# M217_W16_HW11Sols - HOMEWORK 11 SOLUTIONS Part A(15 points...

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HOMEWORK 11, SOLUTIONSPart A (15 points)Solve the following problems from the book:Section 7.2:10, 38.Section 7.3:10, 16, 40.Section 7.4:42, 48, 50.Section 7.5:24, 28.Solution.7.2.10detλ+ 30-40λ+ 102-7λ-3=(λ+ 3)detλ+ 10-7λ-3-4det0λ+ 12-7=(λ+ 3)(λ+ 1)(λ-3)-4(-2)(λ+ 1)=λ2+λ2-λ-1 = (λ2-1)(λ-1)=(λ-1)2(λ+ 1)So the eigenvalues areλ= 1(with multiplicity 2) andλ=-1(with multiplicity 1).7.2.38We know (by Example 7 of the book) that the characteristic polynomial ofAis given byλ2-tr(A)λ+ det(A);on the other hand, the well-known Vieta formulas for the coefficients of polynomials in termsof their roots (which are also mentioned in the book) imply that that ifλ1, λ2are the rootsof this polynomial, then the polynomial isλ2-(λ1+λ2)λ+λ1λ2thus we are looking for two numbers,λ1andλ2, whose sum equals -5 (the negative of thetrace ofA) and whose product equals -14 (the determinant ofA). Clearly the numbers-7and2fit the bill.Solution.7.3.10detλ-1-100λ-1000λ=λ(λ-1)2,so the eigenvalues are0, with algebraic multiplicity1, and1with algebraic multiplicity2.We now find the corresponding eigenspaces. We haveE0= ker-1-100-10000= span001,
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HOMEWORK 11, SOLUTIONS2which has dimension1(and hence the geometric and algebraic multiplicities of0coincide),andE1= ker0-10000001= span100,which also has dimension1(and hence the geometric and algebraic multiplicities of1do notcoincide). Therefore the matrix is not diagonalizable.7.3.16detλ-1-100λ+ 11-2-2λ=(λ-1)detλ+ 11-2λ+ det01-2λ=(λ-1)(λ2+λ+ 2) + 2 =λ3+λ=λ(λ2+ 1),whose only real root is0, with algebraic multiplicity1(the other roots are±1, and so theydon’t count). Therefore we can tell right away that the matrix will not be diagonalizable.Now for the only relevant eigenspace we haveker-1-10011-2-20= span-11-17.3.40detλ-1a0λ-1= (λ-1)2,so the only eigenvalue is1, with algebraic multiplicity2.Now to get the correspondingeigenspace, we look atker0a00=span("10#);ifa6= 0span("10#,"01#);ifa= 0so if we want the matrix to be diagonalizable, it better be the case thata= 0, so the algebraicand geometric multiplicities of the only available eigenvalue coincide.Solution.7.4.42We start by picking a basis for our vector space, of which the most reasonable option seemsto beB=1000,0100,0010,0001therefore theB-matrix forLisA=000001-100-1100000,so thatdet(λI4-A) =λ2((λ-1)2-1) =λ2(λ2-2λ) =λ3(λ-2),
HOMEWORK 11, SOLUTIONS3so the eigenvalues are0, with algebraic multiplicity3, and2with algebraic multiplicity1.Now to calculate the corresponding eigenspaces, the computation forλ= 0is:ker00000-11001-100000= span1000,0001,0110hence the corresponding eigenspace isE0= span1000,0001,0110, whereas forλ= 2we getker2000011001100002= span01-10,thus the actual eigenspace isE2= span01-10.