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Here are the key steps to find the eigenvalues of the given matrix: 1) Write the characteristic equation: det(A - λI) = 0 2) Expand the determinant: (1-λ)(-2-λ) - 4 = 0 3) Simplify and factor: λ(λ + 1)(λ + 2) = 0 4) Find the roots: λ1 = 0, λ2 = -1, λ3 = -2 Therefore, the eigenvalues of the given matrix are -1 and -2.

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Matrix representation with dimensions, matrices defined for different rows and columns.

Matrices are added/subtracted by corresponding elements; operations are valid if matrices share dimensions.

The additive inverse of a matrix A achieves A + (-A) = 0, providing examples of additive inverses.

Scalar multiplication involves multiplying each element of the matrix by a scalar k.

Properties addressed include associative, commutative, and distributive laws of matrix addition, along with operations on augmented matrices.

Introduces the augmented matrix and determines solution consistency based on ranks; discusses parameters for unique solutions.

Defines eigenvalues/eigenvectors; discusses characteristics, multiplicity, and relations with the trace of a matrix.

Eigenvectors associated with distinct eigenvalues are linearly independent; introduces linear dependence relations.

Matrix satisfies own characteristic equation; provides examples illustrating verification of the theorem.

Describes the diagonalization process of a matrix through similarity transformation and orthogonal matrices.

Examples and calculations showing diagonalization of matrix A using orthogonal transformations.

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11 12 13 14 21 22 23 24 31 32 33 34 1 2 3 4m m m m a a a a a a a a a a a a a a a a          Row 1 Row 2 Row 3 Row m Column 1 Column 2 Column 3 Column 4
A matrix of m rows and n columns is called a matrix with dimensions m x n. 2 3 4 1.) 1 1 2           3 8 9 2.) 2 5 6 7 8          10 3.) 7       4.) 3 4 2 X 3 3 X 3 2 X 1 1 X 2
3 5 1 1.) 4 4 0            3 0 2.) 0 3       1 2 3 3.) 0 1 8 0 0 1          4.) 2    5 5.)         6.) 3 3 X 2 2 X 2 3 X 3 1 X 2 2 X 1 1 X 1
To add matrices, we add the corresponding elements. They must have the same dimensions. 5 0 6 3 4 1 2 3 A B               A + B 5 6 0 3 4 2 1 3           1 3 6 4       
2 1 3 0 0 0 2.) 1 0 1 0 0 0            2 1 3 1 0 1      When a zero matrix is added to another matrix of the same dimension, that same matrix is obtained.
To subtract matrices, we subtract the corresponding elements. The matrices must have the same dimensions. 1 2 1 1 3.) 2 0 1 3 3 1 2 3                     1 1 2 ( 1) 2 1 0 3 3 2 1 3                0 3 3 3 5 4           
4 1 6 5 1.) 6 3 7 3              1 3 2 2 1 5 2.) 4 0 5 6 4 3             2 6 13 0         1 4 7 2 4 8        
ADDITIVE INVERSE OF A MATRIX: 1 0 2 3 1 5 A       1 0 2 3 1 5 A         
Find the additive inverse: 2 1 5 6 4 3      2 1 5 6 4 3        
Scalar Multiplication: 1 2 3 1 2 3 4 5 6 k           We multiply each element of the matrix by scalar k. 1 2 3 1 2 3 4 5 6 k k k k k k k k k           
3 0 1.) 3 4 5       9 0 12 15        2 1 2 2.) 5 4 1 0 5 x y x          2 5 10 5 20 5 5 0 25 5 x y x         
• Associative Property of Addition (A+B)+C = A+(B+C) • Commutative Property of Addition A+B = B+A • Distributive Property of Addition and Subtraction S(A+B) = SA+SB S(A-B) = SA-SB • NOTE: Multiplication is not included!!!
• The following operations applied to the augmented matrix [A|b], yield an equivalent linear system – Interchanges: The order of two rows/columns can be changed – Scaling: Multiplying a row/column by a nonzero constant – Sum: The row can be replaced by the sum of that row and a nonzero multiple of any other row. One can use ERO and ECO to find the Rank as follows: EROminimum # of rows with at least one nonzero entry or ECOminimum # of columns with at least one nonzero entry
Math for CS Lecture 2 16 nnnnnn nn nn bxaxaxa bxaxaxa bxaxaxa        2211 22222121 11212111                                      nnnnnn n n b b b x x x aaa aaa aaa      2 1 2 1 21 22221 11211  (1)
bxA  Each side of the equation bAxAA 11   Can be multiplied by A-1 : Due to the definition of A-1: xxIxAA 1 Therefore the solution of (2) is: (2) bAx 1 
• A-1 does not exist for every A. • The linear system of equations A·x=b has a solution, or said to be consistent if Rank{A}=Rank{A|b} • A system is inconsistent when Rank{A}<Rank{A|b} Rank{A} is the maximum number of linearly independent columns or rows of A. Rank can be found by using ERO (Elementary Row Oparations) or ECO (Elementary column operations).
