Chapter 7 Canonical Forms - Duke University

1 Chapter 7 Canonical Eigenvalues and EigenvectorsDefinition a vector space over the fieldFand letTbe a linearoperator onV. AneigenvalueofTis a scalar Fsuch that there exists a non-zero vectorv VwithTv= v. Any vectorvsuch thatTv= vis called aneigenvectorofTassociated with the eigenvalue value .Definition (T)of a linear operatorT:V Vis the set ofall scalars such that the operator(T I)is not `2be the Hilbert space of infinite square-summable se-quences andT:V Vbe the right-shift operator defined byT(v1,v2,..) = (0,v1,v2,..).SinceTis not invertible, it follows that the scalar0is in the spectrum ofT. But, itis not an eigenvalue becauseTv= 0impliesv= 0and an eigenvector must be anon-zero vector. In fact, this operator does not have any finite-dimensional spaces, things are quite a bit the matrix representation of a linear operator on a finite-dimensional vector spaceV, and let be a scalar.

2 The following are equivalent:1. is an eigenvalue ofA2. the operator(A I)is singular113114 Chapter 7. Canonical (A I) = , we show the first and third are equivalent. If is an eigenvalue ofA,then there exists a vectorv Vsuch thatAv= v. Therefore,(A I)v= 0and(A I)is singular. Likewise, if(A I)v= 0for somev Vand F, thenAv= v. To show the second and third are equivalent, we note that the determinantof a matrix is zero iff it is last criterion is important. It implies that every eigenvalue is a root of thepolynomial A( ),det( I A)called thecharacteristic polynomialofA. The equationdet(A I) = 0is calledthe characteristic equation ofA. The spectrum (A)is given by the roots of thecharacteristic polynomial A( ).LetAbe a matrix over the field of real or complex numbers.

3 A nonzero vectorvis called aright eigenvectorfor the eigenvalue ifAv= v. It is called alefteigenvectorifvHA= be an eigenvalue of the matrixA. Theeigenspaceassociatedwith is the setE ={v V|Av= v}. Thealgebraic multiplicityof isthe multiplicity of the zero att= in the characteristic polynomial A(t). Thegeometric multiplicityof an eigenvalue is equal to dimension of the eigenspaceE or nullity(A tI).Theorem the eigenvalues of ann nmatrix are all distinct, then theeigenvectors ofAare linearly will prove the slightly stronger statement: if 1, 2,.., kare distincteigenvalues with eigenvectorsv1,v2,..,vk, then the eigenvectors are linearly in-dependent. Suppose thatk i=1civi= 0for scalarsc1,c2,..,ck. Notice that one can annihilatevjfrom this equation bymultiplying both sides by(A jI).

4 So, multiplying both sides by a product EIGENVALUES AND EIGENVECTORS115these matrices givesk j=1,j6=m(A jI)k i=1cjvi=(k j=1,j6=m(A jI))cmvm=cmk j=1,j6=m( m j) = all eigenvalues are distinct, we must conclude thatcm= 0. Since the choiceofmwas arbitrary, it follows thatc1,c2,..,ckare all zero. Therefore, the vectorsv1,v2,..,vkare linearly a linear operator on a finite-dimensional vector spaceV. The operatorTisdiagonalizableif there exists a basisBforVsuch that eachbasis vector is an eigenvector ofT,[T]B= 10 00 2 n Similarly, a matrixAis diagonalizable if there exists an invertible matrixSsuchthatA=S S 1where is a diagonal ann nmatrix hasnlinearly independent eigenvectors, then itis that then nmatrixAhasnlinearly independent eigenvectors,which we denote byv1.

5 ,vn. Let the eigenvalue ofvibe denoted by iso thatAvj= jvj, j= 1,.., matrix form, we haveA[v1 vn]=[Av1 Avn]=[ 1v1 nvn].116 Chapter 7. Canonical FORMSWe can rewrite the last matrix on the right as[ 1v1 nvn]=[v1 vn] 1 n =S .whereS=[v1 vn]and = 1 n ,Combining these two equations, we obtain the equalityAS=S .Since the eigenvectors are linearly independent, the matrixSis full rank and henceinvertible. We can therefore writeA=S S 1 =S is, the matrixAis type of the transformation fromAto arises in a variety of there exists an invertible matrixTsuch thatA=TBT 1,then matricesAandBare said to similar, then they have the same eigenvalues. Similar matricescan be considered representations of the same linear operator using different ann nHermitian matrix ( ,AH=A).

