Solution.
1.
A=(1−824) Solve equation for finding eigenvalues:
∣A−λI∣=0
∣∣1−λ−824−λ∣∣=λ2−5λ+20. From here we have eigenvaluesλ1=25−55i,λ2=25+55i.
Solve equation for finding the first eigenvector
(A−λ1I)v1=0.
Hence
(2−3+55i−8223+55i)→(1016−3−55i0) then x2=x2,x1=163+55i.
So, v1=(163+55i1).
Solve equation for finding the second eigenvector
(A−λ2I)v2=0.
Hence
(2−3−55i−8223−55i)→(1016−3+55i0) then x2=x2,x1=163−55i.
So, v2=(163−55i1).
2.
B=(a−bba) Solve equation for finding eigenvalues:
∣B−λI∣=0
∣∣a−λ−bba−λ∣∣=a2−λ2+b2. From here we have eigenvaluesλ1=a2+b2,λ2=−a2+b2.
Solve equation for finding the first eigenvector
(B−λ1I)v1=0.
Hence
(a−a2+b2−bba−a2+b2)→(b0−a−a2+b22ba2+b2) then x2=2ba2+b2,x1=2aa2+b2+2(a2+b2).
So, v1=(2aa2+b2+2(a2+b2)2ba2+b2).
Solve equation for finding the second eigenvector
(B−λ2I)v2=0.
Hence
(a+a2+b2−bba+a2+b2)→(−b0a+a2+b2−2ba2+b2) then x2=−2ba2+b2,x1=2aa2+b2−2(a2+b2).
So, v2=(2aa2+b2−2(a2+b2)−2ba2+b2).
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