Remark 1 By definition, an eigenvector must be a nonzero vector, but eigenvalue could be zero.

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1 Sec 5 Eigenvectors and Eigenvalues In this chapter, vector means column vector Definition An eigenvector of an n n matrix A is a nonzero vector x such that A x λ x for some scalar λ A scalar λ is called an eigenvalue of A if there is a nontrivial solution x of A x λ x; such an x is called an eigenvector corresponding to λ Remark By definition, an eigenvector must be a nonzero vector, but eigenvalue could be zero 3 Ex Let A and u, then A u u Therefore is an eigenvalue of A and u is an eigenvector of A corresponding to the eigenvalue Remark is not the only eigenvector of A corresponding to the eigenvalue For example, 6 is another eigenvector of A corresponding to In fact, if u is an eigenvector of A corresponding 3 to an eigenvalue λ, then so is any nonzero multiple c u of u Ex Let B 6 5 Is u 6 5 an eigenvector of B? How about v 3? Remark 3 A matrix may have more than one eigenvalue Ex3 In Ex above, we observed that λ 4 is an eigenvalue of B However, w an eigenvector of B corresponding to the eigenvalue 6 B w 5 7 7, since 7 7 w is Question Suppose λ is an eigenvalue of A to λ? How do we find all the eigenvectors corresponding

2 Definition Let A be an n n square matrix and λ be an eigenvalue of A Nul(A λi n ) is called the eigenspace of A corresponding to λ Therefore, the eigenspace of A corresponding to λ is a subspace of R n The geometric multiplicity of λ is the dimension of the eigenspace of A corresponding to λ Note that the geometric multiplicity of an eigenvalue is always greater than or equal to Remark 4 Any nonzero vector in the eigenspace corresponding to λ is an eigenvector corresponding to λ Ex4 Let A 6, then is an eigenvalue of A with an eigenvector Find a 8 basis for the eigenspace of A corresponding to So every eigenvector of A corresponding to can be written as a linear combination of and Conversely every linear combination of these two vectors except the zero vector is an eigenvector corresponding to Theorem If v, v,, v r are eigenvectors that correspond to distinct eigenvalues λ, λ,, λ r of an n n matrix A, then the set { v, v,, v r } is linearly independent 6 6 Ex5 Go back to Ex above, where B We saw that and are eigenvectors 5 { 5 } 6 of B corresponding to 4 and 7, respectively So the set, is linearly independent 5 So, how can we find all the eigenvalues of a given matrix? That is the main subject in next section

3 3 Sec 5 The Characteristic Equation 3 Ex Let A We want to find all the eigenvalues of A By definition, λ is an eigenvalue 3 6 of A if and only if A x λ x for some nonzero vector x, which is the same as saying that the homogeneous system (A λi ) x has a nontrivial solution Therefore by the Invertible Matrix Theorem, we see that λ is an eigenvalue of A if and only if the matrix A λi is singular, that is, if and only if det(a λi ) Now A λi and hence λ λ det(a λi ) ( λ)( 6 λ) 9 λ + 4λ Therefore and, and only they, are eigenvalues of A In general, to find all the eigenvalues of a given n n matrix A, First, find A λi n ; Second, compute det(a λi n ), which is a polynomial in λ of degree n; 3 Third, finally solve the equation det(a λi n ) The solution set of this equation is exactly the set of eigenvalues of A Definition The polynomial det(a λi n ) is called the characteristic polynomial of A The equation det(a λi n ) is called the characteristic equation of A 5 6 Ex Find all the eigenvalues of A In general, the eigenvalues of a triangular matrix are precisely the diagonal entries of the matrix Ex3 True or false? Some 3 3 matrices can have 4 distinct eigenvalues Ex4 Give an example of 3 3 matrix M having, 3, and 4 as eigenvalues Let v, v, and v 3 be eigenvectors of M corresponding to eigenvalues, 3, and 4, respectively Show that { v, v, v 3 } is a basis for R 3

