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Linear Algebra
Eigenvalues, Eigenvectors
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Eigenvalues and Eigenvectors
If A is an n x n matrix and x is a vector in , then
there is usually no general geometric relationship
between the vector x and the vector Ax. However,
there are often certain nonzero vectors x such that x
and Ax are scalar multiples of one another. Suchvectors arise naturally in the study of vibrations,
electrical systems, chemical reactions, quantum
mechanics, mechanical stress, economics and
geometry.
nR
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Eigenvalues, Eigenvectors
If A is an n n, then a non zero vector xin Rnis called an eigenvectorof Aif Axis a scalarmultiple of x; that is
for some scalar . The scalar is called aneigenvalue of A, and x is called theeigenvectorof A corresponding to .
Eigen values are also called the propervalues or characteristic values they arealso called the Latent roots.
Ax x
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Graphical Interpretation
Eigen values and Eigen vectors have also a useful
graphical interpretation in R2 and R3. If is the
eigenvalue of A corresponding to x then Ax=x so
that the multiplication by A dilates x, contracts x, or
reverses the direction of x, depending upon the value.
Dilation(>1) Contraction(0<
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Finding Eigen values
To find the eigenvalues of nn matrix A we
write Ax=x as
For to be the eigenvalue ,there must be a
nonzero solution of the above equation.
However above equation will have a nonzerosolution if and only if
( ) 0
Ax x
I A x
det( ) 0I A
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Characteristic Equation
det(I-A)=0 is called the characteristic
equation of A; the scalars satisfying this
equation are called the eigenvalues of A.
When expanded , the determinant det(I-A) isa polynomial in called the characteristic
polynomialin A.
Example: Find the eigenvalues and
eigenvectors of the matrix
3 2
1 0A
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Example 1(Eigenvalues)
Solution:
The characteristic polynomial in A.
Characteristic Equation:
1 0 3 2
0 1 1 0
3 2
1
I A
I A
23 2det( ) 3 21
I A
2
3 2 0
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Example 1(Eigenvectors)
The solution of this equation are =1 and =2;
these are the eigenvalues of A.
Since Ax=x
Eigenvector corresponding to =1
Putting =1 gives
1
2
3 20
1
x
x
1
2
2 20
1 1
x
x
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Example 1(Eigenvectors)
The above matrix gives two equations
which gives the general solution in the form
1 2x x
1 2
1 2
2 2 0
0
x x
x x
2
1x 1 1 xLet
1
Then
The eigenvector corresponding to is
1
1
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Example 1(Eigenvectors)
Similarly 12
1
2
1 2
1 2
1 2
2 1
3 20
1
Putting 2
1 2
0givesequations1 2
2 0
2 0
Thesolution of aboveequations is
2
1 2
Eigenvector correspponding to 2is
2
1
x
x
x
x
x x
x x
x x
let x then x
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Multiple Eigenvalues
The order Mof an eigenvalue as a root of
characteristic polynomial is called the
algebraic multiplicityof .
The number m of linearly independenteigenvectors corresponding to is called the
geometric multiplicityof .
Thus m is the dimension of the eigenspace
corresponding to this .
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Multiple Eigenvalues
The characteristic polynomial has a degree
n, the sum of all algebraic multiplicity must
equal n.
In general m M The defectof is defined as
Example :Find the eigenvalues and
eigenvectors of the matrix
M m
2 2 3
2 1 6
1 2 0
A
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Example 2(1/7)
Eigenvalues
The characteristic polynomial
have three latent root (eigenvalues) as
2 2 3
2 1 6
1 2
det( ) 0
A I
A I
3 2 21 45 0
1 2 35, 3
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Example 2(2/7)
The algebraic multiplicity of =-3 is 2
M-3=2
and that of =5 is M5=1
Eigenvectors: To find the eigenvectors apply
the gauss elimination to the system (A-I)x=0
first =5 and =-3.
