linearly independent associated eigenvectors. Nevertheless, it is a (somewhat disparaging)fact that eigenvalues can have fewer linearly independent eigenvectors than their multiplicitysuggests.Example:Find the eigenvalues and associated eigenvectors of the matrixA=2-112.We computedet(A-λI)=2-λ-112-λ=(λ-2)2+ 1=λ2-4λ+ 5.The roots of this polynomial areλ1= 2+iandλ2= 2-i; that is, the eigenvaluesare not real numbers.This is a common occurrence, and we can press on tofind the eigenvectors just as we have in the past with real eigenvalues. To findeigenvectors associated withλ1= 2 +i, we look forxsatisfying(A-(2 +i)I)x=0⇒-i-11-ix1x2=00⇒-ix1-x2x1-ix2=00⇒x1=ix2.Thus all eigenvectors associated withλ1= 2+iare scalar multiples ofu1= (i,1).Proceeding withλ2= 2-i, we have(A-(2-i)I)x=0⇒i-11ix1x2=00⇒ix1-x2x1+ix2=00⇒x1=-ix2,which shows all eigenvectors associated withλ2= 2-ito be scalar multiples ofu2= (-i,1).Notice thatu2, the eigenvector associated with the eigenvalueλ2= 2-iin the lastexample, is the complex conjugate ofu1, the eigenvector associated with the eigenvalueλ1= 2 +i. It is indeed a fact that, ifA∈ Mn×n(R) has a nonreal eigenvalueλ1=λ+iμwith corresponding eigenvectorξ1, then it also has eigenvalueλ2=λ-iμwith correspondingeigenvectorξ2=¯ξ1.5
This is the end of the preview.
access the rest of the document.