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Title: Diagonalization of symmetric matrices
Description: Linear algebra course
Description: Linear algebra course
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Diagonalization of
Symmetric
Matrices
8
...
In Chapter 6, we saw that diagonalization of a square matrix may not be possible
if some of the roots of its characteristic polynomial are complex or if the geometric
multiplicity of an eigenvalue is less than the algebraic multiplicity of that eigenvalue
...
Before considering why this works, we give three examples
...
Thus, the resulting diagonal matrix is
D
=
[
-1
+
0
V2
0
]
-1 - V2
EXAMPLE 1
Y2,
For /l1 = -1 +
we have
(continued)
[
1
/
1
A /l =
-1 -
l
A
-1 - Y2,
[
1 +
/
/l2 =
_
1
1
]
[
Y2
-
0
-
o
-
Y2
]
{ [ 1 4}
...
1 -
+
Hence, A 1
...
That is,
[ 1 [ -1 1
1+
·
1
1
= (1 +
Y2)(1 - Y2) + 1(1) = 1 - 2 +
1
= 0
Hence, if we normalized the columns of P, we would find that A is diagonalized by
an orthogonal matrix
...
)
EXAMPLE2
Diagonali'e the symmetric A =
[� : -n
Show that A can be diagonali,ed by an
_
orthogonal matrix
...
For /l1 = 4, we get
[� -�] [� !I
{ [�]}
1
0
A -Ail=
Thus, a basis focthe eigenspace is
For /l2 = 3, we get
-3
-2
0
-
-3
0
[� -�]- [� 1 !]
0
A-A2l =
0
0
-2
-2
-2
0
0
EXAMPLE2
(continued)
{[-:]}·
Thus, a basis for the eigenspace is
For A3 =
-1, we get
{[:]}·
[H [-:l
1 [4 OJ
-2]
[-�
Thus, a basis for the eigenspace is
Observe that the vectors V 1 =
V2 =
·
and Vl =
[:]
fonn an orthogonal set
...
/2
1/
...
/2
1/
...
-1
-4
5
Diagonalize the symmetric A =
-2
an orthogonal matrix
...
Show that A can be diagonalized by
8
-2
-4
5-A
-2
-2
-2
8 -A
2
= -A(A - 9)
The eigenvalues are At = 9 with algebraic multiplicity 2 and A2 = 0 with algebraic
multiplicity
1
...
However, observe
that these vectors are not orthogonal to each other
...
We can do this by applying the Gram-Schmidt Procedure to this set
...
Then St = Span{Vi} and
EXAMPLE3
(continued)
{V1, v2} is an orthogonal basis for the eigenspace of /l1
...
<2 is
Observe that
find that
v3 is orthogonal to v1
0
0
[i13) =
and i12
...
;-fI
...
Definition
A matrix
Orthogonally
matrix
A
is said to be orthogonally diagonalizable if there exists an orthogonal
P and a diagonal matrix D such that
Diagonalizable
Pr AP= D
Remark
Two matrices A and Bare said to be orthogonally similar if there exists an orthogonal
P such that pr AP = B
...
In particular, pr AP = Dis equivalent
to p-1 AP = D for an orthogonal matrix P
...
Lemma 1
An n x n matrix
A is symmetric if and only if
x (Ay) = (Ax) y
·
fo r all x,Y E JR
...
For any
x, y
E
JR!!,
x
...
(Ay) =(Ax)
...
1
...
Hence,
(xTA)7 = Ax or ATx = Ax for all x E lltn
...
1
...
Since
In Examples 1-3, we saw that the basis vectors of distinct eigenspaces are or
thogonal
...
We now
use Lemma 1 to prove that is always true
...
If
v1
and
Proof: By definition of eigenvalues and eigenvectors,
corresponding to distinct
Av1 = tl1v1
and
Av2 = tl2v2
...
Note that this theorem applies only to eigenvectors that correspond to different
eigenvalues
...
Thus, as in Example 3, if an eigenvalue has al
gebraic multiplicity greater than 1, it may be necessary to apply the Gram-Schmidt
Procedure to find an orthogonal basis for its eigenspace
...
The proof of this theorem requires properties of complex numbers and hence is
postponed until Chapter 9
...
5
...
We can now prove that every symmetric matrix is orthogonally diagonalizable
...
Lemma4
tl
...
1
T
=
p AP
Suppose that
[
where
A1 is an (n -
On-I,!
n
1) x (n - 1) symmetric matrix and Om, is the m xn zero matrix
...
, applying the Gram-Schmidt
Procedure, and normalizing the vectors, we can produce an orthonormal basis 13
lV1, W2,
...
n
...
1, so we have
Since 13 is orthonormal, we get
So, all other entries in the first column are 0
...
Moreover, the (n symmetric since pTAP is symmetric
...
Then there exists an orthogonal matrix
P and diagonal matrix D such that pT AP = D
...
Proof: The proof is by induction on n
...
Now suppose the result is true
for (n - 1) x (n - 1) symmetric matrices, and consider an n x n symmetric matrix
A
...
Then, by our hypothesis, there is an
(n - 1) x (n - 1) orthogonal matrix Pi such that
PfA1P1=Di
where D i is an (n - 1) x (n - 1) diagonal matrix
...
Hence, P2 is orthogonal
...
EXERCISE 1
Orthogonally diagonalize A=
EXERCISE2
Orthogonally diagonalize A=
[-� -n
[-�
-1
�
-1
-
]
Oi,11-i
p2
Ai
�]
=2
·
Remarks
1
...
We will see why this definition makes
sense in Section 8
...
2
...
That is, every or
thogonally diagonalizable matrix is symmetric
...
Hence, we can say that a matrix is orthogonally diagonalizable if
and only if it is symmetric
...
1
Practice Problems
Al Determine which of the following matrices are
symmetric
...
(a) A=
-2
-1
0
-1
-
(b) A=
3
1
-1
[ � -71
[5 )
(c) A
=
(d) A=
(e) A=
3
-3
�[ �i
[ � - � = �]
[ ! : -� ]
1
0
1
-2
0
-2
_
Homework Problems
Bl Determine which of the following matrices are
symmetric
...
-2
(g)
(h)
-4
-2
-2
2
2
1
-2
-1
-2
-2
-2
(i)
2
(e)
0
(0 A=l-1� 11 -ii4
2
[
A= � -2 11
[
-1
A=
1
A= H -1-5 -_;1l
Computer Problems
Cl Use a computer to determine the eigenvalues and a
basis of orthonormal eigenvectors of the following
4[
...
9 0
...
95 0
...
0
...
0
...
0
...
0
...
0
...
0
...
0
...
0
...
0
...
(c)
1
...
Deter
mine which of the following is symmetric
...
calculating
the
eigenvalues
S
S
2t
t
-
-2t
2 t
-t
-t
+
-
and S
t
l
t
of
S
l ),
explore how
the eigenvalues of S (t) change as t varies
...
(-0
...
05), (0
...
(b)
[2 +
C2 Let S (t) denote the symmetric matrix
A
A-1
D3 Prove that if
then
is an invertible symmetric matrix,
is orthogonally diagonalizable
Title: Diagonalization of symmetric matrices
Description: Linear algebra course
Description: Linear algebra course