Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors The subject of eigenvalues and eigenvectors will take up most of the rest of the course. We will again be working with square matrices. Eigenvalues are special numbers associated with a matrix and eigenvectors are special vectors. Eigenvectors and eigenvalues A matrix A acts on vectors x like a function does, with input x and output Ax. Eigenvectors are vectors for which Ax is parallel to x. In other words: Ax = λx. In this equation, x is an eigenvector of A and λ is an eigenvalue of A. Eigenvalue 0 If the eigenvalue λ equals 0 then Ax = 0x = 0. Vectors with eigenvalue 0 make up the nullspace of A; if A is singular, then λ = 0 is an eigenvalue of A. Examples Suppose P is the matrix of a projection onto a plane. For any x in the plane Px = x, so x is an eigenvector with eigenvalue 1. A vector x perpendicular to the plane has Px = 0, so this is an eigenvector with eigenvalue λ = 0. The eigenvectors of P span the whole space (but this is not true for every matrix). � � � � 0 1 1 The matrix B = has an eigenvector x = with eigenvalue 1 1 0 1 � � 1 and another eigenvector x = with eigenvalue −1. These eigenvectors −1 span the space. They are perpendicular because B = B T (as we will prove). det( A − λI ) = 0 An n by n matrix will have n eigenvalues, and their sum will be the sum of the diagonal entries of the matrix: a11 + a22 + · · · + ann . This sum is the trace of the matrix. For a two by two matrix, if we know one eigenvalue we can use this fact to find the second. Can we solve Ax = λx for the eigenvalues and eigenvectors of A? Both λ and x are unknown; we need to be clever to solve this problem: = λx ( A − λI )x = 0 Ax In order for λ to be an eigenvector, A − λI must be singular. In other words, det( A − λI ) = 0. We can solve this characteristic equation for λ to get n solutions. If we’re lucky, the solutions are distinct. If not, we have one or more repeated eigenvalues. Once we’ve found an eigenvalue λ, we can use elimination to find the nullspace of A − λI. The vectors in that nullspace are eigenvectors of A with eigenvalue λ. Calculating eigenvalues and eigenvectors � � 3 1 Let A = . Then: 1 3 det( A − λI ) = � � 3−λ � � 1 � 1 �� 3−λ � = (3 − λ )2 − 1 = λ2 − 6λ + 8. Note that the coefficient 6 is the trace (sum of diagonal entries) and 8 is the determinant of A. In general, the eigenvalues of a two by two matrix are the solutions to: λ2 − trace( A) · λ + det A = 0. Just as the trace is the sum of the eigenvalues of a matrix, the product of the eigenvalues of equals its determinant. � any matrix � 3 1 For A = , the eigenvalues are λ1 = 4 and λ2 = 2. We find the 1 3 � � � � 1 −1 1 eigenvector x1 = for λ1 = 4 in the nullspace of A − λ1 I = . 1 1 −1 � � 1 1 x2 will be in the nullspace of A − 2I = . The nullspace is an entire 1 1 � � −1 line; x2 could be any vector on that line. A natural choice is x2 = . 1 � � 0 1 Note that these eigenvectors are the same as those of B = . Adding 1 0 � � 0 1 3I to the matrix B = added 3 to each of its eigenvalues and did not 1 0 change its eigenvectors, because Ax = ( B + 3I )x = λx + 3x = (λ + 3)x. A caution Similarly, if Ax = λx and Bx = αx, ( A + B)x = (λ + α)x. It would be nice if the eigenvalues of a matrix sum were always the sums of the eigenvalues, but this is only true if A and B have the same eigenvectors. The eigenvalues of the product AB aren’t usually equal to the products λ( A)λ(b), either. Complex eigenvalues � � 0 −1 The matrix Q = rotates every vector in the plane by 90◦ . It has 1 0 trace 0 = λ1 + λ2 and determinant 1 = λ1 · λ2 . Its only real eigenvector is the zero vector; any other vector’s direction changes when it is multiplied by Q. How will this affect our eigenvalue calculation? � � � − λ −1 � � � det( A − λI ) = � 1 −λ � = λ2 + 1. det( A − λI ) = 0 has solutions λ1 = i and λ2 = −i. If a matrix has a complex eigenvalue a + bi then the complex conjugate a − bi is also an eigenvalue of that matrix. Symmetric matrices have real eigenvalues. For antisymmetric matrices like Q, for which A T = − A, all eigenvalues are imaginary (λ = bi). Triangular matrices and repeated eigenvalues � � 3 1 For triangular matrices such as A = , the eigenvalues are exactly the 0 3 entries on the diagonal. In this case, the eigenvalues are 3 and 3: � � � 3−λ 1 �� � det( A − λ det I ) = � 3−λ � 0 � � = (3 − λ)(3 − λ) = ( a11 − λ)( a22 − λ) = 0, so λ1 = 3 and λ2 = 3. To find the eigenvectors, solve: � � 0 1 ( A − λI )x = x=0 0 0 � � 1 . There is no independent eigenvector x2 . to get x1 = 0