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1 Math 20 Chapter 5 Eigenvalues and Eigenvectors Eigenvalues and Eigenvectors. Definition: A scalar λ is called an eigenvalue of the n n matrix A is there is a nontrivial solution x of Ax = λx. Such an x is called an eigenvector corresponding to the eigenvalue λ. 2. What does this mean geometrically? Suppose that A is the standard matrix for a linear transformation T : R n R n. Then if Ax = λx, it follows that T (x) = λx. This means that if x is an eigenvector of A, then the image of x under the transformation T is a scalar multiple of x and the scalar involved is the corresponding eigenvalue λ. In other words, the image of x is parallel to x. 3. Note that an eigenvector cannot be 0, but an eigenvalue can be Suppose that 0 is an eigenvalue of A. What does that say about A? There must be some nontrivial vector x for which Ax = 0x = 0 which implies that A is not invertible which implies a whole lot of things given our Invertible Matrix Theorem. 5. Invertible Matrix Theorem Again: The n n matrix A is invertible if and only if 0 is not an eigenvalue of A. 6. Definition: The eigenspace of the n n matrix A corresponding to the eigenvalue λ of A is the set of all eigenvectors of A corresponding to λ. 7. We re not used to analyzing equations like Ax = λx where the unknown vector x appears on both sides of the equation. Let s find an equivalent equation in standard form. Ax = λx Ax λx = 0 Ax λix = 0 (A λi)x = 0 8. Thus x is an eigenvector of A corresponding to the eigenvalue λ if and only if x and λ satisfy (A λi)x = It follows that the eigenspace of λ is the null space of the matrix A λi and hence is a subspace of R n. 0. Later in Chapter 5, we will find out that it is useful to find a set of linearly independent eigenvectors for a given matrix. The following theorem provides one way of doing so. See page 307 for a proof of this theorem.. Theorem 2: If v,..., v r are eigenvectors that correspond to distinct eigenvalues λ,..., λ r of an n n matrix A, then the set {v,..., v r } is linearly independent.

2 2 Determinants. Recall that if λ is an eigenvalue of the n n matrix A, then there is a nontrivial solution x to the equation Ax = λx or, equivalently, to the equation (A λi)x = 0. (We call this nontrivial solution x an eigenvector corresponding to λ.) 2. Note that this second equation has a nontrivial solution if and only if the matrix A λi is not invertible. Why? If the matrix is not invertible, then it does not have a pivot position in each column (by the Invertible Matrix Theorem) which implies that the homogeneous system has at least one free variable which implies that the homogeneous system has a nontrivial solution. Conversely, if the matrix is invertible, then the only solution is the trivial solution. 3. To find the eigenvalues of A we need a condition on λ that is equivalent to the equation (A λi)x = 0 having a nontrivial solution. This is where determinants come in. 4. We skipped Chapter 3, which is all about determinants, so here s a recap of just what we need to know about them. [ ] a b 5. Formula: The determinant of the 2 2 matrix A = is c d deta = ad bc. 6. Formula: The determinant of the 3 3 matrix A = a a 2 a 3 a 2 a 22 a 23 is a 3 a 32 a 33 deta = a a 22 a 33 + a 2 a 23 a 3 + a 3 a 2 a 32 a 3 a 22 a 3 a 32 a 23 a a 33 a 2 a 2. See page 9 for a useful way of remembering this formula. 7. Theorem: The determinant of an n n matrix A is 0 if and only if the matrix A is not invertible. 8. That s useful! We re looking for values of λ for which the equation (A λi)x = 0 has a nontrivial solution. This happens if and only if the matrix A λi is not invertible. This happens if and only if the determinant of A λi is 0. This leads us to the characteristic equation of A. 3 The Characteristic Equation. Theorem: A scalar λ is an eigenvalue of an n n matrix A if and only if λ satisfies the characteristic equation det(a λi) = It can be shown that if A is an n n matrix, then det(a λi) is a polynomial in the variable λ of degree n. We call this polynomial the characteristic polynomial of A.

3 3. Example: Consider the matrix A = To find the eigenvalues of A, we must compute det(a λi), set this expression equal to 0, and solve for λ. Note that A λi = λ λ 0 = 3 λ λ λ λ Since this is a 3 3 matrix, we can use the formula given above to find its determinant. det(a λi) = (3 λ)( λ)(2 λ) + (6)(6)(0) + ( 8)(0)(0) (0)( λ)( 8) (0)(6)(3 λ) ( λ)(0)(6) = λ(3 λ)(2 λ) Setting this equal to 0 and solving for λ, we get that λ = 0, 2, or 3. These are the three eigenvalues of A. 4. Note that A is a triangular matrix. (A triangular matrix has the property that either all of its entries below the main diagonal are 0 or all of its entries above the main diagonal are 0.) It turned out that the eigenvalues of A were the entries on the main diagonal of A. This is true for any triangular matrix, but is generally not true for matrices that are not triangular. 5. Theorem : The eigenvalues of a triangular matrix are the entries on its main diagonal. 6. In the above example, the characteristic polynomial turned out to be λ(λ 3)(λ 2). Each of the factors λ, λ 3, and λ 2 appeared precisely once in this factorization. Suppose the characteristic function had turned out to be λ(λ 3) 2. In this case, the factor λ 3 would appear twice and so we would say that the corresponding eigenvalue, 3, has multiplicity Definition: In general, the multiplicity of an eigenvalue l is the number of times the factor λ l appears in the characteristic polynomial. 4 Finding Eigenvectors Example (Continued): Let us now find the eigenvectors of the matrix A = We have to take each of its three eigenvalues 0, 2, and 3 in turn. 2. To find the eigenvectors corresponding to the eigenvalue 0, we need to solve the equation (A λi)x = 0 where λ = 0. That is, we need to solve (A λi)x = 0 (A 0I)x = 0 Ax = x = Row reducing the augmented matrix, we find that x = x x 2 = x x 3

