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Solutions Manual Linear Algebra 4th Edition Spence, Stephen Friedberg, Arnold Insel

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Linear Algebra, 4th Edition Lawrence E. Spence, Stephen H. Friedberg, Arnold J. Insel Instructor Solution Manual
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  [ Lawrence E. Spence, Stephen H. Friedberg, Arnold J. Insel Instructor Solution Manual ] Linear Algebra, 4th Edition PDF file ISBN-13: 978-0130084514 To download the Solutions Manual for this book, simply send message with the title and Edition to: skyusbook"@"gmail.com (@ with no quotation marks) This Service is NOT free   You can see sample below ontact ema: syusoo@gma.co  Chapter 7 Canonical Forms 7.1 The Jordan Canonical Form I 1. (a) Yes. It comes directly from the definition.(b) No. If   x  is a generalized eigenvector, we can find the smallest positiveinteger  p  such that  ( T   −  λI  )  p ( x )  =  0. Thus  y  =  ( T   −  λI  )  p − 1 ≠  0 isan eigenvector with respect to the eigenvalue  λ . Hence  λ  must be aneigenvalue.(c) No. To apply the theorems in this section, the characteristic polyno-mial should split. For example, the matrix   0  − 11 0    over  R  has noeigenvalues.(d) Yes. This is a result of Theorem 7.6.(e) No. The identity mapping  I  2  from  C 2 to  C 2 has two cycles for theeigenvalue 1.(f) No. The basis β  i  may not consisting of a union of cycles. For example,the transformation  T  ( a,b )  =  ( a + b,b )  has only one eigenvalue 1. Thegeneralized eigenspace  K  1  = F 2 . If  β   =  {( 1 , 1 ) , ( 1 , − 1 )} , then the matrix representation would be [ T  ] β  = 12   3  − 11 1   , which is not a Jordan form.(g) Yes. Let  α  be the standard basis. Then  [ L J  ] α  =  J   is a Jordan form.(h) Yes. This is Theorem 7.2.240 ontact ema: syusoo@gma.co  2. Compute the characteristic polynomial to find eigenvalues as what we didbefore. For each  λ , find a basis for  K  λ  consisting of a union of disjointcycles by computing bases for the null space of   ( A − λI  )  p for each  p . Writedown the matrix  S   whose columns consist of these cycles of generalizedeigenvectors. Then we will get the Jordan canonical form  J   =  S  − 1 AS  .When the matrix is diagonalizable, the Jordan canonical form should bethe diagonal matrix sismilar to  A . For more detail, please see the examplesin the textbook. On the other hand, these results were computed byWolfram Alpha. For example, the final question need the command below. JordanDecomposition[{{2,1,0,0},{0,2,1,0},{0,0,3,0},{0,1,-1,3}}] (a) S   =   1  − 11 0   ,J   =   2 10 2  . (b) S   =   − 1 21 3  ,J   =   − 1 00 4  . (c) S   =  1 1 13 1 20 1 0  ,J   =  − 1 0 00 2 10 0 2  . (d) S   =  1 0 0 10 1 0 10 0 0 10  − 1 1 0  ,J   =  2 1 0 00 2 0 00 0 3 00 0 0 3  . 3. Pick one basis  β   and write down the matrix representation  [ T  ] β . Thendo the same thing in the previous exercises. Again, we denote the Jordancanonical form by  J   and the matrix consisting of Jordan canonical basisby  S  . The Jordan canonical basis is the set of vector in  V    correspondingthose column vectors of   S   in  F n .