Eigen Values and Eigen Vectors - GATE Study Material in PDF

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1 Eigen Values and Eigen Vectors - GATE Study Material in PDF Some of the most used concepts in Linear Algebra and Matrix Algebra are Eigen Values and Eigen Vectors. This makes Eigen Values and Eigen Vectors a very important concept in Engineering Mathematics if you are appearing for GATE Download this free GATE Study Material in PDF. Also useful for IES, BSNL, DRDO, BARC and other competitive exams. A non zero vector X such that AX = λx is said to be an Eigen vector (or)characteristic vector and corresponding values of the scalar λ is said to be Eigen values (or)latent roots (or) characteristic roots. I. e. AX = λx [A λi]x = 0 system of homogenous linear equations Since X is a non zero vector, the system should have a non zero solution hence ρ(a λi) < n A λi = 0 Characteristic equation of A a 11 λ a 12 a 13. a 1n ie a 21 : a 22 λ a 23. a 2n = 0 a n1 a n2 a n3. a nn λ = ( 1) n [a 0 λ n + a 1 λ n 1 + a 2 λ n 2 +. a n ] = 0 So λ = λ 1, λ 2, λ 3 λ n are Eigen values Properties of Eigen Values and Eigen Vectors: 1 P a g e

2 i. For Diagonal, Upper, lower and scalar matrices the diagonal elements becomes the Eigen value ii. Let A = [ a b c d ] then characteristic equation is given by λ2 trace(a)λ + A = 0 a 11 a 12 a 13 iii. Let A = [ a 21 a 22 a 23 ] then characteristic equation is given by a 31 a 32 a 33 λ 3 + trace(a)λ 2 (M 11 + M 22 + M 33 )λ + A = 0 Here M 11, M 22 and M 33 are the minors of the elements a 11, a 22 and a 33 respectively. iv. For even order coefficient of λ n = 1 and for odd order the coefficient of λ n = 1. v. The constant term in characteristic equation is always represents the determinant of a matrix. vi. The sum of Eigen values of a matrix is equal to the trace of a matrix whereas product of eigen values is equal to Determinant of a matrix. vii. If λ is Eigen value of a Matrix A then i. Kλ is Eigen value of (KA) ii. λ K is the Eigen value of A K iii. (λ ± K) are the Eigen values of (A ± K) a 0 + a 1 λ + a 2 λ 2 is an Eigen value of (a 0 + a 1 A + a 2 A 2 ) viii. If λ is an Eigen values of a non singular matrix A, then i. 1 λ is the Eigen value of (A 1 ) 2 P a g e ii. A λ is the Eigen value of (AdjA) iii. If λ is an Eigen value of an orthogonal matrix then 1 becomes the λ another Eigen value of the same matrix iv. Eigen values of A and A T are same ix. If a + b (or)a + b is an Eigen value of a matrix A then (a ib)(or)a + b is also Eigen value of matrix A. x. The Eigen vectors corresponds to distinct Eigen values of a matrix are linearly independent

3 xi. The Eigen values of a symmetric matrix the Eigen values are either zero (or) purely imaginary. xii. The Eigen values of an orthogonal matrix are of unit modulus i. e. λ = 1 Cauley Hamilton Theorem: Every Square matrix satisfies its own characteristic equation i. e. A λi = ( 1) n [a 0 λ n + a 1 λ n 1 + a 2 λ n 2 +. a n ] is the characteristic polynomial then a 0 A n + a 1 A n 1 + a 2 A n 2 + a n I = 0 Example 1: Find the Eigen values and Eigen vectors of the matrix given below A = [ ] Solution: Trace of A = = 4 Determinant of A = A = (2 2) (4 4) = 12 Characteristic equation, A λi = 0 λ 2 Trace (A) λ + A = 0 λ 2 4λ 12 = 0 (λ 6) (λ + 2) = 0 λ = 2 (or) 6 Hence Eigen values are λ 1 = 2 and λ 2 = 6 Case i If λ = 2 then AX = λx [ ] [x 1 ] = 2 [ x 1 ] 3 P a g e

4 2x = 2x 1 x 1 + = 0 4x = 2 x 1 + = 0 Let x 1 = k 1 then = k 1 X = [ k 1 k 1 ] = k 1 [ 1 1 ] Case ii If λ = 6 then AX = λx [ ] [x 1 ] = 6 [ x 1 ] 2x = 6x 1 x 1 = 4x = 6 x 1 = If x 1 = k 2 then = k 2 X = [ k 2 k 2 ] = k 2 [ 1 1 ] Since the Eigen values are different, the two Eigen vectors are linearly independent. Example If [1 0 1] T is an Eigen vector of a matrix [ 1 2 1]. Then find the corresponding Eigen value of the matrix. Solution: We know that AX = λx Here A = [ 1 2 1] and X = [1 0 1] T = [ 0 ] P a g e

5 [ 1 2 1] [ 0 ] = λ [ 0 ] [ ] = λ [ 0 ] [ 0 ] = λ [ 0 ] 1 1 λ = 1 Hence corresponding Eigen value is λ = 1 5 P a g e