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Matrices

LINEAR ALGEBRA

2023

LINEAR ALGEBRA

Linear Algebra: Illustration of a few Engineering Examples with Application of Linear Algebra (Lecture by the Respective Domain Faculty members) - Eigenvalues and Eigenvectors of a real matrix – Characteristic equation – Properties of Eigenvalues and Eigenvectors – Cayley-Hamilton theorem – Diagonalization of matrices – Reduction of a quadratic form to canonical form by orthogonal transformation – Nature of quadratic forms. Linear Algebra, mathematical discipline that deals with vectors and matrices and, more generally, with vector spaces and linear transformations. Unlike other parts of mathematics that are frequently invigorated by new ideas and unsolved problems, linear algebra is very well understood. Its value lies in its many applications, from mathematical physics to modern algebra and coding theory. Recapitulations: Matrix: Matrix is an array of numbers in row and columns. A matrix with ‘𝑚’ rows and ‘𝑛’ columns is called as 𝑚 × 𝑛 matrix. Square Matrix: If number of rows and columns are equal, then the matrix is called as square matrix. 1 2 3 Ex. 𝐴 = [4 5 6] 7 8 9 The elements 1, 5, 9 are called as leading diagonal elements. Unit Matrix: A matrix which has unity for all its diagonal elements is called as unit matrix or identity matrix of order ‘𝑛’ and is denoted by 𝐼. 1 0 0 Ex. 𝐼 = [0 1 0] 0 0 1 Eigenvalues and Eigenvectors: Let A be a square matrix, if there exist a scalar 𝜆 and a nonzero column matrix 𝑋 such that (𝐴 − 𝜆𝐼)𝑋 = 0, then 𝜆 is called an Eigenvalue of 𝐴 and 𝑋 is called Eigenvector corresponding to Eigenvalue of 𝐴. Characteristic Equations:

• The characteristic equation of 3 × 3 matrix A is 𝜆^3 − 𝑠 1 𝜆^2 + 𝑠 2 𝜆 − 𝑠 3 = 0, where 𝑠 1 = sum of leading diagonal elements, 𝑠 2 = sum of minor of leading diagonal elements,

[0]

[ 1 5 − 𝜆 1 ] [𝑥 2 ] = [0]

3 1 1 − 𝜆 𝑥 3 0 i.e., Case i). If 𝜆 = 3, (A) becomes (1 − 𝜆)𝑥 1 + 𝑥 2 + 3𝑥 3 = 0 𝑥 1 + (5 − 𝜆)𝑥 2

• 𝑥 3 = 0 } ---------- (A) 3𝑥 1 + 𝑥 2 + (1 − 𝜆)𝑥 3 = 0 (−2)𝑥 1 + 𝑥 2 + 3𝑥 3 = 0 -------------(1) 𝑥 1 + (2)𝑥 2 + 𝑥 3 = 0 ------------- (2) 3𝑥 1 + 𝑥 2 + (−2)𝑥 3 = 0 ------------- (3) Solving these equations 𝑥 1 𝑥 2 𝑥 3 = 1 − 6 3 − (−2) −4 − 1 𝑥 1 𝑥 2 𝑥 3 = = −5 5 − −5 − Eigenvector for 𝜆 = 3 is 𝑋 1 = [ 5] or [ 1]. −5 − Case ii). If 𝜆 = 6, (A) becomes

3𝑥 1 + 𝑥 2 + (−5)𝑥 3 = 0 ------------- (6) Solving these equations 𝑥 1 𝑥 2 𝑥 3 = 1 − (−3) 3 − (−5) 5 − 1 𝑥 1 𝑥 2 𝑥 3 = = 4 8 4 4 1 Eigenvector for 𝜆 = 6 is 𝑋 2 = [ 8] or [ 2] 4 1 Case iii). If 𝜆 = −2, (A) becomes 3 𝑥 1 + 𝑥 2 + 3𝑥 3 = 0 ------------- (7) 𝑥 1 + (7)𝑥 2 + 𝑥 3 = 0 ------------- (8) 3𝑥 1 + 𝑥 2 + (3)𝑥 3 = 0 ------------- (9) Solving these equations 𝑥 1 𝑥 2 𝑥 3 = 1 − 21 3 − (3) 21 − 1 𝑥 1 𝑥 2 𝑥 3 = = −20 0 20 −20 − Eigenvector for 𝜆 = −2 is 𝑋 3 = [ 0 ] or [ 0]. 20 1 8 −6 2