Putting =5 gives1
2
3
7 2 3
( ) 2 4 6 0
1 2 5
x
A I x x
x
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Example 2(3/7)
Performing Gauss elimination on the matrix
above gives1 2 5
0 1 2
0 0 0
A I
1 2 3
2 3
2 5 0
2 0
x x x
x x
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Example 2(4/7)
The above matrix can be written in the
equation form which gives
1 2 3
2 3
3 2 1
1
(2 5 )
2
Let 1 then 2 and 1
The eigen vector corresponding to 5
1
2
1
x x x
x x
x x x
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Example 2(5/7)
Putting =-3 and performing Gauss
Elimination gives
1 2 3
3 0 0 0
0 0 0
A I
1 2 3
1 2 3
2 3 0
2 3
x x x
or
x x x
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Example 2(6/7)
Or
So there are two linearly independent vectors
corresponding to
1
0
3
0
1
2
32
3
2
1
xx
x
x
x
3
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Example 2(7/7)
Since there are two Linear Independent
vectors corresponding to =-3 the geometric
multiplicity of
m-3=2m5=1
The defect of =-3 is
-3 =M-3m-3-3=0
5=0
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Example:
Find the Eigenvalues and Eigenvectors of the matrix
Solution:
The algebraic multiplicity is
00
10A
0
1
10
01
00
10IA
001
det 2
IA
021
20 MM
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Example (Contd)
To find Eigenvector
The solution is
The Eigenvector is
The geometric multiplicity is
Thus
0
0
00
10
2
1
x
x
12 .0xx
0
1
12
1
xx
x
10 mm
00 Mm
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Theorems
If A is an nn triangular matrix (upper
triangular, lower triangular or diagonal) then
the eigenvalues of A are entries on main
diagonal of A. Example :
1 2 3
1 0 0
2
21 0
3
1 5 8
4
TheEigenvaluesare
1 2 1
, ,2 3 4
A
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Theorems
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Theorems
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Example
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Theorems
If k is a positive integer, is an eigenvalue of
a matrix A, and x is a corresponding
eigenvector, then k is an eigenvalue of Ak
and x is a corresponding eigenvector.
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Example:
Eigenvalues of are:
1282,11 7327
1
1 2 3
0 0 2
If 1 2 1
1 0 3then find Eigenvalues for if eigenvalues
for are 1, 2
A
7A
A
7A
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Theorems
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Theorems
If x is an eigenvector of a matrix A to an
eigenvalue ,so is kxwith any k 0.
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Example
Let
The eigenvalues of A are 3, 0, and 2. Theeigenvalues of B are 4 and 1.
What does it mean for a matrix to have an eigenvalueof 0 ?
This happens if and only if the equation has
a nontrivial solution. Equivalently, Ax=0 has anontrivial soltion if and only if A is not invertible. Thus
0 is an eigenvalue of A if and only if A is notinvertible.
435
012
004
and
200
600
863
BA
0xxA
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Diagonalization
The Eigenvector Problem:
Given an matrix A, does there exist abasis for consisting of eigenvectors of A?
The Diagonalization Problem:Given an matrix A, does there exist aninvertible matrix P such that isa diagonal matrix?
Apparently above two problems are different butthey are equivalent thats why diagonalizationproblem is considered in the discussion ofeigenvectors.
n nnR
n n1D P AP
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Diagonalizable Matrix
Definition:
A square matrix A is said to bediagonalizable if there exist an invertiblematrix P such that is a diagonalmatrix.
Theorem:
If A is an matrix then the following are
equivalent: A is diagonalizable
A has n linearly independent eigenvectors.
1D P AP
n n
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Procedure for Diagonalization
1. Find eigenvalues of A.
2. Find n linearly independent eigenvectors
of A, say,
3. Form matrix P having asits columns.
4. Construct . The matrix D will be
diagonal with as itssuccessive diagonal entries, where is the
eigenvalue corresponding to .
1 2 3 4, , , ,..., .np p p p p
1 2 3 4, , , ,..., np p p p p
1D P AP
1 2 3 4, , , ,...,
n
i
, 1, 2,...,ip i n
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Example 1 (1/10)
Find a matrix P that diagonalizes
0 0 2
1 2 1
1 0 3
A
Step 1
Find the eigenvalues of A
Solution:
1 0 0 0 0 2
0 1 0 1 2 1
0 0 1 1 0 3
I A
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0 0 0 0 2
0 0 1 2 1
0 0 1 0 3
0 2 1 2 1
1 0 3
0 2
det( ) 1 2 1
1 0 3
I A
Example 1 (2/10)
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The characteristic equation of A is3 25 8 4 0
In factored form
2
1 2 0 Verify!