Math for CS Lecture 2 19                   5 4 42 21 2 1 x x       00 21 Rank{A}=1 Rank{A|b}=2 > Rank{A} ERO:Multiply the first row with -2 and add to the second row       3 4 0 2 0 1
Math for CS Lecture 2 20 • The system has a unique solution if Rank{A}=Rank{A|b}= n, where n is the order of the system.
Math for CS Lecture 2 21 • If Rank{A}=n Det{A}  0  A-1 exists  Unique solution                    2 4 11 21 2 1 x x
• If Rank{A}=m<n Det{A} = 0  A is singular so not invertible infinite number of solutions (n-m free variables) under-determined system                   8 4 42 21 2 1 x x Consistent so solvable Rank{A}=Rank{A|b}=1
• A nonzero vector x is an eigenvector (or characteristic vector) of a square matrix A if there exists a scalar λ such that Ax = λx. Then λ is an eigen value (or characteristic value) of A. Note: The zero vector can not be an eigenvector even though A0 = λ0. But λ = 0 can be an eigen value. Eigenvalues and Eigenvectors
2 2 4 1 3 6 2 4 2 0 : 3 6 1 0 2 0 0, 0 1 0 , , 0 . Show x is aneigenvector for A Solution Ax But for x Thus xis aneigenvector of A and is aneigenvalue                                                 Example:
Eigenvalues Let x be an eigenvector of the matrix A. Then there must exist an eigenvalue λ such that Ax = λx or, equivalently, Ax - λx = 0 or (A – λI)x = 0 If we define a new matrix B = A – λI, then Bx = 0 If B has an inverse then x = B-10 = 0. But an eigenvector cannot be zero. Thus, it follows that x will be an eigenvector of A if and only if B does not have an inverse, or equivalently det(B)=0, or det(A – λI) = 0 This is called the characteristic equation of A. Its roots determine the eigenvalues of A.
Example 1: Find the eigenvalues of two eigenvalues: 1,  2 Note: The roots of the characteristic equation can be repeated. That is, λ1 = λ2 =…= λk. If that happens, the eigenvalue is said to be of multiplicity k. Example 2: Find the eigenvalues of λ = 2 is an eigenvector of multiplicity 3.          51 122 A )2)(1(23 12)5)(2( 51 122 2           AI Eigenvalues: examples            200 020 012 A 0)2( 200 020 012 3           AI
Example 1 (cont.):                 00 41 41 123 )1(:1 AI 0, 1 4 ,404 2 1 1 2121               tt x x txtxxx x                 00 31 31 124 )2(:2 AI 0, 1 3 2 1 2              ss x x x Eigenvectors To each distinct eigenvalue of a matrix A there will correspond at least one eigenvector which can be found by solving the appropriate set of homogenous equations. If λi is an eigenvalue then the corresponding eigenvector xi is the solution of (A – λiI)xi = 0
Example 2 (cont.): Find the eigenvectors of Recall that λ = 2 is an eigenvector of multiplicity 3. Solve the homogeneous linear system represented by Let . The eigenvectors of  = 2 are of the form s and t not both zero.                                  0 0 0 000 000 010 )2( 3 2 1 x x x AI x txsx  31 , , 1 0 0 0 0 1 0 3 2 1                                             ts t s x x x x Eigenvectors            200 020 012 A
Properties of Eigenvalues and Eigenvectors Definition: The trace of a matrix A, designated by tr(A), is the sum of the elements on the main diagonal. Property 1: The sum of the eigenvalues of a matrix equals the trace of the matrix. Property 2: A matrix is singular if and only if it has a zero eigenvalue. Property 3: The eigenvalues of an upper (or lower) triangular matrix are the elements on the main diagonal. Property 4: If λ is an eigenvalue of A and A is invertible, then 1/λ is an eigenvalue of matrix A-1.
Properties of Eigenvalues and Eigenvectors Property 5: If λ is an eigenvalue of A then kλ is an eigenvalue of kA where k is any arbitrary scalar. Property 6: If λ is an eigenvalue of A then λk is an eigenvalue of Ak for any positive integer k. Property 8: If λ is an eigenvalue of A then λ is an eigenvalue of AT. Property 9: The product of the eigenvalues (counting multiplicity) of a matrix equals the determinant of the matrix.