4 4 Definition Let A and B be n n matrices A is said to be similar to B if there is an invertible matrix P such that P AP B When A is similar to B, then we write A B 4 Ex5 A is similar to B because with P (so P 4 3 ) P AP B Remark If A is similar to B, then B is similar to A (ie, A B B A) A is similar to A itself (A A) 3 If A is similar to B and B is similar to C, then A is similar to C (A B and B C A C) Theorem If A and B are similar n n matrices, then they have the same characteristic polynomials and hence the same eigenvalues Proof Summary Suppose an n n matrix A is given To find all eigenvalues of A, solve the equation det(a λi n ) ; the roots are precisely the eigenvalues For each eigenvalue λ, to get an eigenvector corresponding to λ, find Nul(A λi n ) (the eigenspace corresponding to λ) by solving the associated homogeneous system (A λi n ) x ; any nonzero vector inside Nul(A λi n ) is an eigenvector corresponding to λ

5 5 Sec 53 Diagonalization Motivation: In many practical applications, we need to compute A k for large k This requires a lot of computations for general n n matrix A When A is diagonal, however, the computation is quite simple Ex If D, then D 3 In general, D 9 n n 3 n Ex Compute A 7, where A 4 It s almost impossible to compute A directly Consider a matrix P, then P is invertible with P Then P AP 7 4 In other words, A is similar to a diagonal matrix D 5 RHS: D Take the th power of the both sides: LHS: (P AP ) (P AP )(P AP ) (P AP ) P }{{} A(P P )A(P P )A (P P )AP, which times reduces to P A P Therefore we conclude that P A P A P 5 3 P , or Similarly, one can compute A n 5 n 3 n 5 n 3 n 5 n + 3 n 5 n + 3 n for general n Question Given A, is it always possible to find an invertible matrix P such that P AP is diagonal? The answer is no We will see an example later

6 6 Definition A square matrix A is said to be diagonalizable if A is similar to a diagonal matrix, that is, if P AP D for some invertible matrix P and some diagonal matrix D 7 Ex3 The matrix A in Ex is diagonalizable 4 Question If A is diagonalizable, how can we find a matrix P such that P AP is diagonal? The Diagonalization Theorem, Part An n n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors In fact, P AP D, with D diagonal, if and only if the columns of P are n linearly independent eigenvectors of A In this case, the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors (ie columns) in P Remark So if an n n matrix A has n distinct eigenvalues, then A is diagonalizable 7 Ex4 Go back to Ex and see how we can obtain the matrix P A, the eigenvalues of A are the zeros of the characteristic polynomial of A: det(a λi ) (7 λ)( λ) λ 8λ + 5 (λ 3)(λ 5) Find an eigenvector corresponding to λ 5: Find an eigenvector corresponding to λ 3: Therefore, we can take P Ex5 Diagonalize (that is, find an invertible matrix P and a diagonal matrix D such that P AP 3 3 D) A 3 5 3, if possible 3 3

7 7 Ex6 Show that B is not diagonalizable Proof Proof Now we give another characterization of diagonalizable matrices The Diagonalization Theorem, Part Let A be an n n matrix with distinct eigenvalues λ,, λ p a For k p, the dimension of the eigenspace corresponding to λ k (that is, the geometric multiplicity of λ k ) is always greater than equal to and less than or equal to the algebraic multiplicity of the eigenvalue λ k as a zero of the characteristic polynomial of A b The matrix A is diagonalizable if and only if the geometric multiplicity of λ k equals the algebraic multiplicity of λ k for each k c If B j (resp B k ) is a basis for the eigenspace corresponding to λ j (resp λ k ), then the set B j B k is linearly independent Ex7 Using the theorem above, show again that the matrix in Ex5 is diagonalizable, and that the matrix in Ex6 is not diagonalizable Ex8 Check if is diagonalizable C 3 3

8 8 Ex9 (Application to the Fibonacci sequence) Consider the sequence defined by a, a, a n a n + a n for n 3 We are interested in the closed form of a n Note that for n 3 an an a n a n Similarly, if n is large enough, then an an so an a n an a n 3 a n 3 and 3 an 3 a n 4 a n an a n 3 an 3 n a a a n 4 n n Now let s compute The characteristic polynomial is and eigenvalues are Thus the matrix is diagonalizable (why?) One eigenvector corresponding to λ + 5 is 5 Similarly, + 5 is an eigenvector corresponding to λ 5 Let P then P with P λ P λ n n λ Finally, P P λ λ n λ n λ n λ n 5 λ n λ n λ n 3 λ n 3 and hence ( ) a n (λ n λ n + λ n λ n ) n ( ) n 5 5 after little calculation + 5 5