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Eigen value and eigen vector

This document provides information about eigenvalues and eigenvectors. It defines eigenvalues and eigenvectors as scalars (λ) and vectors (x) that satisfy the equation Ax = λx, where A is a matrix. It discusses properties of eigenvalues including that the sum of eigenvalues is the trace of A, and the product is the determinant. The characteristic equation is defined as det(A - λI) = 0, where the roots are the eigenvalues. Cayley-Hamilton theorem states that every matrix satisfies its own characteristic equation. Examples are given to demonstrate Cayley-Hamilton theorem.

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Eigen Value And Eigen Vector GUIDED BY: MANSI K. DESAI
Group Members:- SR NO. NAME ENROLLMENT NUMBER 1. RANA PAYAL MAHESHBHAI 151100106072 2. PATEL RUTVIJ GANESHBHAI 151100106073 3. SAVALIYAAKSHAY JAYANTILAL 151100106074 4. TANDEL PAYALBENVITHTHALBHAI 151100106075 5. TANDEL SNEHAL MAHENDRABHAI 151100106076 6. VAGHELA SAHIL PREMJIBHAI 151100106077 7. EHSANULLAHAYDIN 151100106078 8. SHUAIB KOHEE 151100106079 9. WAHEDULLAH EHSAS 151100106080
Eigenvalues and Eigenvectors • If A is an n x n matrix and λ is a scalar for which Ax = λx has a nontrivial solution x ∈ ℜⁿ, then λ is an eigenvalue of A and x is a corresponding eigenvector of A. – Ax=λx=λIx – (A-λI)x=0 • The matrix (A-λI ) is called the characteristic matrix of a where I is the Unit matrix. – • The equation det (A-λI )= 0 is called characteristic equation of A and the roots of this equation are called the eigenvalues of the matrix A.The set of all eigenvectors is called the eigenspace of A corresponding to λ.The set of all eigenvalues of a is called spectrum of A.
Characteristic Equation ▪ If A is any square matrix of order n, we can form the matrix , where is the nth order unit matrix. ▪ The determinant of this matrix equated to zero, ▪ i.e., is called the characteristic equation of A. 0 λa...aa ............ a...λaa a...aλa λA nnn2n1 2n2221 1n1211      I
• On expanding the determinant, we get • where k’s are expressible in terms of the elements a •The roots of this equation are called Characteristic roots or latent roots or eigen values of the matrix A. •X = is called an eigen vector or latent vector 0k...λkλkλ1)( n 2n 2 1n 1 nn               4 2 1 ... x x x
Properties of Eigen Values:- 1. The sum of the eigen values of a matrix is the sum of the elements of the principal diagonal. 2.The product of the eigen values of a matrix A is equal to its determinant. 3. If is an eigen value of a matrix A, then 1/ is the eigen value of A-1 . 4.If is an eigen value of an orthogonal matrix, then 1/ is also its eigen value. 
PROPERTY 1:- If λ1, λ2,…, λn are the eigen values of A, then i. k λ1, k λ2,…,k λn are the eigen values of the matrix kA, where k is a non – zero scalar. ii. are the eigen values of the inverse matrix A-1. iii. are the eigen values of Ap, where p is any positive integer.
Algebraic & Geometric Multiplicity ▪ If the eigenvalue λ of the equation det(A-λI)=0 is repeated n times then n is called the algebraic multiplicity of λ.The number of linearly independent eigenvectors is the difference between the number of unknowns and the rank of the corresponding matrix A- λI and is known as geometric multiplicity of eigenvalue λ.
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
 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) Note 1:- Premultiplying equation (1) by A-1 , we have 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.
Example 1:- 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      
                                                                                  222121 212221 212222 211 121 112 655 565 556 655 565 556 211 121 112 211 121 112 23 2 AAA A To verify Cayley – Hamilton theorem, we have to show that A3 – 6A2 +9A – 4I = 0 … (1) Now,
