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Eigenvalues and
Eigenvectors
6
...
Eigenvector
that L(V)
Eigenvalue
The pair A
...
=
A non-zero vector v E
JR11 such
A
...
is called an eigenvalue of L
...
The pairing of eigenvalues and eigenvectors is not one-to-one
...
2
...
In Chapter 9 we will consider the case where we allow eigenvalues and
eigenvectors to be complex
...
It is
natural because L(O)
esting to consider
=
0 for every
linear transformation, so it is completely uninter
0 as an eigenvector
...
In particular, we will
see that we often want to look for a basis of JR11 that contains eigenvectors of a linear
transformation
...
Determine which of the following vectors are eigenvectors of Land give
the corresponding eigenvalues:
Solution:
So,
[]
To test whether
[ �]
[
!
]
[
rn =;J
...
i, so
[ � ] 2[=�]
=
[�]
17
-15
r
r
[ J J J rJ
l
20
-18
=
3
-28
l
-34 f
...
[17 -15] [3] [ -9]
L(3 4)=
...
L(l ,3) =
[!]
1)
is also an eigenvector of L with eigenvalue 2
...
Since
So,
...
Eigenvectors and Eigenvalues of Projections and Reflections in JR3
1
...
If vis orthogonal to it, then proj,lV) =
0
= Ov, so vis an eigen
vector of proji1 with corresponding eigenvalue 0
...
For an arbitrary vector a
is a multiple of rt, so that a is definitely
a multiple of it or orthogonal to it
...
On the other hand, perp,z(it)
(continued)
eigenvalue 0
...
3
...
For
vorthogonal to ii, refl11(V)
=
1 v
...
EXAMPLE3
Eigenvectors and Eigenvalues of Rotations in JR2
...
der the rotation Re
integer multiple of
such that Re(V)
=
1r
...
cose
...
sme
[
cose
l
...
This linear transformation has no real
eigenvalues or real eigenvectors
...
[
EXERCISE 1
Let Re : IR3
cose
--
- sine
IR3 denote the rotation in IR3 with matrix sine
cose
0
0
H
Derermine
any real eigenvectors of Re and the corresponding eigenvalues
...
However, in many applications of these ideas, it
is a matrix A that is given
...
Definition
Suppose that A is an n x n matrix
...
In Example 1, we saw that
/l is called
[ �] [!]
and
an eigenvalue of
A
...
Finding Eigenvectors and Eigenvalues
If eigenvectors and eigenvalues are going to be of any use, we need a systematic
method for finding them
...
A
is given; then a non-zero
This condition can be rewritten
Av-M=o
(A
- ,i)v
=
0,
but this would be incorrect because A is
a matrix and /l is a number, so their difference is not defined
...
Then the
eigenvector condition can be rewritten
(A - ,U)v=
0
The eigenvector vis thus any non-trivial solution (since it cannot be the zero vector)
of the homogeneous system of linear equations with coefficient matrix (A -tl/)
...
Hence, for tl to be an eigenvalue, we must have det(A-tl/) = 0
...
Theorem 1
Suppose that A is an
n x n matrix
...
Observe that the set of all eigenvectors corresponding to an eigenvalue tl is just
the nullspace of A -tl/, excluding the zero vector
...
We make the
following definition
...
n x n matrix
A
...
Hence, the dimension of the eigenspace
must be at least 1
...
17-tl
-15
20
-18 -,{
]
(You should set up your calculations like this: you will need A - tll later when you
...
) Then
det(A -tl/)
=
1
17-,{
20
-15
_
18
_
tl
I
= (17 - tl)(-18 - tl) - (-15)20
= tl
...
These are all of the eigenvalues of A
...
3
[ i4]
-3
3
{[ i4]}
...
W riting A - ill and row reducing gives
A
_
so that the general solution of (A-ill)v
of A corresponding to il
eigenspace for il
=
-3
=
-15
-15
20
20
=
are v
is v
=
=
t E R Thus, all eigenvectors
t
=
for any non-zero value oft, and the
is Span
A_ 21
The general solution of (A - ill)V
{[ � ]}
0
t
We repeat the process for the eigenvalue il
Span
1
�
=
=
[
15
20
0 is v
- 15
-20
t
=
=
2:
] [
1
�
0
-1
0
]
[ �],
t E JR, so the eigenspace for il
...
This moti
vates the following definition
...
Then C(il)
Characteristic Polynomial
polynomial