Eigen value and vectors

Eigen value and vectors

• 1.
  Math 20 Chapter 5Eigenvalues and Eigenvectors 1 Eigenvalues and Eigenvectors 1. 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 : Rn → Rn . 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 0. 4. 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 = 0. 9. It follows that the eigenspace of λ is the null space of the matrix A − λI and hence is a subspace of Rn . 10. 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. 11. Theorem 2: If v1, . . . , vr are eigenvectors that correspond to distinct eigenvalues λ1, . . . , λr of an n × n matrix A, then the set {v1, . . . , vr} is linearly independent.
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  2 Determinants 1. Recallthat 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. 5. Formula: The determinant of the 2 × 2 matrix A = a b c d is detA = ad − bc. 6. Formula: The determinant of the 3 × 3 matrix A =   a11 a12 a13 a21 a22 a23 a31 a32 a33   is detA = a11a22a33 + a12a23a31 + a13a21a32 − a31a22a13 − a32a23a11 − a33a21a12. See page 191 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 1. 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) = 0. 2. 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.
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  3. Example: Considerthe matrix A =   3 6 −8 0 0 6 0 0 2  . 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 =   3 6 −8 0 0 6 0 0 2   −   λ 0 0 0 λ 0 0 0 λ   =   3 − λ 6 −8 0 −λ 6 0 0 2 − 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 1: 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 2. 7. Definition: In general, the multiplicity of an eigenvalue is the number of times the factor λ − appears in the characteristic polynomial. 4 Finding Eigenvectors 1. Example (Continued): Let us now find the eigenvectors of the matrix A =   3 6 −8 0 0 6 0 0 2  . 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 = 0   3 6 −8 0 0 6 0 0 2   x = 0 Row reducing the augmented matrix, we find that x =   x1 x2 x3   = x2   −2 1 0   .
• 4.
  This tells usthat the eigenvectors corresponding to the eigenvalue 0 are precisely the set of scalar multiples of the vector   −2 1 0  . In other words, the eigenspace corresponding to the eigenvalue 0 is Span      −2 1 0      . 3. 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 = 0     3 6 −8 0 0 6 0 0 2   −   2 0 0 0 2 0 0 0 2     x = 0   1 6 −8 0 −2 6 0 0 0   x = 0 Row reducing the augmented matrix, we find that x =   x1 x2 x3   = x3   −10 3 1   . This tells us that the eigenvectors corresponding to the eigenvalue 2 are precisely the set of scalar multiples of the vector   −10 3 1  . In other words, the eigenspace corresponding to the eigenvalue 2 is Span      −10 3 1      . 4. I’ll let you find the eigenvectors corresponding to the eigenvalue 3. 5 Similar Matrices 1. 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 = PBP−1 . 2. Similar matrices have at least one useful property, as seen in the following theorem. See page 315 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 = 2 0 0 1 and B = 1 0 0 1 . If these two matrices were similar, then there would exist an invertible matrix P such that A = PBP−1 . Since B is the identity matrix, this means that A = PIP−1 = PP−1 = I. Since A is not the identity matrix, we have a contradiction, and so A and B cannot be similar.
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  5. We canalso 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 1. 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 = PDP−1 . 2. Why is this useful? Suppose you wanted to find A3 . If A is diagonalizable, then A3 = (PDP−1 )3 = (PDP−1 )(PDP−1 )(PDP−1 ) = PDP−1 PDP−1 PDP−1 = PD(PP−1 )D(PP−1 )DP−1 = PDDDP−1 = PD3 P−1 . In general, if A = PDP−1 , then Ak = PDk P−1 . 3. Why is this useful? Because powers of diagonal matrices are relatively easy to compute. For example, if D =   7 0 0 0 −2 0 0 0 3  , then D3 =   73 0 0 0 (−2)3 0 0 0 33   . This means that finding Ak involves only two matrix multiplications instead of the k matrix multipli- cations 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 321 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 v1, v2, . . . , vn are linearly independent eigenvectors of A and λ1, λ2, . . . , λn are their corre- sponding eigenvalues, then A = PDP−1 , where P = v1 · · · vn and D =      λ1 0 · · · 0 0 λ2 · · · 0 ... ... ... 0 0 · · · λn      (c) If A = PDP−1 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.
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  6. What canwe 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 321 for examples of the Diagonalization Theorem in action.