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           
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
Example 2:- 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      
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-
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                                                                    
                                       173 143 110 AAdj. 173 143 110 3 1 3)(AAdj. 3 113 110 121 ANow,
Similarity of Matrix ▪ IfA & B are two square matrices of order n then B is said to be similar to A, if there exists a non-singular matrix P such that, B= P-1AP 1. Similarity matrices is an equivalence relation. 2. Similarity matrices have the same determinant. 3. Similar matrices have the same characteristic polynomial and hence the same eigenvalues. If x is an eigenvector corresponding to the eigenvalue λ, then P-1x is an eigenvector of B corresponding to the eigenvalue λ where B= P-1AP.
Diagonalization ▪ A matrix A is said to be diagonalizable if it is similar to diagonal matrix. ▪ A matrix A is diagonalizable if there exists an invertible matrix P such that P-1AP=D where D is a diagonal matrix, also known as spectral matrix.The matrix P is then said to diagonalize A of transform A to diagonal form and is known as modal matrix.
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 ofA into a square matrix.
Example:- Reduce the matrix A = to diagonal form by similarity transformation. Hence find A3. Solution:- Characteristic equation is => λ = 1, 2, 3 Hence eigenvalues of A are 1, 2, 3.             300 120 211 0             λ-300 1λ-20 21λ1-
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
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
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
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
                                                   300 020 001 200 21-0 311 300 120 211 2 1 00 11-0 2 1 11 AMM 1 Now, since D = M-1AM => A = MDM-1 A2 = (MDM-1) (MDM-1) = MD2M-1 [since M-1M = I]
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
Orthogonally Similar Matrices ▪ If A & B are two square matrices of order n then B is said to be orthogonally similar to A, if there exists orthogonal matrix P such that B= P-1AP Since P is orthogonal, P-1=PT B= P-1AP=PTAP 1. A real symmetric of order n has n mutually orthogonal real eigenvectors. 2. Any two eigenvectors corresponding to two distinct eigenvalues of a real symmetric matrix are orthogonal.
Diagonalises the matrix A = by means of an orthogonal transformation. Solution:- Characteristic equation of A is 32 Example :- 204 060 402 66,2,λ 0λ)16(6λ)λ)(2λ)(6(2 0 λ204 0λ60 40λ2      
I                                                  1 1 2 3 1 1 2 3 1 3 2 1 3 1 1 2 3 1 1 1 x when λ = -2,let X = x be the eigen vector 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
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
                               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
                                                          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 .
Quadratic Forms DEFINITION:- A homogeneous polynomial of second degree in any number of variables is called a quadratic form. For example, ax2 + 2hxy +by2 ax2 + by2 + cz2 + 2hxy + 2gyz + 2fzx and ax2 + by2 + cz2 + dw2 +2hxy +2gyz + 2fzx + 2lxw + 2myw + 2nzw are quadratic forms in two, three and four variables
In n – variables x1,x2,…,xn, the general quadratic form is In the expansion, the co-efficient of xixj = (bij + bji). 38    n 1j n 1i jiijjiij bbwhere,xxb ).b(b 2 1 awherexxaxxb baandaawherebb2aSuppose jiijijji n 1j n 1i ijji n 1j n 1i ij iiiijiijijijij      
