2 (10) (a 1,1 λ)v 1 + a v 2 0 (11) a 2,1 v 1 + (a 2,2 λ)v 2 0 (12) (M λi) v 0 (13) where I is the identity matrix (the product of any vector by I leaves the vector unchanged) 1 0 I (14) 0 1 Equations 11 and 12 form a homogeneous system of linear equations which will have v 1 v 2 0 as its only solution unless the determinant of the matrix of coefficients vanishes. If det(m λi) 0 the two equations are linearly dependent (they are basically the same equation.) One can then substitute the value found for λ in either equation and use it to find the values of v 1 and v 2, which of course will be determined only up to a common constant of proportionality. (As a check that everything has been done correctly, one can substitute the value found for v 1, v 2 in the other equation and see that it is also satisfied.) The determinant The determinant of a 2 2 matrix is simply the product a 1,1 a 2,2 of the two elements on the principal diagonal minus the product a a 2,1 of the two elements on the other diagonal. So det(m λi) is given by det(m λi) (a 1,1 λ)(a 2,2 λ) a a 2,1 (15) 3
Setting this equal to zero produces an algebraic equation of the second degree, which one can solve to find the two eigenvalues λ 1, λ 2. The rule for a determinant of a 3 3 matrix is more complicated. One way to calculate it consists in repeating the first two columns, getting a 3 5 table of numbers. One then adds the three products of the elements along the diagonals parallel to the principal diagonal of the matrix, from which one subtracts the three products of the elements along the diagonals running the other way. A specific example, which includes the variable λ, will clarify the concept. Say that the matrix is 2 1 3 M 1 5 4 (16) 3 4 6 (You should note the symmetry under inversion about the principal diagonal: a j,i a i,j. Matrices with this symmetry are called symmetric. If the matrix elements are complex numbers the symmetry generalizes to a j,i a i,j and the matrix is called Hermitian. Real symmetric matrices are also Hermitian. Most matrices appearing in physical problems are Hermitian and as such have special mathematical properties.) Subtracting λi from the matrix M we get 2 λ 1 3 M λi 1 5 λ 4 (17) 3 4 6 λ We replicate now the first two columns getting: 2 λ 1 3 2 λ 1 1 5 λ 4 1 5 λ (18) 3 4 6 λ 3 4 Taking the products of the elements along the diagonals, as explained above, we find det(m λi) (2 λ)( 5 λ)(6 λ) + (1)( 4)(3) + (3)(1)( 4) (3)( 5 λ)(3) (2 λ)( 4)( 4) (1)(1)(6 λ) (19) This is a polynomial of the third degree in λ. (We could carry out the multiplications and express it as aλ 3 +bλ 2 +cλ+d but this would serve no purpose.) 4
Setting it equal to zero would give an algebraic equation of the third degree. While these equations can be solved analytically, the result is quite unwieldy and it is very unlikely that you would run into a problem where one is asked to find the eigenvalues of a 3 3 matrix, unless there is some symmetry which simplifies the expression for det(m λi). As illustration consider an example like that was just given, but with a matrix 2 1 3 M 1 5 1 (20) 3 1 2 Inspection of this matrix shows that it is symmetric under interchange of row and column 1 with row and column 3. We calculate the determinant of M λi. We go straight to the 5 column formula which now reads 2 λ 1 3 2 λ 1 1 5 λ 1 1 5 λ (21) 3 1 2 λ 3 1 We have det(m λi) (2 λ)( 5 λ)(2 λ) + (1)(1)(3) + (3)(1)(1) This expression factorizes into (3)( 5 λ)(3) (2 λ)(1)(1) (1)(1)(2 λ) (2 λ) 2 (5 + λ) 2(2 λ) + 9(5 + λ) + 6 λ 3 λ 2 + 27λ + 27 (22) (λ + 1)( λ 2 + 27) (23) Setting it equal to zero we get the three eigenvalues λ 1 and λ ± 27 ±3 3. 5