Linearly independent eigenvectors Theorem: Eigenvectors corresponding to distinct (that is, different) eigenvalues are linearly independent. Theorem: If λ is an eigenvalue of multiplicity k of an n  n matrix A then the number of linearly independent eigenvectors of A associated with λ is given by m = n - r(A- λI). Furthermore, 1 ≤ m ≤ k. Example 2 (cont.): The eigenvectors of  = 2 are of the form s and t not both zero.  = 2 has two linearly independent eigenvectors , 1 0 0 0 0 1 0 3 2 1                                             ts t s x x x x
LINEAR INDEPENDENCE • Definition: A set of vectors {v1, …, vp} in is said to be linearly independent if the vector equation has only the trivial solution. The set {v1, …, vp} is said to be linearly dependent if there exist weights c1, …, cp, not all zero, such that n 1 1 2 2 v v ... v 0p p x x x    1 1 2 2 v v ... v 0p p c c c   
• Equation (1) is called a linear dependence relation among v1, …, vp when the weights are not all zero. A set is linearly dependent if and only if it is not linearly independent. Example 1: Let , , and . 1 1 v 2 3          2 4 v 5 6          3 2 v 1 0         
a. Determine if the set {v1, v2, v3} is linearly independent. b. If possible, find a linear dependence relation among v1, v2, and v3.  Solution: We must determine if there is a nontrivial solution of the following equation. 1 2 3 1 4 2 0 2 5 1 0 3 6 0 0 x x x                                         
 Row operations on the associated augmented matrix show that .  x1 and x2 are basic variables, and x3 is free.  Each nonzero value of x3 determines a nontrivial solution of (1).  Hence, v1, v2, v3 are linearly dependent. 1 4 2 0 1 4 2 0 2 5 1 0 0 3 3 0 3 6 0 0 0 0 0 0                     :
b. To find a linear dependence relation among v1, v2, and v3, row reduce the augmented matrix and write the new system: • Thus, , , and x3 is free. • Choose any nonzero value for x3—say, . • Then and . 1 0 2 0 0 1 1 0 0 0 0 0          1 3 2 3 2 0 0 0 0 x x x x      1 3 2x x 2 3 x x  3 5x  1 10x  2 5x  
CAYLEY HAMILTON THEOREM Every square matrix satisfies its own characteristic equation. Let A = [aij]n×n be a square matrix then, nnnn2n1n n22221 n11211 a...aa ................ a...aa a...aa A              
Let the characteristic polynomial of A be  (λ) Then, The characteristic equation is             11 12 1n 21 22 2n n1 n2 nn φ(λ) = A - λI a - λ a ... a a a - λ ... a = ... ... ... ... a a ... a - λ | A - λI|=0
Note 1:- Premultiplying equation (1) by A-1 , we have  n n-1 n-2 0 1 2 n n n-1 n-2 0 1 2 n We are to prove that p λ +p λ +p λ +...+p = 0 p A +p A +p A +...+p I= 0 ...(1) I  n-1 n-2 n-3 -1 0 1 2 n-1 n -1 n-1 n-2 n-3 0 1 2 n-1 n 0 =p A +p A +p A +...+p +p A 1 A =- [p A +p A +p A +...+p I] p
This result gives the inverse of A in terms of (n-1) powers of A and is considered as a practical method for the computation of the inverse of the large matrices. Note 2:- If m is a positive integer such that m > n then any positive integral power Am of A is linearly expressible in terms of those of lower degree.
Verify Cayley – Hamilton theorem for the matrix A = . Hence compute A-1 . Solution:- The characteristic equation of A is              211 121 112 tion)simplifica(on049λ6λλor 0 λ211 1λ21 11λ2 i.e.,0λIA 23       Example 1:-
To verify Cayley – Hamilton theorem, we have to show that A3 – 6A2 +9A – 4I = 0 … (1) Now,                                                                                   222121 212221 212222 211 121 112 655 565 556 655 565 556 211 121 112 211 121 112 23 2 AAA A
A3 -6A2 +9A – 4I = 0 = - 6 + 9 -4 = This verifies Cayley – Hamilton theorem.               222121 212221 212222              655 565 556              211 121 112           100 010 001 0 000 000 000           
44 Now, pre – multiplying both sides of (1) by A-1 , we have A2 – 6A +9I – 4 A-1 = 0 => 4 A-1 = A2 – 6 A +9I                                                                    311 131 113 4 1 311 131 113 100 010 001 9 211 121 112 6 655 565 556 4 1 1 A A
45 Given find Adj A by using Cayley – Hamilton theorem. Solution:- The characteristic equation of the given matrix A is               113 110 121 A tion)simplifica(on035λ3λλor 0 λ113 1λ10 1-2λ1 i.e.,0λIA 23       Example 2:-
46 By Cayley – Hamilton theorem, A should satisfy A3 – 3A2 + 5A + 3I = 0 Pre – multiplying by A-1 , we get A2 – 3A +5I +3A-1 = 0                                                         339 330 363 3A 146 223 452 113 110 121 113 110 121 A.AANow, (1)...5I)3A(A 3 1 A 2 21-
47 AAAAdj. A AAdj. Athat,knowWe 173 143 110 3 1 500 050 005 339 330 363 146 223 452 3 1 AFrom(1), 1 1 1                                                                    
48                                        173 143 110 AAdj. 173 143 110 3 1 3)(AAdj. 3 113 110 121 ANow,
49 DIAGONALISATION OF A MATRIX Diagonalisation of a matrix A is the process of reduction A to a diagonal form. If A is related to D by a similarity transformation, such that D = M-1AM then A is reduced to the diagonal matrix D through modal matrix M. D is also called spectral matrix of A.