Hence every quadratic form can be written as     getweform,matrixin formsquadraticofexamplessaidabovethewritingNow .x,...,x,xXandaAwhere symmetric,alwaysisAmatrixthethatso AX,X'xxa n21ij ji n 1j n 1i ij                 y x bh ha y][xby2hxyax(i) 22
                                                   w z y x dnml ncgf mgbh lfha wzyx 2nzw2myw2lxwzx2f2gyz2hxydw2czbyax(iii) z y x cgf gbh fha zyx2fzx2gyz2hxyczbyax(ii) 222 222
Two Theorems On Quadratic Form Theorem(1): A quadratic form can always be expressed with respect to a given coordinate system as where A is a unique symmetric matrix. Theorem2: Two symmetric matrices A and B represent the same quadratic form if and only if B=PTAP where P is a non-singular matrix. AxxY T 
Nature of Quadratic Form A real quadratic form X’AX in n variables is said to be i. Positive definite if all the eigen values ofA > 0. ii. Negative definite if all the eigen values of A < 0. iii. Positive semi definite if all the eigen values ofA 0 and at least one eigen value = 0. iv. Negative semi definite if all the eigen values of A 0 and at least one eigen value = 0. v. Indefinite if some of the eigen values ofA are + ve and others – ve.
Find the nature of the following quadratic forms i. x2 + 5y2 + z2 + 2xy + 2yz + 6zx ii. 3x2 + 5y2 + 3z2 – 2yz + 2zx – 2xy Solution:- i. The matrix of the quadratic form is 43 Example :-            113 151 311 A
The eigen values of A are -2, 3, 6. Two of these eigen values being positive and one being negative, the given quadratric form is indefinite. ii. The matrix of the quadratic form is The eigen values of A are 2, 3, 6. All these eigen values being positive, the given quadratic form is positive definite.               311 151 113 A
Linear Transformation of a Quadratic Form ▪ Let X’AX be a quadratic form in n- variables and let X = PY ….. (1) where P is a non – singular matrix, be the non – singular transformation. ▪ From (1), X’ = (PY)’ =Y’P’ and hence X’AX =Y’P’APY =Y’(P’AP)Y =Y’BY …. (2) where B = P’AP.
Therefore,Y’BY is also a quadratic form in n- variables. Hence it is a linear transformation of the quadratic form X’AX under the linear transformation X = PY and B = P’AP. Note. (i) Here B = (P’AP)’ = P’AP = B (ii) ρ(B) = ρ(A) Therefore, A and B are congruent matrices.
Reduce 3x2 + 3z2 + 4xy + 8xz + 8yz into canonical form. Or Diagonalises the quadratic form 3x2 + 3z2 + 4xy + 8xz + 8yz by linear transformations and write the linear transformation. Or Reduce the quadratic form 3x2 + 3z2 + 4xy + 8xz + 8yz into the sum of squares. 47 Example:-
Solution:- The given quadratic form can be written as X’AX where X = [x, y, z]’ and the symmetric matrix A = Let us reduce A into diagonal matrix. We know tat A = I3AI3.           344 402 423                                          100 010 001 344 402 423 100 010 001 344 402 423
                                            21 31OperatingR ( 2 / 3),R ( 4 / 3) (for A onL.H.S.andpre factor on R.H.S.), we get 3 2 4 1 0 0 1 0 0 4 4 2 0 1 0 A 0 1 0 3 3 3 0 0 1 4 7 4 0 0 1 3 3 3                                                      100 010 3 4 3 2 1 A 10 3 4 01 3 2 001 3 7 3 4 0 3 4 3 4 0 423 getweR.H.S),onfactorpostandL.H.S.onA(for 4/3)(C2/3),(COperating 3121
                                       100 010 3 4 3 2 1 A 112 01 3 2 001 100 3 4 3 4 0 003 getwe(1),ROperating 32 APP'1, 3 4 3,Diagor 100 110 2 3 2 1 A 112 01 3 2 001 100 0 3 4 0 003 getwe(1),COperating 32                                              
The canonical form of the given quadratic form is Here ρ(A) = 3, index = 1, signature = 1 – (2) = -1. Note:- In this problem the non-singular transformation which reduces the given quadratic form into the canonical form is X = PY. i.e.,   2 3 2 2 2 1 3 2 1 321 yy 3 4 3y y y y 100 0 3 4 0 003 yyyAP)Y(P'Y'                                                          3 2 1 112 01 3 2 001 y y y z y x
Thank you 

