Class Notes (1,100,000)
CA (630,000)
UTSG (50,000)
MAT (4,000)
MAT224H1 (100)
Lecture

# Diagonalization

Department
Mathematics
Course Code
MAT224H1
Professor
Martin, Burda

This 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:VVbe alinear operator. Suppose that its charac-
teristic polynomial factors as pT(x)=±(xλ1)n1·. . . ·(xλk)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 Eλini.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
www.notesolution.com