# 3.2 The Characteristic Equation of a Matrix

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1 32 The Characteristic Equation of a Matrix Let A be a 2 2 matrix; for example A = 2 8 If v is a vector in R 2, eg v = [2, 3], then we can think of the components of v as the entries of a column vector (ie a 2 1 matrix) Thus [2, 3] 2 3 If we multiply this vector on the left by the matrix A, we get another column vector with two entries : A 2 = (2) + 8(3) = = (2) + ( 3)(3) 3 So multiplication on the left by the 2 2 matrix A is a function sending the set of 2 1 column vectors to itself - or, if we wish, we can think of it as a function from the set of vectors in R 2 to itself Note: In fact this function is an example of a linear transformation from R 2 into itself Linear transformations are functions which have certain interesting geometric properties Basically they are functions which can be represented in this way by matrices In general, if v is a column vector with two entries, then Av is a another vector (with two entries), which typically does not resemble v at all For example if v = 1 2 then Av = = 18 3 However, suppose v = 8 3 Then Av = = 4 15 = ie A 8 3 = 5 8 3, or 47

2 Multiplying the vector 8 (on the left) by the matrix 3 is the same as multiplying it by 5 with corre- Terminology: 8 is called an eigenvector for the matrix A = 3 sponding eigenvalue Definition 321 Let A be a n n matrix, and let v be a non-zero column vector with n entries (so not all of the entries of v are zero) Then v is called an eigenvector for A if A v = λ v, where λ is some real number In this situation λ is called an eigenvalue for A, and v is said to correspond to λ Note: λ is the symbol for the Greek letter lambda It is conventional to use this symbol to denote an eigenvalue Example 322 If A = and v = 1, then 2 Av = = 3 = 3 1 = 3v Thus 1 is an eigenvector for the matrix 1 1 corresponding to the eigenvalue Question: Given a n n matrix A, how can we find its eigenvalues and eigenvectors? 48

3 Answer: We are looking for column vectors v and real numbers λ satisfying Av = λ v ie λ v Av = = λi n v Av = = (λi n A) }{{} v = a n n matrix This may be regarded as a system of linear equations in which the coefficient matrix is λi n A and the variables are the n entries of the column vector v, which we can denote by,, x n We are looking for solutions to (λi n A) x n = This system always has at least one solution : namely = x 2 = = x n = - all entries of v are zero However this solution does not give an eigenvector since eigenvectors must be non-zero The system can have additional solutions only if det(λi n A) = (otherwise if the square matrix λi n A is invertible, the system will have = x 2 = = x n = as its unique solution) Conclusion: The eigenvalues of A are those values of λ for which det(λi n A) = Example 323 Let A = 1 8 Find all eigenvalues of A and find an eigenvector 4 2 corresponding to each eigenvalue 49

4 Solution: We need to find all values of λ for which det(λi 2 A) = λi 2 A = λ = λ 1 8 λ 4 2 = λ λ + 2 det(λi 2 A) = (λ 1)(λ + 2) 8( 4) = λ 2 1λ + 2λ = λ 2 8λ + 12 So det(λi 2 A) is a polynomial of degree 2 in λ The eigenvalues of A are those values of λ for which det(λi 2 A) = ie λ 2 8λ + 12 = = (λ 6)(λ 2) =, λ = 6 or λ = 2 Eigenvalues of A : 6, 2 To find an eigenvector of A corresponding to λ = 6, we need a vector x y for which A x y = 6 x y ie 1 8 x 4 2 y 1x 8y = 4x 2y = 6 x y = 6x 6y = 1x 8y = 6x and 4x 2y = 6y Both of these equations say x 2y = ; hence any non-zero vector x y in which x = 2y is 5

5 an eigenvector for A corresponding to the eigenvalue 6 For example we can take y = 1, x = 2 to obtain the eigenvector 2 1 Exercises: 1 Show that = Find an eigenvector for A corresponding to the other eigenvalue λ = 2 Definition 324 Let A be a square matrix (n n) The characteristic polynomial of A is the determinant of the n n matrix λi n A This is a polynomial of degree n in λ Example 325 (a) Let A = 4 1 Then 2 1 λi 2 A = λ = λ λ = λ λ 1 det(λi 2 A) = (λ 4)(λ 1) 1( 2) = λ 2 5λ + 6 Characteristic Polynomial of A: λ 2 5λ (b) Let B = λ λi 3 B = λ λ = λ λ λ + 2 We can calculate det(λi 3 B) using cofactor expansion along the first row det(λi 3 B) = (λ 5)[(λ + 1)(λ + 2) ()(8)] ( 6)[(λ + 2) 8( 1)] + ( 2)[() ( 1)(λ + 1)] = (λ 5)(λ 2 + 3λ + 2) + 6(8) 2(λ + 1) = λ 3 2λ 2 13λ λ 2 = λ 3 2λ 2 15λ