Thus, the eigenvalues of A are
1 2 31, 2
Example 1 (3/10)
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Step 2
Find the corresponding eigenvectors
By definition1
2
3
x
x
x
x
is an eigenvector of A corresponding to if and
only if x is a non-trivial solution of
that is
0I A x
1
2
3
0 2 0
1 2 1 0
1 0 3 0
x
x
x
Example 1 (4/10)
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If , then21
2
3
2 0 2 0
1 0 1 0
1 0 1 0
x
x
x
Solving this system yields
1 3 x x x2,,x3 (free variables)
The eigenvectors of A corresponding to
are nonzero vectors of the form 2
0 1 0
0 0 1
0 1 0
s s
t t s t
s s
x
Example 1 (5/10)
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As
are linearly independent, so they form the basisfor eigenspace corresponding to 2
1 0
0 , 1
1 0
If , then1
1
2
3
1 0 2 0
1 1 1 0
1 0 2 0
x
x
x
Example 1 (6/10)
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Solving this system yields
1 32 x x x3 (free variable)
The eigenvectors corresponding to are
nonzero vectors of the form
1
2 2
1
1
s
s s
s
x So2
1
1
2 3x x
=>
1 2 3
1 0 2
0 , = 1 , 1
1 0 1
p p p
Example 1 (7/10)
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Step 3 It is easy to verify that are
linearly independent so that
diagonalizes A.
Step 4 The resulting diagonal matrix is
1 2 3{ , , }p p p
1 0 20 1 1
1 0 1
P
1
2 0 0
0 2 0
0 0 1
D P AP
Example 1 (8/10)
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To verify that
1
2 0 0
0 2 0
0 0 1
A P P
We can compute
PA DP Verify it for the given case
Example 1 (9/10)
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Remark:
There is no preferred order for the columns of
P. Since the ithdiagonal entry of is an
eigenvalue corresponding to ith
column vectorof P, changing the order of columns of P will
change the order of eigenvalues on the
diagonal of .
1P AP
1P AP
1 2 0
0 1 1
1 1 0
P
For example
1
2 0 0
0 1 0
0 0 2
D P AP=>
Example 1 (10/10)
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Example 2 (1/3)
Diagonalize
The characteristic equation of A is
Thus, is the only eigenvalue of A, and
corresponding eigenvectors are the solutions of
3 2
2 1
A
1
( ) 0 I A x
2( 1) 0.
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Example (3/3)
Since A does not have two linearly
independent eigenvectors i.e. the eigenspace
is 1-dimensional, therefore, A is not
diagoalizable.
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Conditions for Diagonalizability
Theorem 1:
If are eigenvectors of A
corresponding to distinct eigenvalues
,thenare linearly independent.
Theorem 2:
If an matrix A has ndistinct eigenvalues,
then A is diagonalizable.
1 2 3 4, , , ,..., nv v v v v
n n
1 2 3 4, , , ,..., n 1 2 3, , ,..., nv v v v
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Example
Whether A is
diagonalizable or not?0 1 0
0 0 1
4 17 8
A
The eigenvalues of A are
1 2 34, 2 3, 2 3
Since A has three distinct eigenvalues, so A can
be diagonalized. We can verify that
1
4 0 0
0 2 3 0
0 0 2 3
D P AP
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LU-Factorization
With Gaussian elimination and Gauss-Jordan
elimination, a linear system is solved by
operating systematically on the augmented
matrix.Another approach is based on factoring the
coefficient matrix into a product of lower (L)
and upper (U) triangular matrix (LU-
Decomposition). LU-decomposition speeds up the solution of
Ax b
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S l i Li S t b
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Step 3
Use (2) to rewrite (1) as and solve for y
Step 4
Substitute yin (2) and solve for x
L y = b
x b
y
Multiplication by A
Solving Linear Systems by
LU-Factorization
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LU Factorization
The LU factorization is motivated by the fairlycommon industrial and business problem of solving asequence of equations, all with the same coefficientmatrix A:
(1)
When A is invertible, one could compute and thencompute and so on. However, it is moreefficient to solve the first equation in (1) by row
reduction and obtain an LU factorization of A at thesame time. Thereafter, the remaining equations in (1)are solved with LU factorization.
