Diagonalization of matrix

Diagonalization of matrix

1. DEPARTMENT OF PHYSICS EIGEN VALUES, EIGEN VECTORS, & DIAGONALIZATION Submitted by :-Vandana (M.Sc-1 st .)

2. CONTENTS  Eigen values  Eigen vectors  Similarity Transformation  Diagonalization of a matrix

3. EIGEN VALUES To find eigen values first we find the characteristic equation given by | A- λ I | = 0 Now the roots of the equation or values of λ are called eigen values.

4. EXAMPLE Let A be a square matrix of order 3  3 1 1 3 A= 1 5 1 3 1 1 According to Cayley Hamilton theorem | A- λ I | = Where λ is any scalar, and I is the 3×3 unit matrix

5. The determinant of these matrix equated to zero i.e. 1 - λ 1 3 1 5 - λ 1 =0 3 1 1- λ On expanding the determinant, we get the equation i.e. λ³-7 λ² +36=0 On solving these equations we get, λ=-2,3,6 These are called eigen values.

6. PROPERTIES OF EIGEN VALUES 1. Any square matrix A and its transpose A ՚ have the same eigen values. 2. The sum of the eigen values of a matrix is equal to the trace of the matrix. 3. The product of the eigen values of a matrix A is equal to the determinant of A. 4. The eigen values of a triangular matrix are just the diagonal elements of the matrix. 5. The eigen values of an idempotent matrix are either zero or unity.

7. 6. The sum of the eigen values of a matrix is the sum of the elements of the principal diagonal. 7. If λ is an eigen value of a matrix A then 1/ λ is the eigen value of Aˉ ¹ . 8. If λ is an eigen value of an orthogonal matrix then 1/ λ is also its eigen value. 9. If λ₁ , λ₂ , λ₃ ,……., λ ո are the eigen values of a matrix A then A ͫ has the eigen values λ₁ ͫ , λ₂ ͫ ,…….., λ ո ͫ .

8. EIGEN VECTOR An eigen vector of a square matrix A is a non zero vector X such that for some number λ , we have, AX= λ X AX- λ IX=0 [A- λ I]X=0 Where 0 represents the zero vector.

9. EXAMPLE Now eigen vector corresponding to λ =-2,3,6 For λ =-2 3x ₁ +x ₂ +3x ₃ =0 x ₁ +7x ₂ +x ₃ =0 3x ₁ +x ₂ +3x ₃ =0 By solving these equations, we get x ₁=1, x₂=0, x₃=-1

10. For λ =3 -2x ₁+x₂+3x₃=0 x₁+2x₂+x₃=0 3x₁+x₂-2x₃=0 By solving these equations, we get X₁=1, x₂=-1, x₃=1

11. For λ =6 -5x₁+x₂+3x₃=0 x₁-x₂+x₃=0 3x₁+x₂-5x₃=0 By solving these equations, we get x₁=1, x₂=2, x₃=1 Hence λ=-2(1,0,-1) λ=3(1,-1,1) λ=6(1,2,1) Hence corresponding to each eigen value there exist an eigen vector.

12. PROPERTIES OF EIGEN VECTORS 1. The eigen vector X of a matrix A is not unique. 2. If λ₁ , λ₂ ,…… λ ո be distinct eigen values of an n ᵡ n matrix then corresponding to eigen vectors X ₁, X₂,……..X ո form a linearly independent set. 3. If two or more eigen values are equal it may or may not be possible to get linearly independent eigen vectors corresponding to the equal roots. 4. Two eigen vectors X ₁ and X₂ are called orthogonal vectors if X₁ ՚ X₂=0. 5. Eigen vectors of a symmetric matrix corresponding to different eigen values are orthogonal.

13. NORMALISED FORM OF A VECTORS To find normalised form of a we divide each b c element by √ a ²+b²+c². For example:- Normalised form of 1 1/3 2 = 2/3 2 2/3

14. ORTHOGONAL VECTORS Two vectors X and Y are said to be orthogonal if X ₁ ᵀX ₂=X₂ ᵀx ₁=0. Let a matrix A is 1 0 -1 1 2 1 2 2 3 According to Cayley Hamilton theorem |A- λ I|=0

15. Characteristic equation is 1- λ 0 -1 1 2- λ 1 =0 2 2 3- λ λ =1,2,3 are three distinct eigen values of A. For λ =1 -x₃=0 x₁+x₂+x₃=0 2x₁+2x₂+2x₃=0

16. 1 X ₁ = k -1 For λ =2 2 X₂ = K -1 -2 For λ =3

17. For λ =3 1 X₃=k -1 -2 X₁ ᵀX ₂=3≠0, X₂ ᵀx ₃=7≠0, X₃ ᵀX ₁=2≠0 Thus there are three distinct eigen vectors. So X₁, X₂, X₃ are not orthogonal eigen vectors.

18. SIMILARITY TRANSFORMATION Let A and B be 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 ˉ ¹ AP This equation is called a similar transformation.

19. DIAGONALIZATION OF A MATRIX Diagonalization of a matrix A is the process of reduction of A to a diagonal form ‘D’. If A is related to D by a similarity transformation such that D=P ˉ ¹ AP then A is reduced to the diagonal matrix D through model matrix P. D is also called spectral matrix of A.

20. EXAMPLE Let a matrix 6 -2 2 A= -2 3 -1 2 -1 3 By solving this matrix by Cayley Hamilton th ͫ |A- λ I|=0 We get λ =2,2,8 These are called eigen values.

21. Eigen vector for λ =2 X₁= 1 1 Eigen vector for λ =2 again 1 X₂= 3 1

22. Eigen vector for λ =8 2 X₃= -1 1 Then power of a matrix is 0 1 2 P= 1 3 -1 1 1 1 4 1 -7 Pˉ ¹ = -1/6 -2 -2 2 -2 1 -1

23. Then 2 0 O D = P ˉ ¹ AP= 0 2 0 0 0 1 Thus D is the diagonal matrix.

24. REFERENCES ● H.K Dass, Dr. Rama Verma,”Mathematical Physics”, S. Chand & Company PVT.LTD., New Delhi. ● Dr. B.S. Grewal”Higher Engineering Mathematics”, Khanna Publishers, New Delhi, 2007.