4 This tells us that the eigenvectors corresponding to the eigenvalue 0 are precisely the set of scalar multiples of the vector 2. In other words, the eigenspace corresponding to the eigenvalue 0 is 0 Span To find the eigenvectors corresponding to the eigenvalue 2, we need to solve the equation (A λi)x = 0 where λ = 2. That is, we need to solve (A λi)x = 0 (A 2I)x = x = x = Row reducing the augmented matrix, we find that x = x x 2 = x x 3 This tells us that the eigenvectors corresponding to the eigenvalue 2 are precisely the set of scalar multiples of the vector 0 3. In other words, the eigenspace corresponding to the eigenvalue 2 is 0 Span I ll let you find the eigenvectors corresponding to the eigenvalue 3. 5 Similar Matrices. Definition: The n n matrices A and B are said to be similar if there is an invertible n n matrix P such that A = P BP. 2. Similar matrices have at least one useful property, as seen in the following theorem. See page 35 for a proof of this theorem. 3. Theorem 4: If n n matrices are similar, then they have the same characteristic polynomial and hence the same eigenvalues (with the same multiplicities). 4. Note that if the n n matrices A and B are row equivalent, then they [ are ] not necessarily [ ] similar. For a simple counterexample, consider the row equivalent matrices A = and B =. If these two 0 0 matrices were similar, then there would exist an invertible matrix P such that A = P BP. Since B is the identity matrix, this means that A = P IP = P P = I. Since A is not the identity matrix, we have a contradiction, and so A and B cannot be similar.

5 5. We can also use Theorem 4 to show that row equivalent matrices are not necessarily similar: Similar matrices have the same eigenvalues but row equivalent matrices often do not have the same eigenvalues. (Imagine scaling a row of a triangular matrix. This would change one of the matrix s diagonal entries which changes its eigenvalues. Thus we would get a row equivalent matrix with different eigenvalues, so the two matrices could not be similar by Theorem 4.) 6 Diagonalization. Definition: A square matrix A is said to be diagonalizable if it is similar to a diagonal matrix. In other words, a diagonal matrix A has the property that there exists an invertible matrix P and a diagonal matrix D such that A = P DP. 2. Why is this useful? Suppose you wanted to find A 3. If A is diagonalizable, then A 3 = (P DP ) 3 = (P DP )(P DP )(P DP ) = P DP P DP P DP = P D(P P )D(P P )DP = P DDDP = P D 3 P. In general, if A = P DP, then A k = P D k P. 3. Why is this useful? Because powers of diagonal matrices are relatively easy to compute. For example, if D = 0 2 0, then D 3 = 0 ( 2) This means that finding A k involves only two matrix multiplications instead of the k matrix multiplications that would be necessary to multiply A by itself k times. 4. It turns out that an n n matrix is diagonalizable if and only it has n linearly independent eigenvectors. That s what the following theorem says. See page 32 for a proof of this theorem. 5. Theorem 5 (The Diagonalization Theorem): (a) An n n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors. (b) If v, v 2,..., v n are linearly independent eigenvectors of A and λ, λ 2,..., λ n are their corresponding eigenvalues, then A = P DP, where P = [ v v n ] and λ λ 2 0 D = λ n (c) If A = P DP and D is a diagonal matrix, then the columns of P must be linearly independent eigenvectors of A and the diagonal entries of D must be their corresponding eigenvalues.

6 6. What can we make of this theorem? If we can find n linearly independent eigenvectors for an n n matrix A, then we know the matrix is diagonalizable. Furthermore, we can use those eigenvectors and their corresponding eigenvalues to find the invertible matrix P and diagonal matrix D necessary to show that A is diagonalizable. 7. Theorem 4 told us that similar matrices have the same eigenvalues (with the same multiplicities). So if A is similar to a diagonal matrix D (that is, if A is diagonalizable), then the eigenvalues of D must be the eigenvalues of A. Since D is a diagonal matrix (and hence triangular), the eigenvalues of D must lie on its main diagonal. Since these are the eigenvalues of A as well, the eigenvalues of A must be the entries on the main diagonal of D. This confirms that the choice of D given in the theorem makes sense. 8. See your class notes or Example 3 on page 32 for examples of the Diagonalization Theorem in action.

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