(a) Pick  β   to be the standard basis { 1 ,x,x 2 } and get [ T  ] β  =  − 2  − 1 00 2  − 20 0 2  and S   =  1  − 1  14 0 4 00 0  − 2  ,J   =  − 2 0 00 2 10 0 2  . 241 ontact ema: syusoo@gma.co  (b) Pick  β   to be the basis { 1 ,t,t 2 ,e t ,te t } and get [ T  ] β  =  0 1 0 0 00 0 2 0 00 0 0 0 00 0 0 1 10 0 0 0 1  and S   =  1 0 0 0 00 1 0 0 00 0  12  0 00 0 0 1 00 0 0 0 1  ,J   =  0 1 0 0 00 0 1 0 00 0 0 0 00 0 0 1 10 0 0 0 1  . (c) Pick  β   to be the standard basis { 1 00 0  ,  0 10 0  ,  0 01 0  ,  0 00 1 } and get [ T  ] β  =  1 0 1 00 1 0 10 0 1 00 0 0 1  and S   =  1 0 0 00 0 1 00 1 0 00 0 0 1  ,J   =  1 1 0 00 1 0 00 0 1 10 0 0 1  . (d) Pick  β   to be the standard basis { 1 00 0  ,  0 10 0  ,  0 01 0  ,  0 00 1 } and get [ T  ] β  =  3 0 0 00 2 1 00 1 2 00 0 0 3  and S   =  1 0 0 00 0 0 10 1 1 00  − 1 1 0  ,J   =  3 0 0 00 3 0 00 0 3 00 0 0 1  . 242 ontact ema: syusoo@gma.co  4. We may observe that  W   =  span ( γ  )  is  ( T   − λI  ) -invariant by the definitionof a cycle. Thus for all  w  ∈  W  , we have T  ( w )  =  ( T   − λI  )( w ) + λI  ( w )  =  ( T   − λI  )( w ) + λw  ∈  W. 5. If   x  is an element of in two cycles, which is said to be  γ  1  and  γ  2  withoutlose of generality, we may find the smallest ingeter  q   such that ( T   − λI  ) q ( x )  =  0 . This means that the initial eigenvectors of   γ  1  and  γ  2  are both  ( T   − λI  ) q − 1 ( x ) . This is a contradiction. Hence all cycles are disjoint.6. (a) Use the fact that  T  ( x )  =  0 if only if   ( − T  )( x )  =  − 0  =  0.(b) Use the fact that  ( − T  ) k =  ( − 1 ) k T  .(c) It comes from the fact ( λI  V    − T  ) k =  [ − ( T   − λI  V   )] k and the previous argument.7. (a) If   U  k ( x )  =  0, then  U  k + 1 ( x )  =  U  k ( U  ( x ))  =  0.(b) We know  U  m + 1 ( V   )  =  U  m ( U  ( V   ))  ⊂  U  m ( V   ) . With the assumptionrank ( U  m )  =  rank ( U  m + 1 ) , we know that U  m + 1 ( V   )  =  U  m ( V   ) . This means U  ( U  m ( V   ))  =  U  m ( V   ) and so  U  k ( V   )  =  U  m ( V   )  for all integer  k  ≥  m .(c) The assumption rank ( U  m )  =  rank ( U  m + 1 )  impliesnull ( U  m )  =  null ( U  m + 1 ) by Dimension Theorem. This means  N  ( U  m )  =  N  ( U  m + 1 )  by theprevious argument. If   U  m + 2 ( x )  =  0, then  U  ( x )  is an element in N  ( U  m + 1 )  =  N  ( U  m ) . Hence we have  U  m ( U  ( x ))  =  0 and thus  x  is anelement in  N  ( U  m + 1 ) . This means that  N  ( U  m + 2 )  ⊂  N  ( U  m + 1 )  andso they are actually the same. Doing this inductively, we know that N  ( U  m )  =  N  ( U  k )  for all integer  k  ≥  m .(d) By the definition of   K  λ , we know K  λ  =  ∪  p ≥ 1 N  (( T   − λI  )  p ) . But by the previous argument we know that N  (( T   − λI  ) m )  =  N  (( T   − λI  ) k ) for all integer  k  ≥  m  and the set is increasing as  k  increases. Soactually  K  λ  is  N  (( T   − λI  ) m ) .243 ontact ema: syusoo@gma.co
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