50 REDUCTION OF A MATRIX TO DIAGONAL FORM If a square matrix A of order n has n linearly independent eigen vectors then a matrix B can be found such that B-1AB is a diagonal matrix. Note:- The matrix B which diagonalises A is called the modal matrix of A and is obtained by grouping the eigen vectors of A into a square matrix.
51 Similarity of matrices:- A square matrix B of order n is said to be a similar to a square matrix A of order n if B = M-1AM for some non singular matrix M. This transformation of a matrix A by a non – singular matrix M to B is called a similarity transformation. Note:- If the matrix B is similar to matrix A, then B has the same eigen values as A.
52 Reduce the matrix A = to diagonal form by similarity transformation. Hence find A3. Solution:- Characteristic equation is => λ = 1, 2, 3 Hence eigen values of A are 1, 2, 3.             300 120 211 0             λ-300 1λ-20 21λ1- Example:-
53 Corresponding to λ = 1, let X1 = be the eigen vector then           3 2 1 x x x                                                  0 0 1 kX x0x,kx 02x 0xx 02xx 0 0 0 x x x 200 110 210 0X)I(A 11 3211 3 32 32 3 2 1 1
54 Corresponding to λ = 2, let X2 = be the eigen vector then,           3 2 1 x x x                                                  0 1- 1 kX x-kx,kx 0x 0x 02xxx 0 0 0 x x x 100 100 211- 0X)(A 22 32221 3 3 321 3 2 1 2 0, I2
55 Corresponding to λ = 3, let X3 = be the eigen vector then,           3 2 1 x x x                                                  2 2- 3 kX xk-x,kx 0x 02xxx 0 0 0 x x x 000 11-0 212- 0X)(A 33 13332 3 321 3 2 1 3 3 2 2 3 , 2 I3 k x
56 Hence modal matrix is                                         2 1 00 11-0 2 1- 11 M MAdj. M 1-00 220 122- MAdj. 2M 200 21-0 311 M 1
57 Now, since D = M-1AM => A = MDM-1 A2 = (MDM-1) (MDM-1) = MD2M-1 [since M-1M = I]                                                    300 020 001 200 21-0 311 300 120 211 2 1 00 11-0 2 1 11 AMM 1
58 Similarly, A3 = MD3M-1 = A3 =                                                2700 19-80 327-1 2 1 00 11-0 2 1 11 2700 080 001 200 21-0 311
59 ORTHOGONAL TRANSFORMATION OF A SYMMETRIC MATRIX TO DIAGONAL FORM A square matrix A with real elements is said to be orthogonal if AA’ = I = A’A. But AA-1 = I = A-1A, it follows that A is orthogonal if A’ = A-1. Diagonalisation by orthogonal transformation is possible only for a real symmetric matrix.
60 If A is a real symmetric matrix then eigen vectors of A will be not only linearly independent but also pairwise orthogonal. If we normalise each eigen vector and use them to form the normalised modal matrix N then it can be proved that N is an orthogonal matrix.
61 The similarity transformation M-1AM = D takes the form N’AN = D since N-1 = N’ by a property of orthogonal matrix. Transforming A into D by means of the transformation N’AN = D is called as orthogonal reduction or othogonal transformation. Note:- To normalise eigen vector Xr, divide each element of Xr, by the square root of the sum of the squares of all the elements of Xr.
62 Diagonalise the matrix A = by means of an orthogonal transformation. Solution:- Characteristic equation of A is 204 060 402 66,2,λ 0λ)16(6λ)λ)(2λ)(6(2 0 λ204 0λ60 40λ2       Example :-
63 I                                                  1 1 2 3 1 1 2 3 1 3 2 1 3 1 1 2 3 1 1 1 x whenλ = -2,let X = x betheeigenvector x then (A + 2 )X = 0 4 0 4 x 0 0 8 0 x = 0 4 0 4 x 0 4x + 4x = 0 ...(1) 8x = 0 ...(2) 4x + 4x = 0 ...(3) x = k ,x = 0,x = -k 1 X = k 0 -1
64 2 2I 0                                          1 2 3 1 2 3 1 3 1 3 1 3 2 2 2 3 x whenλ = 6,let X = x betheeigenvector x then (A -6 )X = 0 -4 0 4 x 0 0 0 x = 0 4 0 -4 x 0 4x +4x = 0 4x - 4x = 0 x = x and x isarbitrary x must be so chosen that X and X are orthogonal among th .1 emselves and also each is orthogonal with X
65                                2 3 3 1 3 2 3 1 α Let X = 0 and let X = β 1 γ Since X is orthogonal to X α - γ = 0 ...(4) X is orthogonal to X α + γ = 0 ...(5) Solving (4)and(5), we get α = γ = 0 and β is arbitrary. 0 Taking β =1, X = 1 0 1 1 0 Modal matrix is M = 0 0 1 -1 1         0
66                                                           The normalised modal matrix is 1 1 0 2 2 N = 0 0 1 1 1 - 0 2 2 1 1 0 - 1 1 02 2 2 0 4 2 2 1 1 D =N'AN = 0 0 6 0 0 0 1 2 2 4 0 2 1 1 - 00 1 0 2 2 -2 0 0 D = 0 6 0 which is the required diagonal matrix 0 0 6 .