i.e., (8 − 𝜆)𝑥 1 − 6𝑥 2 + 2𝑥 3 = 0 −6𝑥 1 + (7 − 𝜆)𝑥 2 − 4𝑥 3 = 0}------------- (A) 2𝑥 1 − 4𝑥 2 + (3 − 𝜆)𝑥 3 = 0 Case i). If 𝜆 = 0, (A) becomes (8)𝑥 1 − 6𝑥 2 + 2𝑥 3 = 0 ------------- (1) −6𝑥 1 + (7)𝑥 2 − 4𝑥 3 = 0 ------------- (2) 2𝑥 1 − 4𝑥 2 + (3)𝑥 3 = 0 ------------- (3) Solving these equations 𝑥 1 𝑥 2 𝑥 3 = 24 − 14 −12 − (−32) 56 − 36 𝑥 1 𝑥 2 𝑥 3 = = 10 20 20 10 1 Eigenvector for 𝜆 = 0 is 𝑋 1 = [ 20] or [2]. 20 2 Case ii). If 𝜆 = 3, (A) becomes 5𝑥 1 − 6𝑥 2 + 2𝑥 3 = 0 ------------- (4) −6𝑥 1 + (4)𝑥 2 − 4𝑥 3 = 0 ------------- (5) 2𝑥 1 − 4𝑥 2 + (0)𝑥 3 = 0 ------------- (6) Solving these equations 𝑥 1 𝑥 2 𝑥 3 = = 24 − 8 −12 − (−20) 20 − 36 𝑥 1 𝑥 2 𝑥 3

Eigenvector for 𝜆 = 3 is 𝑋 1 = [ 8 ] or [ 1 ]. −16 − Case iii). If 𝜆 = 15, (A) becomes −7𝑥 1 − 6𝑥 2 + 2𝑥 3 = 0 ------------- (7) −6𝑥 1 − 8𝑥 2 − 4𝑥 3 = 0 ------------- (8) 2𝑥 1 − 4𝑥 2 − 12𝑥 3 = 0 ------------- (9) Solving these equations 𝑥 1 𝑥 2 𝑥 3 = 24 − (−16) −12 − (28) 56 − 36 𝑥 1 𝑥 2 𝑥 3 = = 40 −40 20 40 2 Eigenvector for 𝜆 = 15 is 𝑋 1 = [ −40] or [−2]. 20 1 Practice Problems

1. Find the Eigenvalues and Eigenvectors of the matrices: (a) [ 4 3 ] (b) [ 1 4 ] 2 9 3 2 7 −2 0
2. Find the Eigenvalues and Eigenvectors of the matrix[−2 6 −2]. 0 −2 5

2. If −2 and 6 are the two Eigenvalues of 𝐴 = [1 5 1]. Find the third Eigenvalue of 3 1 1 𝐴. 1 1 3 Solution. Let 𝐴 = [1 5 1]. 3 1 1 Let 𝜆 1 , 𝜆 2 , 𝜆 3 are the Eigenvalues of 𝐴. We know that, Sum of Eigenvalues = Sum of main diagonals. 𝜆 1 + 𝜆 2 + 𝜆 3 = 𝑎 11 + 𝑎 22 + 𝑎 33 −2 + 6 + 𝜆 3 = 1 + 5 + 1 𝜆 3 = 3. Therefore, the Eigenvalues of 𝐴 are −2, 6, 3. Practice Problems 2 2 1
3. Two Eigenvalues of the matrix 𝐴 = [1 3 1] are equal to 1 each. Find the third 1 2 2 Eigenvalue.
   2. If 2, −1,−2 are the Eigenvalues of the matrix 𝐴, then what is the Eigenvalues of the matrix 𝐴 2 − 2𝐼. Cayley-Hamilton Theorem Every square matrix satisfies its own characteristic equation. Uses. To calculate 1). the positive integral powers of 𝐴 and 2). the inverse of a non-singular square matrix 𝐴. Problems 1 3 7

1. Verify Cayley-Hamilton theorem for the matrix [4 2 3] and also use it to find 𝐴 − . 1 2 1 1 3 7 Solution. Given 𝐴 = [4 2 3]

2 − 4𝐴 − 20𝐼 − 35𝐴 − = 0 𝐴 − = [𝐴 2 − 4𝐴 − 20𝐼]. 20 23 23 1 3 7 35 0 0 𝐴 − = [[15 22 37] − 4 [4 2 3] − 20 [ 0 35 0 ]] 10 9 14 1 2 1 0 0 35 20 23 23 4 12 28 20 0 0 = [[15 22 37] − [16 8 12] − [ 0 20 0 ]] 10 9 14 4 8 4 0 0 20 −4 11 − = [−1−6 25 ]. 6 1 − 2 1 1

