Department

MathematicsCourse Code

MAT224H1Professor

Martin, BurdaThis

**preview**shows half of the first page. to view the full**3 pages of the document.**Lecture 8notes byY. Burda

1Diagonalization of operators, continued

Last time weshowed that eigenvectors corresponding to diﬀerenteigenvalues

are linearly independent. Now wewill use it to get acriterion for when an

operator is diagonalizable.

Theorem. Let T:V→Vbe alinear operator. Suppose that its charac-

teristic polynomial factors as pT(x)=±(x−λ1)n1·. . . ·(x−λk)nk.Then

it is diagonalizable if and only if dim Eλi=nifor every i,i.e. geometric

multiplicity coincides with algebraic multiplicity for alleigenvalues.

Proof. For one direction wesuppose that the operator Tis diagonalizable

and try to provethat dim Eλi=ni.Since Tis diagonalizable, there is a

basis v1,...,vnin whichits matrix is diagonal:

∗0··· 0

0∗....

.

.

.

.

.......0

0. . . 0∗

Since weknowthe characteristic polynomial ofthis matrix, weget that n1

of the entries on the diagonal are equalto λ1,n2of the entries on the diagonal

are equal to λ2and so on. But this also means that n1of the vectors v·in the

basis are eigenvectors with eigenvalue λ1,n2of them are eigenvectors with

eigenvalue λ2and so on. Thus dim Eλi≥ni.Since wealready provedthe

other inequalityin the previous lecture, weget dim Eλi=ni.

For the other direction weassume that dim Eλi=nifor all iand try to

ﬁnd abasis of eigenvectors for T.

For eachilet vi

1,...,vi

nibeabasis for Eλi.Weclaim that if wetakeall

these vectors together (for all i), they will form abasis for V.Indeed, there

are n1+n2+. . . +nk=nvectors in this set, so all wehaveto checkis that

they are linearly independent.

Assume then that alinear combination of them is equal to zero:

c1

1v1

1+. . . +c1

n1v1

n1+

+c2

1v2

1+. . . +c2

n2v2

n2+

+..................+

+ck

1vk

1+. . . +ck

nkvk

nk=0

1

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