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Introduces eigenvalues and eigenvectors concepts; slide 2 lists group members.

Defines eigenvalues and eigenvectors; discusses characteristic matrix and equation.

Expands on properties of eigenvalues, focusing on algebraic and geometric multiplicities.

Cayley-Hamilton theorem states matrices satisfy their characteristic equations; includes practical applications.

Demonstrates theorem through an example, verifying that A satisfies its characteristic polynomial.

Discusses matrix similarity, diagonalization, and conditions like the existence of eigenvectors.

Explains how to reduce a matrix to diagonal form using eigenvalues and transformation techniques.

Defines orthogonal similarity and implications on eigenvectors of symmetric matrices.

Introduces quadratic forms, nature based on eigenvalues, and conditions for definiteness.

Analyzes specific quadratic forms, determines their nature based on calculated eigenvalues.

Discusses linear transformations on quadratic forms and demonstrates with examples, providing canonical transformations.

Eigen value and eigen vector

  • 1.
    Eigen Value AndEigen Vector GUIDED BY: MANSI K. DESAI
  • 2.
    Group Members:- SR NO.NAME ENROLLMENT NUMBER 1. RANA PAYAL MAHESHBHAI 151100106072 2. PATEL RUTVIJ GANESHBHAI 151100106073 3. SAVALIYAAKSHAY JAYANTILAL 151100106074 4. TANDEL PAYALBENVITHTHALBHAI 151100106075 5. TANDEL SNEHAL MAHENDRABHAI 151100106076 6. VAGHELA SAHIL PREMJIBHAI 151100106077 7. EHSANULLAHAYDIN 151100106078 8. SHUAIB KOHEE 151100106079 9. WAHEDULLAH EHSAS 151100106080
  • 3.
    Eigenvalues and Eigenvectors •If A is an n x n matrix and λ is a scalar for which Ax = λx has a nontrivial solution x ∈ ℜⁿ, then λ is an eigenvalue of A and x is a corresponding eigenvector of A. – Ax=λx=λIx – (A-λI)x=0 • The matrix (A-λI ) is called the characteristic matrix of a where I is the Unit matrix. – • The equation det (A-λI )= 0 is called characteristic equation of A and the roots of this equation are called the eigenvalues of the matrix A.The set of all eigenvectors is called the eigenspace of A corresponding to λ.The set of all eigenvalues of a is called spectrum of A.
  • 4.
    Characteristic Equation ▪ IfA is any square matrix of order n, we can form the matrix , where is the nth order unit matrix. ▪ The determinant of this matrix equated to zero, ▪ i.e., is called the characteristic equation of A. 0 λa...aa ............ a...λaa a...aλa λA nnn2n1 2n2221 1n1211      I
  • 5.
    • On expandingthe determinant, we get • where k’s are expressible in terms of the elements a •The roots of this equation are called Characteristic roots or latent roots or eigen values of the matrix A. •X = is called an eigen vector or latent vector 0k...λkλkλ1)( n 2n 2 1n 1 nn               4 2 1 ... x x x
  • 6.
    Properties of EigenValues:- 1. The sum of the eigen values of a matrix is the sum of the elements of the principal diagonal. 2.The product of the eigen values of a matrix A is equal to its determinant. 3. If is an eigen value of a matrix A, then 1/ is the eigen value of A-1 . 4.If is an eigen value of an orthogonal matrix, then 1/ is also its eigen value. 
  • 7.
    PROPERTY 1:- Ifλ1, λ2,…, λn are the eigen values of A, then i. k λ1, k λ2,…,k λn are the eigen values of the matrix kA, where k is a non – zero scalar. ii. are the eigen values of the inverse matrix A-1. iii. are the eigen values of Ap, where p is any positive integer.
  • 8.
    Algebraic & GeometricMultiplicity ▪ If the eigenvalue λ of the equation det(A-λI)=0 is repeated n times then n is called the algebraic multiplicity of λ.The number of linearly independent eigenvectors is the difference between the number of unknowns and the rank of the corresponding matrix A- λI and is known as geometric multiplicity of eigenvalue λ.