6 Then, theeigenvalues ofAare real and the eigenvectors association with distinct eigenvaluesare , we notice thatA=AHimpliesvHAvis real becauses=(vHAv)H=vHAHv=vHAv= APPLICATIONS OF EIGENVALUES117 IfAv= 1v, left multiplication byvHshows thatvHAv= 1vHv= 1 v .Therefore, 1is real. Next, assume thatAw= 2wand 26= 1. Then, we have 1 2wHv=wHAHAv=wA2v= also assume, without loss of generality, that 16= 0. Therefore, if 26= 1, thenwHv= 0and the eigenvectors are Applications of Differential EquationsIt is well known that the solution of the 1st-order linear differential equationddtx(t) =ax(t)is given byx(t) =eatx(0).It turns out that this formula can be extended to coupled differential a diagonalizable matrix and consider the the set of 1st order linear differ-ential equations defined byddtx(t) =Ax(t).

7 Using the decompositionA=S S 1and the substitutionx(t) =Sy(t), we findthatddtx(t) =ddtSy(t)=Sddty(t).andddtx(t) =Ax(t)=ASy(t).118 Chapter 7. Canonical FORMSThis implies thatddty(t) =S 1 ASy(t) = y(t).Solving each individual equation givesyj(t) =e jtyj(0)and we can group them together in matrix form withy(t) =e ty(0).In terms ofx(t), this givesx(t) =Se tS 1x(0).In the next section, we will see this is equal tox(t) =eAtx(0). Functions of a MatrixThe diagonal form of a diagonalizable matrix can be used in a number of ap-plications. One such application is the computation of matrix exponentials. IfA=S S 1thenA2=S S 1S S 1=S 2S 1and, more generally,An=S nS that nis obtained in a straightforward manner as n= n1 nn .This observation drastically simplifies the computation of the matrix exponentialeA,eA= i=0 Aii!

8 =S( i=0 ii!)S 1=Se S 1,wheree = e 1 e n . THE JORDAN FORM119 Theorem ( )be a given polynomial. If is an eigenvalue ofA, whilevis an associated eigenvector, thenp( )is an eigenvalue of the matrixp(A)andvisan eigenvector ofp(A)associated withp( ). (A)v. Then,p(A)v=l k=0pkAkv=l k=0pk kv=p( ) isp(A)v=p( ) matrixAis singular if and only if0is an eigenvalue The Jordan FormNot all matrices are diagonalizable. In particular, ifAhas an eigenvalue whosealgebraic multiplicity is larger than its geometric multiplicity, then that eigenvalueis calleddefective. A matrix with a defective eigenvalue is not ann nmatrix. ThenAis diagonalizable if and only ifthere is a set ofnlinearly independent vectors, each of which is an eigenvector independent eigenvectorsv1.

9 ,vn, then letSbe aninvertible matrix whose columns are therenvectors. ConsiderS 1AS=S 1[Av1 Avn]=S 1[ 1v1 nvn]=S 1S = .Conversely, suppose that there is a similarity matrixSsuch thatS 1AS= is adiagonal matrix. ThenAS=S . This implies thatAtimes theith column ofSis theith diagonal entry of times theith column ofS. That is, theith columnofSis an eigenvector ofAassociated with theith diagonal entry of . SinceSisnonsingular, there are exactlynlinearly independent 7. Canonical FORMSD efinition normal formof any matrixA Cn nwithl nlinearly independent eigenvectors can be written asA=TJT 1,whereTis an invertible matrix andJis the block-diagonal matrixJ= Jm1( 1) Jml( l) .TheJm( )arem mmatrices called Jordan blocks, and they have the formJm( ) = 1 0 00 1 0 0.

10 It is important to note that the eigenvalues 1,.., lare not necessarily distinct( , multiple Jordan blocks may have the same eigenvalue). The Jordan matrixJassociated with any matrixAis unique up to the order of the Jordan , two matrices are similar iff they are both similar to the same every matrix is similar to a Jordan block matrix, one can gain some in-sight by studying Jordan blocks. In fact, Jordan blocks exemplify the way thatmatrices can be degenerate. For example,Jm( )has the single eigenvectore1( ,the standard basis vector) and satisfiesJm(0)ej+1=ejforj= 1,2,..,m , the reason this matrix has only one eigenvector is that left-multiplication by thismatrix shifts all elements in a vector up the Jordan normal form of a matrix can be broken into two , one can identify, for each distinct eigenvalue , thegeneralized eigenspaceG ={v Cn (A I)nv= 0}.