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Engineering Mathematics

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Published in: Mathematics
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Rules for Diagonalization of Matrix and example problems

Devi / Chennai

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Qualification: B.Tech/B.E. ( Saveetha Engineering College, Chennai - 2017)

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  1. Similarity of matrices: A square matrix B of order n is called similar to a square matrix A of order n if B = P-IAP for some non-singular n x n matrix P. This transformation of a matrix P to B is called a similarity transformation. Diagonalisaion of a matrix by means of similarity transformation: If a square matrix A of order n has n linearly independent Eigen vectors, then a Matrix P can be found such that P-IAP is a diagonal matrix. Note: (i) The matrix P which diagonalise A is called modal matrix of A and the resulting diagonal matrix D is known as the spectral matrix of A. (ii) A square matrix of order n can be diagonalised by a similarity transformation if and only if it has n linearly independent Eigenvectors. Working Rule for Diagonalisation : [Similarity transformation] Let A be any square matrix of order n. Step 1. To find the characteristic equation. Step 2. To solve the characteristic equation. Step 3. To find the Eigenvectors. Step 4. Form the model matrix P, its columns the Eigenvectors of A. step 5. Find PI Step 6. Calculate AP step 7. Calculate D = P-IAP 1. 1 Diagonalise the matrix A 1 Solution: 1 by an orthogonal transformation. (Dec. 2001) 2 The characteristic equation of A is 1 31 -k 1 The eigen values are 2 To find the eigen vectors , (A 1-1 1 1 When 1 2 2 11)X 0 0
  2. X2 = 2 2 2 When 1 — 2 2 2 2 2 2 2 10 -k The modal matrix N 10 + + 26 2 10 2US 10 2dS 5 2(5+2J5 2(5+2U5 5 N/S 3 + US The values of is 2US) 265 Similarly the value of is 3 2 2. Diagonalise the matrix A Solution: 2 1 1 1 1 2 1 2 by means of orthogonal transformation. 1 For the given matrix DI —4, D The characteristic equation is 13 + 412 — 1 + 4 — 0 The eigen values are 1, The eigen vectors are given by (A Il)X = ()
  3. 1 2 For A 2 i.e. 1, 2 Solving by cross multiplication method For 2 1, 1 Solving by cross multiplication method 0 For 5 4, 1 Solving by cross multiplication method
  4. 1 The modal matrix is N 2 1 o 1 1 Normalised modal matrix is N To diagonalise by orthogonal transformation, NT AN D 1 o o 1 o o 0 4 D 1 1 2 0 I 2 o 3 o Diagonalise the matrix A (Nov 2006) Solution: 0 by means of orthogonal transformation. 2 Let A 2 0 I o 3 o 1 0 2 For the given matrix DI — 7 , 15, D3 9 The characteristic equation is 13 — 712 + 151 The eigen values are 1, 3, 3 The eigen vectors are given by (A 11)X = () 9
  5. 2 i.e. For A 0 0 1, Solving by cross multiplication method For A 2 0 1, o 2 Solving by cross multiplication method 2 Forl 3, o 0 2 0 0&.. 0 Substituting in 0 if x2 0 1 if , W e get Xl 1
  6. All the three eigen vectors are orthogonal to each other, The normalized modal matrix is Using orthogonal transformation, NT AN D o 0 D o o 1

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