Solutions for Quiz on March 29
We need to diagonalize the matrix,
A
=
1
0
0

1
0
1
1
1
0
.
First we find the eigenvalues. We need to solve
fl
fl
fl
fl
1

λ
0
0

1
0

λ
1
1
1
0

λ
fl
fl
fl
fl
= 0
.
We expand the determinant in cofactors of the first row. This row has only
one nonzero element, so we have
(1

λ
)(

1)
1+1
M
13
= (1

λ
)
fl
fl
fl
fl

λ
1
1

λ
fl
fl
fl
fl
= 0
.
The minor is given by
λ
2

1, so the cherecteristic equation becomes
(1

λ
)(
λ
2

1) = 0
.
This has roots
λ
1
= 1,
λ
2
= 1 and
λ
3
=

1.
Now we need to find the
eigenvectors. We start from the eigenvector corresponding to
λ
3
. We have,
2
0
0

1
1
1
1
1
1
x
1
x
2
x
3
= 0
.
From the first equation, we have 2
x
1
= 0, which means
x
1
= 0. From the
second and third equations, we have, 0+
x
2
+
x
3
= 0, which means
x
2
=

x
3
.
Therefore, the corresponding solution is
α
0

1
1
.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Remark.
This equation can also be solved by row reduction, if you like.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Winter '07
 nataliakomarova
 Linear Algebra, Determinant, Factors, independent variables, linearly independent eigenvectors

Click to edit the document details