# M115A-AssignmentSevenExtraSolutions - MATH 115A -...

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MATH 115A - Assignment Seven Extra SolutionsPaul SkoufranisFebruary 26, 2012For Question 3 of Section 5.1, for each matrixA∈ Mn×n(F), determine all eigenvalues ofAand for eacheigenvalueλofA, find the set of eigenvectors corresponding toλ. If possible, find a basis ofFnconsistingof eigenvectors ofAand determine an invertible matrixQand a diagonal matrixDsuch thatQ-1AQ=D.§5.1 Question 3a)A=1232forF=R.
§5.1 Question 3b)A=0-2-3-11-1225forF=R.Solution:We desire to find the eigenvalues and eigenvector ofA.First we compute the eigenvalues.To find the eigenvalues, we compute the characteristic polynomial:χA(λ) = detλ231λ-11-2-2λ-5=λ(λ-1)(λ-5) + (2)(1)(-2) + (3)(1)(-2)-(3)(λ-1)(-2)-(2)(1)(λ-5)-λ(1)(-2)= (λ3-6λ2+ 5λ)-4-6 + (6λ-6)-(2λ-10) + 2λ=λ3-6λ2+ 11λ-6Therefore, to find the eigenvalues ofA, we need to factorχA.One way is to notice thatχA(1) = 0 (wewould try 1 as 1 divides-6 (see the Rational Roots Theorem)) and then use long division of polynomials.Another way is to notice thatχA(λ) = (λ3-6λ2+ 5λ) + (6λ-6)=λ(λ-5)(λ-1) + 6(λ-1)= (λ-1)(λ(λ-5) + 6)= (λ-1)(λ2-5λ+ 6) = (λ-1)(λ-2)(λ-3)Thus the eigenvalues ofAare 1, 2, and 3.Now we shall compute the eigenspaces. We notice thatE3=ker323121-2-2-2=ker101010000=span({(1,0,-1)})where the second equality comes from a simple row reduction. Therefore{(1,0,-1)}is a basis forE3.Next we notice thatE2=ker223111-2-2-3=ker110001000=span({(1,-1,0)})where the second equality comes from a simple row reduction. Therefore{(1,-1,0)}is a basis forE2.

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