Unformatted text preview: MA 36600 LECTURE NOTES: FRIDAY, APRIL 23 3 It is unclear whether we can ﬁnd a matrix T such that T−1 A T is a diagonal matrix because we only have
one eigenvector. Consider instead the matrix
1 −1 0
1
0
1
0
−1
T=
=⇒
T=
=
.
−1 −1
11
−1 −1
−1
Consider the product
1
0 1 −1
1
0
1
0
2
1
21
−1
J = T AT =
=
=
.
−1 −1 1
3 −1 −1
−1 −1 −2 −3
02 We claim that this is the best we can do i.e., there does not exist a matrix T such that T−1 A T = D is a
diagonal matrix. Assume to the contrary, that such a matrix does exist. Then D = 2 I would be a diagonal
matrix with 2’s along the diagonal. Then we have
T−1 A T = D
=⇒
A = T D T−1 = T (2 I) T−1 = 2 T I T−1 = 2 I = D.
This implies that A = D is a diagonal matrix, which is clearly a contradiction. Hence A is not diagonalizable. ...
View
Full
Document
This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue.
 Spring '09
 EdrayGoins

Click to edit the document details