n21 bx,,bx,bx AAA 1A
,, 21
1
1 bAbA
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E l S l i Li
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1
2
3
2 0 0 1 3 1 2
3 1 0 0 1 3 24 3 7 0 0 1 3
x
x
x
Step 1
Rewriting the system as
Step 2
Defining vector y as y=[y1, y2, y3]T
1 1
2 2
3 3
1 3 1
0 1 3
0 0 1
x y
x y
x y
ExampleSolving Linear
Systems by LU-Factorization
E ample Sol ing Linear
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Equivalently1
1 2
1 2 3
2 2
3 2
4 3 7 3
y
y y
y y y
Solving using forward substitution, we get
1 2 31, 5, 2y y y
Step 3
1
2
3
1 3 1 1
0 1 3 5
0 0 1 2
x
x
x
ExampleSolving Linear
Systems by LU-Factorization
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LU decomposition
Theorem:
If A is a square matrix that can be reduced to
a row-echelon form U without using row
interchanges, then A can be factored asA = LU, where L is a lower triangular matrix.
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Procedure for LU decomposition
Step 1 : Reduce A to a row-echelon form U without
row interchanges, keeping track of the multipliers
used to introduce the leading 1sand the multipliers
used to introduce the zeros below the leading 1s.
Step 2: In each position along the main diagonal ofL, place the reciprocal of the multiplier that introduced
the leading 1 in that position in U.
Step 3: In each position below the main diagonal of
L, place the negative of the multiplier used tointroduce the zero in that position in U.
Step 4: Form the decomposition A=L U
LU Decomposition
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LU Decomposition
Example
Find an LU-decomposition of
Solution: We begin by reducing A to row-echelon form,keeping track of all multipliers.
573
119
026
A
LU Decomposition
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LU Decomposition
Example
573
119
026
573
119
0311 61Multiplier
580
120
0311
3Multiplier
9Multiplier
5802
110
03
11
2
1Multiplier
1002
110
0311
8Multiplier
1002
110
0311
1Multiplier
LU Decomposition
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LU Decomposition
Example
Constructing L from the multipliers yields the LU-
decomposition
1002
1100311
183
029006
LUA
Iterative Solution of Linear
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Iterative Solution of Linear
Systems
Although Gaussian elimination or Gauss-
Jordan elimination is generally the method of
choice for solving a linear system of n
equations in n unknowns, there are otherapproaches to solving linear systems, called
iterativeor indirect methods, that are better in
certain situations.
The Gauss-Seidel method is the mostcommonly used iterative method.
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Gauss-Seidel Method
Basic Procedure:
Algebraically solve each linear equation for xi
Assume an initial guess
Solve for each xiand repeat
Use absolute relative approximate error after each
iteration to check if error is within a prespecified
tolerance.
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A set of nequations in nunknowns:
11313212111 ... bxaxaxaxa nn
22323222121 bxa...xaxaxa nn
nnnnnnn
bxaxaxaxa ...332211
. .
. .
. .
Gauss-Seidel Method (Algorithm)
The system Ax=b is reshaped by solving the firstequation for x1, the second equation for x2, and thethird forx3, and n
thequation forxn.
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Rewriting each equation
11
131321211
a
xaxaxabx nn
nn
nn,nnnn
n
,nn
n,nnn,nn,n,nn
n
nn
a
xaxaxabx
a
xaxaxaxab
x
axaxaxabx
112211
11
12212211111
1
22
232312122
Gauss-Seidel Method (Algorithm)
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Solve for the unknowns
Assume an initial guess forx
n
-n
2
x
x
x
x
1
1
Use rewritten equations to
solve for each value of xi.
Important:Remember to
use the most recent value
of xi. Which means to
apply values calculated to
the calculations remainingin the current iteration.