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Matrices ppt

  • 2.
    11 12 1314 21 22 23 24 31 32 33 34 1 2 3 4m m m m a a a a a a a a a a a a a a a a          Row 1 Row 2 Row 3 Row m Column 1 Column 2 Column 3 Column 4
  • 3.
    A matrix ofm rows and n columns is called a matrix with dimensions m x n. 2 3 4 1.) 1 1 2           3 8 9 2.) 2 5 6 7 8          10 3.) 7       4.) 3 4 2 X 3 3 X 3 2 X 1 1 X 2
  • 4.
    3 5 1 1.) 4 4 0           3 0 2.) 0 3       1 2 3 3.) 0 1 8 0 0 1          4.) 2    5 5.)         6.) 3 3 X 2 2 X 2 3 X 3 1 X 2 2 X 1 1 X 1
  • 6.
    To add matrices,we add the corresponding elements. They must have the same dimensions. 5 0 6 3 4 1 2 3 A B               A + B 5 6 0 3 4 2 1 3           1 3 6 4       
  • 7.
    2 1 30 0 0 2.) 1 0 1 0 0 0            2 1 3 1 0 1      When a zero matrix is added to another matrix of the same dimension, that same matrix is obtained.
  • 8.
    To subtract matrices,we subtract the corresponding elements. The matrices must have the same dimensions. 1 2 1 1 3.) 2 0 1 3 3 1 2 3                     1 1 2 ( 1) 2 1 0 3 3 2 1 3                0 3 3 3 5 4           
  • 9.
    4 1 65 1.) 6 3 7 3              1 3 2 2 1 5 2.) 4 0 5 6 4 3             2 6 13 0         1 4 7 2 4 8        
  • 10.
    ADDITIVE INVERSE OFA MATRIX: 1 0 2 3 1 5 A       1 0 2 3 1 5 A         
  • 11.
    Find the additiveinverse: 2 1 5 6 4 3      2 1 5 6 4 3        
  • 12.
    Scalar Multiplication: 1 23 1 2 3 4 5 6 k           We multiply each element of the matrix by scalar k. 1 2 3 1 2 3 4 5 6 k k k k k k k k k           
  • 13.
    3 0 1.) 3 45       9 0 12 15        2 1 2 2.) 5 4 1 0 5 x y x          2 5 10 5 20 5 5 0 25 5 x y x         
  • 14.
    • Associative Propertyof Addition (A+B)+C = A+(B+C) • Commutative Property of Addition A+B = B+A • Distributive Property of Addition and Subtraction S(A+B) = SA+SB S(A-B) = SA-SB • NOTE: Multiplication is not included!!!
  • 15.
    • The followingoperations applied to the augmented matrix [A|b], yield an equivalent linear system – Interchanges: The order of two rows/columns can be changed – Scaling: Multiplying a row/column by a nonzero constant – Sum: The row can be replaced by the sum of that row and a nonzero multiple of any other row. One can use ERO and ECO to find the Rank as follows: EROminimum # of rows with at least one nonzero entry or ECOminimum # of columns with at least one nonzero entry
  • 16.
    Math for CSLecture 2 16 nnnnnn nn nn bxaxaxa bxaxaxa bxaxaxa        2211 22222121 11212111                                      nnnnnn n n b b b x x x aaa aaa aaa      2 1 2 1 21 22221 11211  (1)
  • 17.
    bxA  Each sideof the equation bAxAA 11   Can be multiplied by A-1 : Due to the definition of A-1: xxIxAA 1 Therefore the solution of (2) is: (2) bAx 1 
  • 18.
    • A-1 doesnot exist for every A. • The linear system of equations A·x=b has a solution, or said to be consistent if Rank{A}=Rank{A|b} • A system is inconsistent when Rank{A}<Rank{A|b} Rank{A} is the maximum number of linearly independent columns or rows of A. Rank can be found by using ERO (Elementary Row Oparations) or ECO (Elementary column operations).
  • 19.
    Math for CSLecture 2 19                   5 4 42 21 2 1 x x       00 21 Rank{A}=1 Rank{A|b}=2 > Rank{A} ERO:Multiply the first row with -2 and add to the second row       3 4 0 2 0 1
  • 20.
    Math for CSLecture 2 20 • The system has a unique solution if Rank{A}=Rank{A|b}= n, where n is the order of the system.
  • 21.
    Math for CSLecture 2 21 • If Rank{A}=n Det{A}  0  A-1 exists  Unique solution                    2 4 11 21 2 1 x x
  • 22.
    • If Rank{A}=m<n Det{A}= 0  A is singular so not invertible infinite number of solutions (n-m free variables) under-determined system                   8 4 42 21 2 1 x x Consistent so solvable Rank{A}=Rank{A|b}=1
  • 23.