  • 9.
    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              
  • 10.
    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
  • 11.
     n n-1n-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) Note 1:- Premultiplying equation (1) by A-1 , we have 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
  • 12.
    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.
  • 13.
    Example 1:- 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      
  • 14.
  • 15.
    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           
  • 16.
    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
  • 17.
    Example 2:- Given findAdj 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      
  • 18.
    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-
  • 19.
  • 20.
  • 21.
    Similarity of Matrix ▪IfA & B are two square matrices of order n then B is said to be similar to A, if there exists a non-singular matrix P such that, B= P-1AP 1. Similarity matrices is an equivalence relation. 2. Similarity matrices have the same determinant. 3. Similar matrices have the same characteristic polynomial and hence the same eigenvalues. If x is an eigenvector corresponding to the eigenvalue λ, then P-1x is an eigenvector of B corresponding to the eigenvalue λ where B= P-1AP.
  • 22.
    Diagonalization ▪ A matrixA is said to be diagonalizable if it is similar to diagonal matrix. ▪ A matrix A is diagonalizable if there exists an invertible matrix P such that P-1AP=D where D is a diagonal matrix, also known as spectral matrix.The matrix P is then said to diagonalize A of transform A to diagonal form and is known as modal matrix.
  • 23.
    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 ofA into a square matrix.
  • 24.
    Example:- Reduce the matrixA = to diagonal form by similarity transformation. Hence find A3. Solution:- Characteristic equation is => λ = 1, 2, 3 Hence eigenvalues of A are 1, 2, 3.             300 120 211 0             λ-300 1λ-20 21λ1-
  • 25.
    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
  • 26.
    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
  • 27.
    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
  • 28.
    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
  • 29.
  • 30.
    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
  • 31.
    Orthogonally Similar Matrices ▪If A & B are two square matrices of order n then B is said to be orthogonally similar to A, if there exists orthogonal matrix P such that B= P-1AP Since P is orthogonal, P-1=PT B= P-1AP=PTAP 1. A real symmetric of order n has n mutually orthogonal real eigenvectors. 2. Any two eigenvectors corresponding to two distinct eigenvalues of a real symmetric matrix are orthogonal.
  • 32.
    Diagonalises the matrixA = by means of an orthogonal transformation. Solution:- Characteristic equation of A is 32 Example :- 204 060 402 66,2,λ 0λ)16(6λ)λ)(2λ)(6(2 0 λ204 0λ60 40λ2      
  • 33.
    I                                                 1 1 2 3 1 1 2 3 1 3 2 1 3 1 1 2 3 1 1 1 x when λ = -2,let X = x be the eigen vector 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
  • 34.
    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
  • 35.
                                  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
  • 36.
                                                             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 .
  • 37.
    Quadratic Forms DEFINITION:- A homogeneouspolynomial of second degree in any number of variables is called a quadratic form. For example, ax2 + 2hxy +by2 ax2 + by2 + cz2 + 2hxy + 2gyz + 2fzx and ax2 + by2 + cz2 + dw2 +2hxy +2gyz + 2fzx + 2lxw + 2myw + 2nzw are quadratic forms in two, three and four variables
  • 38.
