# HW11 Solutions Math 110 Fall 2016 - HW 11 Solutions MATH...

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HW 11 Solutions, MATH 110 with Professor StankovaThe author of this note prefers the nomenclature “Jordan normal form” to “Jordan canonical form.”The reasoning is that the word “canonical” should be reserved for things that are somehow uniqueor independent of choice, but the Jordan normal form clearly depends on a choice of ordering (ofJordan blocks).7.1 #1efh (e) False. Take the identity matrix, for example. (f) False. We need to take a basis ofKλwhich can be partitioned into cycles. (h) True. A previous homework problem showed that forany operatorT, when dim(ker(Tk)) = dim(ker(Tk+1)), we have dim(ker(Tk)) = dim(ker(T))for allk. Thus aftern“steps” it must stabilize since thereVisn-dimensional.7.1 #2 (a) The characteristic polynomial isχ(t) =t2-4t+ 4, sot= 2 is the only eigenvalue.A-2I=-11-11has rank 1, so in particular we have an eigenvector (1,1)t. We want tocomptute the chain terminating here; we solve (A-2I)x= (1,1)tto findx= (0,1)t, so wehave a basisβ2={(1,1)t,(0,1)t}which is a cycle generated by (0,1)t. The Jordan normalform (in this basis) is [T]β=2102.(b) The charactersitic polynomial isχ(t)-t2-3t-4. Thus the eigenvalues aret= 4,-1.A-4Ihas kernel (2,3)tandA+Ihas kernel (1,-1)t. Thus we have a basis of eigenvectorsand in this basisβ, we have [T]β=400-1.(c) The characteristic polynomial isχ(t) =-t3+ 3t2-4.Testing its rational roots, wefind that it factors-(t-2)2(t+ 1), so its eigenvalues aret=-1,2. For the eigenvalue 2,we find that (A-2I) =9-4-521-10-113-1-2, which we can row-reduce to10-101-1000andso has rank 2, with corresponding eigenvector (1,1,1)t. We compute the chain terminatinghere: (A-2I)x= (1,1,1)tgivesx= (1,2,0)t. Thus,β2={(1,1,1)t,(1,2,0)t}isA-cyclicgenerated by (1,2,0)t. Next, for the eigenvalue-1, we findA+I=12-4-521-7-113-11, whichrow reduces to3-10001000so is rank 2.We have (A+I)2= 315-5-724-8-136-2-1, whichrow reduces to3-10001000and has rank 2. An eigenvector for this eigenvalue is (1,3,0)t.Thus, we have a basis of generalized eigenvectorsβ={(1,1,1)t,(1,2,0)t,(1,3,0)t}; so wehave [T]β= 221002000-1.(d) The charactersitic polynomial of the matrix is (t-2)2(t-3)2, so the eigenvalues are1
HW 11 Solutions, MATH 110 with Professor Stankovat= 2,3. Fort= 2, we haveA-2I=01000010001001-11. This matrix has rank 3, so we haveeigenvector (1,0,0,0)t
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