2. Find the characteristic equation of the matrix, 𝐴 = [0 1 0] and hence compute 1 1 2 𝐴 − . Also, find the matrix represented by 𝐴 8 − 5𝐴 7

• 7𝐴 6 − 3𝐴 5
• 𝐴 4 − 5𝐴 3

8𝐴 2 − 2𝐴 + 𝐼. 2 1 1 Solution. Given 𝐴 = [0 1 0] 1 1 2 The Characteristic equation of 𝐴 is 𝜆 3 − 𝑠 1 𝜆 2 + 𝑠 2 𝜆 − 𝑠 3 = 0. 𝑠 1 = 2 + 1 + 2 = 5 1 0 2 1 2 1 (2 − 0) + (4 − 1) + (2 − 0) = 7 𝑠 2 = |1 2| + |1 2| + |0 1| = 2 1 1 𝑠 3 = |𝐴| = |0 1 0| = 2(2 − 0) − 1(0 − 0) + 1(0 − 1) = 3 1 1 2 Therefore, the characteristic equation is 𝜆 3 − 5𝜆 2

• 7𝜆 − 3 = 0. By Cayley-Hamilton theorem 𝐴 3 − 5𝐴 2 + 7𝐴 − 3𝐼 = 0---------------(i)

Multiplying (i) by 𝐴 − , we get 𝐴 2 − 5𝐴 + 7𝐼 − 3𝐴 − = 0 or 𝐴 − = 1 [𝐴 2 − 5𝐴 + 7𝐼] 3 2 1 1 2 1 1 4 + 0 + 1 2 + 1 + 1 2 + 0 + 2 5 4 4 Now, 𝐴 2 = [0 1 0] [0 1 0] = [0 + 0 + 0 0 + 1 + 0 0 + 0 + 0] = [0 1 0] 1 1 2 1 1 2 2 + 0 + 2 1 + 1 + 2 1 + 0 + 4 4 4 5

If a square matrix 𝐴 of order 𝑛 has 𝑛 linearly independent Eigenvectors, then a matrix 𝑃 can be found such that 𝑃 − 𝐴𝑃 is a diagonal matrix. Power of a matrix : Diagonalisation of a matrix is quite useful for obtaining powers of a matrix. 𝜆1𝑛 0 0 i.e., 𝐴 𝑛 = 𝑃𝐷 𝑛 𝑃 − where 𝐷 𝑛 = [ 0 𝜆 2 𝑛 0 ] 0 0 𝜆3𝑛 1 1 3

1. Reduce the matrix 𝐴 = [1 5 1] to the diagonal form. 3 1 1 Soln. The Eigenvalues of 𝐴 (found in Ex 1. Page 2) are -2, 3, 6 and the Eigenvectors are (−1,0,1),(1, −1,1), (1,2,1). Writing these Eigenvectors as the three columns, the required transformation matrix is −1 1 1 𝑃 = [ 0−1 2]. 1 1 1 −1 1 1 𝑎 1 𝑏 1 𝑐 1 To find 𝑃−1, |𝑃| = | 0 −1 2| = |𝑎 2 𝑏 2 𝑐 2 | (say) 1 1 1 𝑎 3 𝑏 3 𝑐 3 −1 2 𝐴 1 = | 11| = −3, 𝐵 1 = 2, 𝐶 1 = 1, 𝐴 2 = 0, 𝐵 2 = −2, 𝐶 2 = 2, 𝐴 3 = 3, 𝐵 3 = 2, 𝐶 3 = 1. Also |𝑃| = 𝑎 1 𝐴 1 + 𝑏 1 𝐵 1 + 𝑐 1 𝐶 1 = −1(−3) + 1(2) + 1(1) = 6 𝐴 1 𝐴 2 𝐴 3 −3 0 3 𝑃−1 = 1 [𝐵 1 𝐵 2 𝐵 3 ] = [ 2 −2 2] |𝑃|

Thus 𝐷 = 𝑃 − 𝐴𝑃 = [ 0 0

3 0].