    • A nonzerovector x is an eigenvector (or characteristic vector) of a square matrix A if there exists a scalar λ such that Ax = λx. Then λ is an eigen value (or characteristic value) of A. Note: The zero vector can not be an eigenvector even though A0 = λ0. But λ = 0 can be an eigen value. Eigenvalues and Eigenvectors
  • 24.
    2 2 4 13 6 2 4 2 0 : 3 6 1 0 2 0 0, 0 1 0 , , 0 . Show x is aneigenvector for A Solution Ax But for x Thus xis aneigenvector of A and is aneigenvalue                                                 Example:
  • 25.
    Eigenvalues Let x bean eigenvector of the matrix A. Then there must exist an eigenvalue λ such that Ax = λx or, equivalently, Ax - λx = 0 or (A – λI)x = 0 If we define a new matrix B = A – λI, then Bx = 0 If B has an inverse then x = B-10 = 0. But an eigenvector cannot be zero. Thus, it follows that x will be an eigenvector of A if and only if B does not have an inverse, or equivalently det(B)=0, or det(A – λI) = 0 This is called the characteristic equation of A. Its roots determine the eigenvalues of A.
  • 26.
    Example 1: Findthe eigenvalues of two eigenvalues: 1,  2 Note: The roots of the characteristic equation can be repeated. That is, λ1 = λ2 =…= λk. If that happens, the eigenvalue is said to be of multiplicity k. Example 2: Find the eigenvalues of λ = 2 is an eigenvector of multiplicity 3.          51 122 A )2)(1(23 12)5)(2( 51 122 2           AI Eigenvalues: examples            200 020 012 A 0)2( 200 020 012 3           AI
  • 27.
    Example 1 (cont.):                00 41 41 123 )1(:1 AI 0, 1 4 ,404 2 1 1 2121               tt x x txtxxx x                 00 31 31 124 )2(:2 AI 0, 1 3 2 1 2              ss x x x Eigenvectors To each distinct eigenvalue of a matrix A there will correspond at least one eigenvector which can be found by solving the appropriate set of homogenous equations. If λi is an eigenvalue then the corresponding eigenvector xi is the solution of (A – λiI)xi = 0
  • 28.
    Example 2 (cont.):Find the eigenvectors of Recall that λ = 2 is an eigenvector of multiplicity 3. Solve the homogeneous linear system represented by Let . The eigenvectors of  = 2 are of the form s and t not both zero.                                  0 0 0 000 000 010 )2( 3 2 1 x x x AI x txsx  31 , , 1 0 0 0 0 1 0 3 2 1                                             ts t s x x x x Eigenvectors            200 020 012 A
  • 29.
    Properties of Eigenvaluesand Eigenvectors Definition: The trace of a matrix A, designated by tr(A), is the sum of the elements on the main diagonal. Property 1: The sum of the eigenvalues of a matrix equals the trace of the matrix. Property 2: A matrix is singular if and only if it has a zero eigenvalue. Property 3: The eigenvalues of an upper (or lower) triangular matrix are the elements on the main diagonal. Property 4: If λ is an eigenvalue of A and A is invertible, then 1/λ is an eigenvalue of matrix A-1.
  • 30.
    Properties of Eigenvaluesand Eigenvectors Property 5: If λ is an eigenvalue of A then kλ is an eigenvalue of kA where k is any arbitrary scalar. Property 6: If λ is an eigenvalue of A then λk is an eigenvalue of Ak for any positive integer k. Property 8: If λ is an eigenvalue of A then λ is an eigenvalue of AT. Property 9: The product of the eigenvalues (counting multiplicity) of a matrix equals the determinant of the matrix.
  • 31.
    Linearly independent eigenvectors Theorem:Eigenvectors corresponding to distinct (that is, different) eigenvalues are linearly independent. Theorem: If λ is an eigenvalue of multiplicity k of an n  n matrix A then the number of linearly independent eigenvectors of A associated with λ is given by m = n - r(A- λI). Furthermore, 1 ≤ m ≤ k. Example 2 (cont.): The eigenvectors of  = 2 are of the form s and t not both zero.  = 2 has two linearly independent eigenvectors , 1 0 0 0 0 1 0 3 2 1                                             ts t s x x x x
  • 32.
    LINEAR INDEPENDENCE • Definition:A set of vectors {v1, …, vp} in is said to be linearly independent if the vector equation has only the trivial solution. The set {v1, …, vp} is said to be linearly dependent if there exist weights c1, …, cp, not all zero, such that n 1 1 2 2 v v ... v 0p p x x x    1 1 2 2 v v ... v 0p p c c c   
  • 33.
    • Equation (1)is called a linear dependence relation among v1, …, vp when the weights are not all zero. A set is linearly dependent if and only if it is not linearly independent. Example 1: Let , , and . 1 1 v 2 3          2 4 v 5 6          3 2 v 1 0         
  • 34.
    a. Determine ifthe set {v1, v2, v3} is linearly independent. b. If possible, find a linear dependence relation among v1, v2, and v3.  Solution: We must determine if there is a nontrivial solution of the following equation. 1 2 3 1 4 2 0 2 5 1 0 3 6 0 0 x x x                                         
  • 35.
