Which does indeed equal the sum of the diagonal entries of A. The sum of the eigenvalues can be found by adding the two values expressed in (**) above: In fact, it can be shown that the eigenvalues of any real, symmetric matrix are real. Being the sum of two squares, this expression is nonnegative, so (**) implies that the eigenvalues are real. Therefore, if b = c, the discriminant becomes ( a − d) 2 + 4 b 2 = ( a − d) 2 + (2 b) 2. The discriminant in (**) can be rewritten as follows: The solutions of this equation-which are the eigenvalues of A-are found by using the quadratic formula: The eigenvalues of A are found by solving the characteristic equation, det ( A − λ I) = 0: What can you say about the matrix A if one of its eigenvalues is 0? Verify that the product of the eigenvalues is equal to the determinant of A.
Verify that the sum of the eigenvalues is equal to the sum of the diagonal entries in A. What can you say about the eigenvalues if b = c (that is, if the matrix A is symmetric)? They are satisfied by any vector x = ( x 1, x 2) T that is a multiple of the vector (2, 3) T that is, the eigenvectors of A corresponding to the eigenvalue λ = −2 are the vectorsĮxample 2: Consider the general 2 x 2 matrixĮxpress the eigenvalues of A in terms of a, b, c, and d. This is equivalent to the “pair” of equationsĪgain, note that these equations are not independent. The eigenvectors corresponding to the eigenvalue λ = −2 are the solutions of the equation A x = −2 x: Consequently, the eigenvectors of A corresponding to the eigenvalue λ = −1 are precisely the vectors and is therefore a multiple of the vector (1, 1) T. Any such vector has the form ( x 1, x 2) T. The equations above are satisfied by all vectors x = ( x 1, x 2) T such that x 2 = x 1. This is equivalent to the pair of equations The eigenvectors corresponding to the eigenvalue λ = −1 are the solutions of the equation A x = −x: Therefore, there are nonzero vectors x such that A x = x (the eigenvectors corresponding to the eigenvalue λ = −1), and there are nonzero vectors x such that A x = −2 x (the eigenvectors corresponding to the eigenvalue λ = −2). In Example 1, the eigenvalues of this matrix were found to be λ = −1 and λ = −2. This process is then repeated for each of the remaining eigenvalues.Įxample 1: Determine the eigenvectors of the matrix Substitute one eigenvalue λ into the equation A x = λ x-or, equivalently, into ( A − λ I) x = 0-and solve for x the resulting nonzero solutons form the set of eigenvectors of A corresponding to the selectd eigenvalue. In order to determine the eigenvectors of a matrix, you must first determine the eigenvalues.