2. If the diagonal form of the matrix 𝐴 , then find 𝐴 4 . Solution: The characteristic equation of 𝐴 is −1 − 𝜆 2 − | 1 2 − 𝜆 1 | = 0 −1 −1 −𝜆 𝜆 3 − 𝜆 2 − 5𝜆 + 5 = 0. Solving, we get 𝜆 as the Eigenvalues of 𝐴. When 𝜆 = 1, the corresponding Eigenvector is given by −2𝑥 + 2𝑦 − 2𝑧 = 0, 𝑥 + 𝑦 + 𝑧 = 0, −𝑥 − 𝑦 − 𝑧 = 0. 𝑥 𝑦 𝑧 Solving the first two equations, we get = = giving the Eigenvector (1,0, −1), 4 0 − when 𝜆 , the corresponding Eigenvector is given by , 𝑥 ,. Solving 2 nd and 3 rd equations. We get 𝑥 𝑦 𝑧 𝑥 𝑦 𝑧 Giving the Eigenvector. Similarly, the Eigenvector corresponding to 𝜆. Writing the three Eigenvectors as the three columns, we get the transformation matrix as 𝑃 𝑏 1 𝑐 1 To find 𝑃 − , |𝑃| 𝑏 2 𝑐 2 | (say 𝑏 3 𝑐 3 𝐴

The symmetric matrix 𝐴 is called the matrix of the quadratic 𝑎𝑛 𝑎𝑛𝑛 form 𝑄. The matrix corresponding to the quadratic form is co-eff 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 1 1 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 1 𝑥 2 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 1 𝑥 3 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 2 𝑥 1 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 22 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 2 𝑥 3 2 [ 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 3 𝑥 1 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 3 𝑥 2 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 3 ] Note: 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 1 𝑥 2 = 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 2 𝑥 1 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 2 𝑥 3 = 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 3 𝑥 2 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 1 𝑥 3 = 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 3 𝑥 1.

1. Write the matrix of the quadratic form 2𝑥 1 2 − 2𝑥 2 2 + 4𝑥 3 2 + 2𝑥 1 𝑥 2 − 6𝑥 1 𝑥 3 + 6𝑥 2 𝑥 3. Soln. Matrix of the quadratic form is 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 1 2 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 1 𝑥 2 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 1 𝑥 3 𝑄 = 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 2 𝑥 1 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 2 2 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 2 𝑥 3 2 (2) (−6) 2 1 − 1 [ 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 3 𝑥 1 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 3 𝑥 2 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 3 ]

(6) = [ 1 −2 3 ].

[(−6)(6)^4 ]

Nature of a Quadratic Form. Let 𝑄 = 𝑋′𝐴𝑋 be a quadratic form in a variables x 1 , x 2 , … … … xn. A real quadratic form 𝑋′𝐴𝑋 in a variable is said to be (i) Positive definite if all the Eigenvalues of A. (ii) negative definite if all the Eigenvalues of A. (iii) Positive semidefinite if all the Eigenvalues of A and at least one Eigenvalue=0. (iv) Negative semidefinite if all the Eigenvalues of A and at least one Eigenvalue=0. (v) Indefinite if some of the Eigenvalues of A are positive and others negative.

1. Reduce the quadratic form 6𝑥^2 + 3𝑦^2 + 3𝑧^2 − 4𝑥𝑦 − 2𝑦𝑧 + 4𝑧𝑥 into canonical form by an orthogonal transformation. Also discuss its nature. Soln. Given 6𝑥 2

• 3𝑦 2
• 3𝑧 2 − 4𝑥𝑦 − 2𝑦𝑧 + 4𝑧𝑥 The matrix of the quadratic form is 𝑐𝑜 − 𝑒𝑓𝑓 𝑥 2 𝑐𝑜 − 𝑒𝑓𝑓 𝑥𝑦 𝑐𝑜 − 𝑒𝑓𝑓 𝑥𝑧 6 −2 2 𝐴 = 𝑐𝑜 − 𝑒𝑓𝑓 𝑦𝑥 𝑐𝑜 − 𝑒𝑓𝑓 𝑦 2 𝑐𝑜 − 𝑒𝑓𝑓 𝑦𝑧 = [−2 3 −1] 2 −1 3 𝑐𝑜 − 𝑒𝑓𝑓 𝑧 2 ] [𝑐𝑜 − 𝑒𝑓𝑓 𝑧𝑥𝑐𝑜 − 𝑒𝑓𝑓 𝑧𝑦 The characteristic equation of A is 𝜆 3 − 𝑠 1 𝜆 2 + 𝑠 2 𝜆 − 𝑠 3 = 0. 2 1