    In n –variables x1,x2,…,xn, the general quadratic form is In the expansion, the co-efficient of xixj = (bij + bji). 38    n 1j n 1i jiijjiij bbwhere,xxb ).b(b 2 1 awherexxaxxb baandaawherebb2aSuppose jiijijji n 1j n 1i ijji n 1j n 1i ij iiiijiijijijij      
  • 39.
    Hence every quadraticform can be written as     getweform,matrixin formsquadraticofexamplessaidabovethewritingNow .x,...,x,xXandaAwhere symmetric,alwaysisAmatrixthethatso AX,X'xxa n21ij ji n 1j n 1i ij                 y x bh ha y][xby2hxyax(i) 22
  • 40.
  • 41.
    Two Theorems OnQuadratic Form Theorem(1): A quadratic form can always be expressed with respect to a given coordinate system as where A is a unique symmetric matrix. Theorem2: Two symmetric matrices A and B represent the same quadratic form if and only if B=PTAP where P is a non-singular matrix. AxxY T 
  • 42.
    Nature of QuadraticForm A real quadratic form X’AX in n variables is said to be i. Positive definite if all the eigen values ofA > 0. ii. Negative definite if all the eigen values of A < 0. iii. Positive semi definite if all the eigen values ofA 0 and at least one eigen value = 0. iv. Negative semi definite if all the eigen values of A 0 and at least one eigen value = 0. v. Indefinite if some of the eigen values ofA are + ve and others – ve.
  • 43.
    Find the natureof the following quadratic forms i. x2 + 5y2 + z2 + 2xy + 2yz + 6zx ii. 3x2 + 5y2 + 3z2 – 2yz + 2zx – 2xy Solution:- i. The matrix of the quadratic form is 43 Example :-            113 151 311 A
  • 44.
    The eigen valuesof A are -2, 3, 6. Two of these eigen values being positive and one being negative, the given quadratric form is indefinite. ii. The matrix of the quadratic form is The eigen values of A are 2, 3, 6. All these eigen values being positive, the given quadratic form is positive definite.               311 151 113 A
  • 45.
    Linear Transformation ofa Quadratic Form ▪ Let X’AX be a quadratic form in n- variables and let X = PY ….. (1) where P is a non – singular matrix, be the non – singular transformation. ▪ From (1), X’ = (PY)’ =Y’P’ and hence X’AX =Y’P’APY =Y’(P’AP)Y =Y’BY …. (2) where B = P’AP.
  • 46.
    Therefore,Y’BY is alsoa quadratic form in n- variables. Hence it is a linear transformation of the quadratic form X’AX under the linear transformation X = PY and B = P’AP. Note. (i) Here B = (P’AP)’ = P’AP = B (ii) ρ(B) = ρ(A) Therefore, A and B are congruent matrices.
  • 47.
    Reduce 3x2 +3z2 + 4xy + 8xz + 8yz into canonical form. Or Diagonalises the quadratic form 3x2 + 3z2 + 4xy + 8xz + 8yz by linear transformations and write the linear transformation. Or Reduce the quadratic form 3x2 + 3z2 + 4xy + 8xz + 8yz into the sum of squares. 47 Example:-
  • 48.
    Solution:- The givenquadratic form can be written as X’AX where X = [x, y, z]’ and the symmetric matrix A = Let us reduce A into diagonal matrix. We know tat A = I3AI3.           344 402 423                                          100 010 001 344 402 423 100 010 001 344 402 423
  • 49.
                                               21 31OperatingR ( 2 / 3),R ( 4 / 3) (for A onL.H.S.andpre factor on R.H.S.), we get 3 2 4 1 0 0 1 0 0 4 4 2 0 1 0 A 0 1 0 3 3 3 0 0 1 4 7 4 0 0 1 3 3 3                                                      100 010 3 4 3 2 1 A 10 3 4 01 3 2 001 3 7 3 4 0 3 4 3 4 0 423 getweR.H.S),onfactorpostandL.H.S.onA(for 4/3)(C2/3),(COperating 3121
  • 50.
  • 51.
    The canonical formof the given quadratic form is Here ρ(A) = 3, index = 1, signature = 1 – (2) = -1. Note:- In this problem the non-singular transformation which reduces the given quadratic form into the canonical form is X = PY. i.e.,   2 3 2 2 2 1 3 2 1 321 yy 3 4 3y y y y 100 0 3 4 0 003 yyyAP)Y(P'Y'                                                          3 2 1 112 01 3 2 001 y y y z y x
  • 52.