     Row operationson the associated augmented matrix show that .  x1 and x2 are basic variables, and x3 is free.  Each nonzero value of x3 determines a nontrivial solution of (1).  Hence, v1, v2, v3 are linearly dependent. 1 4 2 0 1 4 2 0 2 5 1 0 0 3 3 0 3 6 0 0 0 0 0 0                     :
  • 36.
    b. To finda linear dependence relation among v1, v2, and v3, row reduce the augmented matrix and write the new system: • Thus, , , and x3 is free. • Choose any nonzero value for x3—say, . • Then and . 1 0 2 0 0 1 1 0 0 0 0 0          1 3 2 3 2 0 0 0 0 x x x x      1 3 2x x 2 3 x x  3 5x  1 10x  2 5x  
  • 37.
    CAYLEY HAMILTON THEOREM Everysquare matrix satisfies its own characteristic equation. Let A = [aij]n×n be a square matrix then, nnnn2n1n n22221 n11211 a...aa ................ a...aa a...aa A              
  • 38.
    Let the characteristicpolynomial of A be  (λ) Then, The characteristic equation is             11 12 1n 21 22 2n n1 n2 nn φ(λ) = A - λI a - λ a ... a a a - λ ... a = ... ... ... ... a a ... a - λ | A - λI|=0
  • 39.
    Note 1:- Premultiplyingequation (1) by A-1 , we have  n n-1 n-2 0 1 2 n n n-1 n-2 0 1 2 n We are to prove that p λ +p λ +p λ +...+p = 0 p A +p A +p A +...+p I= 0 ...(1) I  n-1 n-2 n-3 -1 0 1 2 n-1 n -1 n-1 n-2 n-3 0 1 2 n-1 n 0 =p A +p A +p A +...+p +p A 1 A =- [p A +p A +p A +...+p I] p
  • 40.
    This result givesthe inverse of A in terms of (n-1) powers of A and is considered as a practical method for the computation of the inverse of the large matrices. Note 2:- If m is a positive integer such that m > n then any positive integral power Am of A is linearly expressible in terms of those of lower degree.
  • 41.
    Verify Cayley –Hamilton theorem for the matrix A = . Hence compute A-1 . Solution:- The characteristic equation of A is              211 121 112 tion)simplifica(on049λ6λλor 0 λ211 1λ21 11λ2 i.e.,0λIA 23       Example 1:-
  • 42.
    To verify Cayley– Hamilton theorem, we have to show that A3 – 6A2 +9A – 4I = 0 … (1) Now,                                                                                   222121 212221 212222 211 121 112 655 565 556 655 565 556 211 121 112 211 121 112 23 2 AAA A
  • 43.
    A3 -6A2 +9A– 4I = 0 = - 6 + 9 -4 = This verifies Cayley – Hamilton theorem.               222121 212221 212222              655 565 556              211 121 112           100 010 001 0 000 000 000           
  • 44.
    44 Now, pre –multiplying both sides of (1) by A-1 , we have A2 – 6A +9I – 4 A-1 = 0 => 4 A-1 = A2 – 6 A +9I                                                                    311 131 113 4 1 311 131 113 100 010 001 9 211 121 112 6 655 565 556 4 1 1 A A
  • 45.
    45 Given find AdjA by using Cayley – Hamilton theorem. Solution:- The characteristic equation of the given matrix A is               113 110 121 A tion)simplifica(on035λ3λλor 0 λ113 1λ10 1-2λ1 i.e.,0λIA 23       Example 2:-
  • 46.
    46 By Cayley –Hamilton theorem, A should satisfy A3 – 3A2 + 5A + 3I = 0 Pre – multiplying by A-1 , we get A2 – 3A +5I +3A-1 = 0                                                         339 330 363 3A 146 223 452 113 110 121 113 110 121 A.AANow, (1)...5I)3A(A 3 1 A 2 21-
  • 47.
  • 48.
  • 49.
    49 DIAGONALISATION OF A MATRIX Diagonalisationof a matrix A is the process of reduction A to a diagonal form. If A is related to D by a similarity transformation, such that D = M-1AM then A is reduced to the diagonal matrix D through modal matrix M. D is also called spectral matrix of A.
  • 50.
    50 REDUCTION OF AMATRIX TO DIAGONAL FORM If a square matrix A of order n has n linearly independent eigen vectors then a matrix B can be found such that B-1AB is a diagonal matrix. Note:- The matrix B which diagonalises A is called the modal matrix of A and is obtained by grouping the eigen vectors of A into a square matrix.
  • 51.
    51 Similarity of matrices:- Asquare matrix B of order n is said to be a similar to a square matrix A of order n if B = M-1AM for some non singular matrix M. This transformation of a matrix A by a non – singular matrix M to B is called a similarity transformation. Note:- If the matrix B is similar to matrix A, then B has the same eigen values as A.
