MATH 2214 Midterm: MATH 2214 Virginia Tech 4.6 Exam
Document Summary
Solutions 4. 6: the characteristic equation is (2 )2 + 1 = 0, which leads to = 2 i. An eigenvector for 2 + i must satisfy. 2x1 + x2 = (2 + i)x1, x1 + 2x2 = (2 + i)x2. This reduces to the single equation x2 = ix1. Therefore we have the eigen- vector (cid:18) 1 i (cid:19) . The eigenvector for 2 i is the complex conjugate, i. e. (cid:18) 1. = (2 )( 1 )( 4 ) 6 9 + 3(2 ) + 9( 1 ) 2( 4 ) This has an integer root = 1. 2 2 2 = 0, which has roots 1 i. We can use the second equation to nd x1 = 3x3, and using this in the remaining two equations yields x2 = 0. An eigenvector for 1 i must satisfy (3 + i)x1 + 2x2 + 9x3 = 0, x1 + ix2 + 3x3 = 0, x1 x2 (3 i)x3 = 0.
