Week 5:[reprint, week 4: 5.4 Eigenvalues and Eigenvectors] + diagonaliization1.5.5 Eigenspaces, Diagonalization—————A vectorv= 0 inRn(or inCn)is aneigenvector with eigenvalueλofann-by-nmatrixAifAv=λv.We re-write the vector equation as (A-λIn)v= 0,which is a homogeneous system with coefmatrix (A-λI),and we wantλso that the system has a non-trivial solution.We see that the eigenvalues are the rootsof thecharacteristic polynomialP(λ) = 0,whereP(λ) = det(A-λI).To find the eigenvectors we find the distinctrootsλ=λi,and for eachisolve (A-λiI)v= 0.—————We collect the terminology and results used for diagonalzation. IfP(λ) writtenin factored form isP(λ) = (λ-λ1)m1. . .(λ-λr)mr,where theλjare the distinct roots, we say thatλjhas (algebraic) multiplicitymj,or thatλjis a simple root (multiplicity 1) ifmj= 1.Letdj= dim(Eλj) be the dimension of the eigenspace (sometimes called
2the geometric multiplicity), then the main facts are1. 1≤dj≤mj;2.Ais non-diagonalizable exactly when there is some eigenvalue (whichmust be a repeated root) that is defective,dj< mj;3.Ais diagonalizable exactly when every eigenvalue is non-defective,dj=mj,forj= 1, . . . , r;4. In particular, ifP(λ) has distinct roots thenAis always diagonalizable(1≤dj≤mj= 1).We recall thatmj- the algebraic multiplicity - is the number of timesthejth distinct eigenvalue is a root of the characteristic polynomial;anddj- the geometric multiplicity - is the dimension of theλjth eigenspace (that is, the nullspace ofA-λjI).We will see that ann-by-nmatrixAisdiagonalizableif and only ifAhasnlinearly independent eigenvectors.First, we have a new definition, matricesAandBaresimilarif there is ann-by-nmatrixPso thatPhas an inverse andB=P-1AP.We also recall that adiagonal matrixD=diag(d1, . . . , dn)is a square matrix with all entries 0 except (possibly) on themain diagonal, where the entries ared1, . . . , dn(in that order).