Gauss-Seidel Method (Algorithm)
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Calculate the Absolute Relative Approximate Error
100new old
i i
x newi
i
x x
x
So when has the answer been found?
The iterations are stopped when the absolute relative
approximate error is less than a pre-specified tolerance for all
unknowns.
Gauss-Seidel Method (Algorithm)
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Gauss-Seidel Method: Example 1
Given the system of equations
15312 321 x-xx
2835 321 xxx
761373 321 xxx
1
0
1
3
2
1
x
x
x
With an initial guess of
The coefficient matrix is:
12 3 5
1 5 33 7 13
A
Will the solution convergeusing the Gauss-Seidel
method?
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Gauss-Seidel Method: Example 1
12 3 5
1 5 3
3 7 13
A
Checking if the coefficient matrix is diagonally dominant
43155232122
aaa
10731313 323133 aaa
8531212 131211 aaa
The inequalities are all true and hence the matrix A is
strictlydiagonally dominant.
Therefore: The solution should converge using the Gauss-
Seidel Method
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Gauss-Seidel Method: Example 1
1
2
3
12 3 5 1
1 5 3 28
3 7 13 76
x
x
x
Rewriting each equation
12
531 321
xxx
5328 312 xxx
13
7376 213
xxx
With an initial guess of
1
0
1
3
2
1
x
x
x
50000.0
12
150311
x
9000.4
5
135.0282
x
0923.3
13
9000.4750000.03763
x
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Gauss-Seidel Method: Example 1
8118.3
7153.314679.0
3
2
1
x
xx
After Iteration #1
14679.0
12
0923.359000.4311
x
7153.35
0923.3314679.0282
x
8118.3
13
900.4714679.03763
x
Substituting the x values into the equations
After Iteration #2
0923.3
9000.4
5000.0
3
2
1
x
x
x
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Gauss-Seidel Method: Example 1
Iteration #2 absolute relative approximate error
x 1
0.14679 0.50000100 240.62%
0.14679
x 2 3.7153 4.9000 100 31.887%3.7153
x 3
3.8118 3.0923100 18.876%
3.8118
The maximum absolute relative error after the first iterationis 240.62%
This is much larger than the maximum absolute relative
error obtained in iteration #1. Is this a problem?
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Gauss-Seidel Method: Example 1
Repeating more iterations, the following values are obtained
1x
2x
3x
Iteration x1 x2 x3
1
23
4
5
6
0.50000
0.146790.74275
0.94675
0.99177
0.99919
67.662
240.6280.23
21.547
4.5394
0.74260
4.900
3.71533.1644
3.0281
3.0034
3.0001
100.00
31.88717.409
4.5012
0.82240
0.11000
3.0923
3.81183.9708
3.9971
4.0001
4.0001
67.662
18.8764.0042
0.65798
0.07499
0.00000
4
3
1
3
2
1
x
x
x
0001.4
0001.3
99919.0
3
2
1
x
x
xThe solution obtained
is close to the exact solution of
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Gauss-Seidel Method: Example 2
Solve the following linear system by Gauss-Seidel method.
1 2 3
1 2 3
1 2 3
3 2
6 4 11 1 5 2 2 9
x x x
x x xx x x
Solution
We can verify that the system is not strictlydiagonally dominant, so the Gauss-Seidel
method does not guaranteed to converge.
1 3 1
6 4 11
5 2 2
A
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Definiteness of Matrices
A matrix is Positive Definite iff xTAx>0 for allx.
A matrix is Negative Definiteiff xTAx0 for some x
and xTAx
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Checking for Definiteness
There are different methods to check thedefiniteness of a matrix. One method is based
on Eigenvalues of the given matrix. If all
eigenvalues of the matrix are positive, then
the matrix would be positive definite. If all the
eigenvalues are negative, then the matrix
would be negative definite.
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Checking for Definiteness -
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Checking for Definiteness
Example
Check the definiteness of the matrix.
12 2
2 10A
Solution
12 2
2 10
12 2
2903
1 2
1
6E E
Since the two diagonal elements are +ve,
therefore matrix A is +ve definite.