  • 52.
    52 Reduce the matrixA = to diagonal form by similarity transformation. Hence find A3. Solution:- Characteristic equation is => λ = 1, 2, 3 Hence eigen values of A are 1, 2, 3.             300 120 211 0             λ-300 1λ-20 21λ1- Example:-
  • 53.
    53 Corresponding to λ= 1, let X1 = be the eigen vector then           3 2 1 x x x                                                  0 0 1 kX x0x,kx 02x 0xx 02xx 0 0 0 x x x 200 110 210 0X)I(A 11 3211 3 32 32 3 2 1 1
  • 54.
    54 Corresponding to λ= 2, let X2 = be the eigen vector then,           3 2 1 x x x                                                  0 1- 1 kX x-kx,kx 0x 0x 02xxx 0 0 0 x x x 100 100 211- 0X)(A 22 32221 3 3 321 3 2 1 2 0, I2
  • 55.
    55 Corresponding to λ= 3, let X3 = be the eigen vector then,           3 2 1 x x x                                                  2 2- 3 kX xk-x,kx 0x 02xxx 0 0 0 x x x 000 11-0 212- 0X)(A 33 13332 3 321 3 2 1 3 3 2 2 3 , 2 I3 k x
  • 56.
    56 Hence modal matrixis                                         2 1 00 11-0 2 1- 11 M MAdj. M 1-00 220 122- MAdj. 2M 200 21-0 311 M 1
  • 57.
    57 Now, since D= M-1AM => A = MDM-1 A2 = (MDM-1) (MDM-1) = MD2M-1 [since M-1M = I]                                                    300 020 001 200 21-0 311 300 120 211 2 1 00 11-0 2 1 11 AMM 1
  • 58.
    58 Similarly, A3 =MD3M-1 = A3 =                                                2700 19-80 327-1 2 1 00 11-0 2 1 11 2700 080 001 200 21-0 311
  • 59.
    59 ORTHOGONAL TRANSFORMATION OF ASYMMETRIC MATRIX TO DIAGONAL FORM A square matrix A with real elements is said to be orthogonal if AA’ = I = A’A. But AA-1 = I = A-1A, it follows that A is orthogonal if A’ = A-1. Diagonalisation by orthogonal transformation is possible only for a real symmetric matrix.
  • 60.
    60 If A isa real symmetric matrix then eigen vectors of A will be not only linearly independent but also pairwise orthogonal. If we normalise each eigen vector and use them to form the normalised modal matrix N then it can be proved that N is an orthogonal matrix.
  • 61.
    61 The similarity transformationM-1AM = D takes the form N’AN = D since N-1 = N’ by a property of orthogonal matrix. Transforming A into D by means of the transformation N’AN = D is called as orthogonal reduction or othogonal transformation. Note:- To normalise eigen vector Xr, divide each element of Xr, by the square root of the sum of the squares of all the elements of Xr.
  • 62.
    62 Diagonalise the matrixA = by means of an orthogonal transformation. Solution:- Characteristic equation of A is 204 060 402 66,2,λ 0λ)16(6λ)λ)(2λ)(6(2 0 λ204 0λ60 40λ2       Example :-
  • 63.
    63 I                                                 1 1 2 3 1 1 2 3 1 3 2 1 3 1 1 2 3 1 1 1 x whenλ = -2,let X = x betheeigenvector x then (A + 2 )X = 0 4 0 4 x 0 0 8 0 x = 0 4 0 4 x 0 4x + 4x = 0 ...(1) 8x = 0 ...(2) 4x + 4x = 0 ...(3) x = k ,x = 0,x = -k 1 X = k 0 -1
  • 64.
    64 2 2I 0                                         1 2 3 1 2 3 1 3 1 3 1 3 2 2 2 3 x whenλ = 6,let X = x betheeigenvector x then (A -6 )X = 0 -4 0 4 x 0 0 0 x = 0 4 0 -4 x 0 4x +4x = 0 4x - 4x = 0 x = x and x isarbitrary x must be so chosen that X and X are orthogonal among th .1 emselves and also each is orthogonal with X
  • 65.
    65                               2 3 3 1 3 2 3 1 α Let X = 0 and let X = β 1 γ Since X is orthogonal to X α - γ = 0 ...(4) X is orthogonal to X α + γ = 0 ...(5) Solving (4)and(5), we get α = γ = 0 and β is arbitrary. 0 Taking β =1, X = 1 0 1 1 0 Modal matrix is M = 0 0 1 -1 1         0
  • 66.
    66                                                          The normalised modal matrix is 1 1 0 2 2 N = 0 0 1 1 1 - 0 2 2 1 1 0 - 1 1 02 2 2 0 4 2 2 1 1 D =N'AN = 0 0 6 0 0 0 1 2 2 4 0 2 1 1 - 00 1 0 2 2 -2 0 0 D = 0 6 0 which is the required diagonal matrix 0 0 6 .