Question

The 2 x 2 matrix 1 = ( 43 II has two distinct real eigenvalues. 1. Give the characteristic polynomial for A in Maple notation in the form t^2 + a*t + b Characteristic polynomial = 2. Find the set of eigenvalues for A, enclosed in braces , ) with the two eigenvalues separated by a comma, like (-4, 7) Set of eigenvalues for A = 5 3. Find one eigenvector for each eigenvalue, using Maple > for vectors, e.g. <5, -6> for the vector c) an eigenvector for the smallest eigenvalue for A = AL an eigenvector for the largest eigenvale for A =

Answer 1

-3 -4 ☺ Solution - Characteristie Bolynomial of A is, det (A- £ I2) * . -3 -4-4 = -f(-4-4) -).(-3) = 44+42 +3 2 +44 +3 .: 1. The Charateristic polynomial is. A^2 + 4* £ + 3 १. Now, we know that the eigen values are the roots of the Charalteristie equation. Now, Charaferis tie equation is given by, ļ+44 +3 = 0 => ť?+34 +4+3=0 - #(++3) +1 (4+3)=0 -> (4+3) (t+!)=0 ++3=0 4+1=0 set=-1 ..4.-3 -3,-1 Thus, the eigen values of A are set of eigen values for A - (-3,-1)

Mew Let, x = ( Josbe the eigen veetor of A corresponding to the eigen value ti=-3 Then, A XI = -3X, => (A+352) X, = 0 3 (3-1)()(0) へい -> 32,+22 .34,-22 a) 32, + X2=0 => 22 = -37, Thus, 1, 37(3) is an eigen veetor Then X If, we take 2,= 1 • (-3) is an of A Corresponding to the eigen value ty=-3.

Leto X2² xz ) (+0) be the eigen veetors of A corresponding to the eigen value t = -1 x4 Then, A X2=-X2 => [A +Iz) x2 =0 57 (3) **)- :) - => 23 4X4 -3x3-324 => 73 +3430 Thus, If, we take x3=1, then X2= => x4=-23 x2 = *s(-1).svo (-) is an eigen veetor of A corresponding to the eigen' value &z=-4. since, -3 is the smallest eigen value foi is the greatest eigen valve. an eigen veetor for the smallest eigen value for <1,-3) A = An eigen rector for the greatest eigen value for Az <1,-1)