6 Characteristic polynomial of B : λ 3 2λ 2 15λ + 36 As we saw in Section 51, the eigenvalues of a matrix A are those values of λ for which det(λi A) = ; ie, the eigenvalues of A are the roots of the characteristic polynomial Example 326 Find the eigenvalues of the matrices A and B of Example 622 (a) A = Characteristic Equation : λ 2 5λ + 6 = = (λ 3)(λ 2) = Eigenvalues of A: λ = 3, λ = (b) B = Characteristic Equation: λ 3 2λ 2 15λ + 36 = To find solutions to this equation we need to factor the characteristic polynomial, which is cubic in λ (in general solving a cubic equation like this is not an easy task unless we can factorize) First we try to find an integer root Fact: The only possible integer roots of a polynomial are factors of its constant term So in this example the only possible candidates for an integer root of the characteristic polynomial p(λ) = λ 3 2λ 2 15λ + 36 are the integer factors of 36 : ie ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36 Try some of these : p(1) = 1 3 2(1) 2 15(1) + 36 p(2) = 2 3 2(2) 2 15(2) + 36 p(3) = 3 3 2(3) 2 15(3) + 36 = = 3 is a root of p(λ), and (λ 3) is a factor of p(λ) Then p(λ) = λ 3 2λ 2 15λ + 36 = (λ 3)(λ 2 + aλ 12) To find a, look at the coefficients of λ 2 (or λ) on the left and right λ 2 : 2 = 3 + a = a = 1 52

7 λ 3 2λ 2 15λ + 36 = (λ 3)(λ 2 + λ 12) = (λ 3)(λ 3)(λ + 4) = (λ 3) 2 (λ + 4) Eigenvalues of B: λ = 3 (occurring twice), λ = 4 We conclude this section by calculating eigenvectors of B corresponding to these eigenvalues Example 327 Let B = From Example 325, the eigenvalues of B are λ = 3 (occurring twice), λ = 4 Find an eigenvector of B corresponding to the eigenvalue λ = 4 Solution: We need a column vector v = x 2 x , with entries not all zero, for which x 2 x 3 = x 2 + 2x 3 ie x 2 8x 3 = 2x 3 4 4x 2 4x 3 x 2 x 3 = 5 + 6x 2 + 2x 3 = 4 x 2 8x 3 = 4x 2 = 2x 3 = 4x x 2 + 2x 3 = 3x 2 8x 3 = + 2x 3 = }{{} system of 3 equations in,x 2,x 3 So we need to solve the system of linear equations with augmented matrix

8 Note: The coefficient matrix here is just B ( 4)I 3 ie To find solutions to the system : R3 R R3 9 R R3 2 R R : RREF The variable x 3 is free : let x 3 = t Then + 2x 3 = = = 2t x x 3 = = x 2 = 8 3 t For example if we take t = 3 we find = 6 and x 2 = 8 Hence v = for B corresponding to λ = 4 Exercise: Check that Bv = 4v is an eigenvector Notes: 1 To find an eigenvector v of a n n matrix A corresponding to the eigenvalue λ : solve the system (A λi n ) x n = ie the system whose coefficient matrix is A λi n and in which the constant term (on the right in each equation) is 54

9 2 If v is an eigenvector of a square matrix A, corresponding to the eigenvalue λ, and if k is a real number, then kv is also an eigenvector of A corresponding to λ, since A(kv) = k(av) = k(λv) = λ(kv) 6 In the above example any (non-zero) scalar multiple of 8 is an eigenvector of A 3 corresponding to λ = 4 (these arise from different choices of value for the free variable t in the solution of the relevant system of equations) Example 328 Find an eigenvector of B corresponding to the eigenvalue λ = 3 Solution: We need to solve the system whose augmented matrix consists of B 3I 3 and a fourth column all of whose entries are zero B 3I 3 = (obtained by subtracting 3 from each of the entries on the main diagonal of B and leaving the other entries unchanged) We apply elementary row operations to the augmented matrix of the system : R R3 R R2 ( 1 4 ) R3 + 3 R R1 3 R : RREF Let x 3 = t Then 5x 3 = = = 5t x 2 + 2x 3 = = x 2 = 2t 55

10 Eigenvectors are given by x 2 x 3 5t = 2t t for t R, t For example of we choose t = 1 we find that v = for B corresponding to λ = 3 (Exercise: Check this) is an eigenvector 56

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