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Eigenvalues and Eigenvectors Section 5.1 Eigenvalues and Eigenvectors At this point in the class, we understand that the operation of matrix multiplication is very different from scalar multiplication. Matrix multiplication is an operation that finds the product of a pair of matrices, whereas scalar multiplication finds the product of a scalar with a matrix. More to the point, scalar multiplication of a scalar k with a matrix X is an easy operation to perform; indeed there's a simple algorithm: multiply each entry of X by the scalar k. Matrix multiplication on the other hand cannot be expressed using a simple algorithm. Indeed, the operation is extremely complex: to multiply a pair of n×n matrices, it takes about n2 operations, a hugely inefficient process. Thus it is surprising that, for any square matrix A, certain matrix multiplications \collapse" (in a sense) into scalar multiplication. As an example, consider the matrices ( ) ( ) ( ) 3 2 −4 10 A = ; w = ; x = : 3 4 −1 15 The product Aw is not particularly interesting: ( ) −14 Aw = ; −16 indeed, there appears to be very little connection between A, w, and Aw. On the other hand, an we see an interesting phenomenon when we calculate Ax: ( ) ( ) 60 10 Ax = = 6 = 6x: 90 15 Forgetting the calculation in the middle, we have Ax = 6x; in a sense, the matrix multiplication Ax collapsed into scalar multiplication 6x. We will see below that the phenomenon described here is more common than one might think. Identifying products like Ax = 6x will allow us to extract a great deal of useful information about A; accordingly, the rest of the section will be devoted to this phenomenon, which we give a name in the next definition. Definition. Let A be an n × n matrix and x any nonzero vector in Rn that satisfies the equation Ax = λx; or alternatively TA(x) = λx for some scalar λ. We call λ an eigenvalue of A (and of the associated linear function TA), and call x an eigenvector corresponding to λ. 1 Section 5.1 In the example above, we saw that ( )( ) ( ) 3 2 10 10 = 6 : 3 4 15 15 In the language of our new definition, λ = 6 is an eigenvalue of ( ) 3 2 A = ; 3 4 and ( ) 10 x = 15 is an eigenvector associated to λ = 6. Geometrically, A \stretches" the vector x by a factor of 6, without changing x's direction. Recall, however, that A did change the direction of the vector w; there is something unique about x with relation to multiplication by A. As mentioned above, the eigenvalues and eigenvectors of a matrix can provide us with a great deal of information about the matrix. This brings us to several questions about the quantities: 1. Are we guaranteed that an n × n matrix A has eigenvalues? 2. What is the maximum number of distinct eigenvalues of an n × n matrix? 3. How do we find the eigenvalues of A? The next theorem answers all of these questions: Theorem 1.5.1. If A is an n × n matrix with real entries, then the number λ (which could be real or complex) is an eigenvalue of A if and only if it satisfies the equation det(λI − A) = 0; which is called the characteristic equation of A. The polynomial p(λ) = det(λI − A) is called the characteristic polynomial of A. Before we answer the questions posed above, let's think about what this theorem says about the matrix ( ) 3 2 A = : 3 4 Theorem 5:1:1 says that we can find the eigenvalues for A by finding the determinant of the matrix λI − A: 2 Section 5.1 Let's find the matrix: ( ) ( ) 1 0 3 2 λI − A = λ − 0 1 3 4 ( ) ( ) λ 0 3 2 = − 0 λ 3 4 ( ) λ − 3 −2 = : −3 λ − 4 Now we need to find the determinant of the matrix above: ( ) λ − 3 −2 det(λI − A) = det −3 λ − 4 = (λ − 3)(λ − 4) − 6 = λ2 − 7λ + 6: Before we proceed, we should note that λ2 − 7λ + 6 = 0 is the characteristic equation of A, and p(λ) = λ2 − 7λ + 6 is its characteristic polynomial; notice that it is a degree 2 polynomial. Now according to the theorem, the numbers λ satisfying the equation above are the eigenvalues of A. Thus finding the roots of the polynomial will give us the desired eigenvalues: the polynomial factors as (λ − 6)(λ − 1) = 0; which means that the roots are λ = 6 and λ = 1: Thus the eigenvalues of the matrix A are λ = 6 and λ = 1. We have already seen that λ = 6 is an eigenvalue of A, since it satisfies the equation ( )( ) ( ) 3 2 10 10 = 6 : 3 4 15 15 We should verify( that λ) = 1 is also an eigenvalue by finding an eigenvector. In other words, we > 2 need a vector x = x1 x2 in R so that Ax = 1x 3 Section 5.1 or ( )( ) ( ) 3 2 x x 1 = 1 : 3 4 x2 x2 This should not be too difficult to do; let's make the calculation: ( )( ) 3 2 x Ax = 1 3 4 x2 ( ) 3x + 2x = 1 2 : 3x2 + 4x2 Now we want Ax = x; so we must have ( ) ( ) 3x + 2x x 1 2 = 1 ; 3x1 + 4x2 x2 or 2x1 + 2x2 = 0 3x1 + 3x2 = 0: We can rewrite this system in matrix equation form as ( ) 2 2 x = 0: 3 3 Now it is easy to see that the coefficient matrix of this equation has determinant 0, so by Theorem 4.8.8, the system has at least one nontrivial solution; choosing, say, x1 = 1 gives us corresponding x2 = −1. So ( ) 1 x = −1 is an eigenvector corresponding to the eigenvalue λ = 1. Key Point. Here are few facts about the characteristic equation of A, along with the answers to the questions posed above: • If A is an n × n matrix, then its characteristic polynomial p(λ) = det(λI − A) is a polynomial of degree n. • The roots of the characteristic equation det(λI − A) = 0 are precisely the eigenvalues of A. Finding the roots of the equation amounts to finding the eigenvalues of A. 4 Section 5.1 • By the Fundamental Theorem of Algebra, a polynomial of degree n has exactly n real or complex roots, which may or may not be repeated. Thus an n × n matrix A has as many as n distinct eigenvalues. • The number 0 is allowed to be an eigenvalue for a matrix, but the vector 0 is not allowed to be an eigenvector, since A0 = k0 is true for any matrix A and any scalar k. • If x is an eigenvector of A with associated eigenvalue λ, so that Ax = λx; then any nonzero scalar multiple kx of eigenvector x is also an eigenvector. To understand why, think about the product A(kx): A(kx) = k(Ax) = k(λx) = λ(kx): • If x and y are eigenvalues associated with the same eigenvector λ, so that Ax = λx and Ay = λy; then the vector x + y is also an eigenvector of A associated with λ, since A(x + y) = Ax + Ay = λx + λy = λ(x + y): Example Given 0 1 5 −2 0 A = @0 −3 0 A ; 1 4 −1 find: 1. The characteristic equation of A. 2. The eigenvalues of A. 3. An eigenvector corresponding to each eigenvalue of A. 1. The characteristic equation of A is given by det(λI − A) = 0; 5 Section 5.1 so let's begin by finding the matrix λI − A: 0 1 0 1 λ 0 0 5 −2 0 λI − A = @0 λ 0A − @0 −3 0 A 0 0 λ 1 4 −1 0 1 λ − 5 2 0 = @ 0 λ + 3 0 A : −1 −4 λ + 1 Next, we need to calculate the determinant det(λI − A); we proceed by cofactor expansion along the last column: 0 1 λ − 5 2 0 det(λI − A) = det @ 0 λ + 3 0 A −1 −4 λ + 1 ( ) = (λ + 1) (λ − 5)(λ + 3) − 2 · 0 = (λ + 1)(λ − 5)(λ + 3) = (λ + 1)(λ2 − 2λ − 15) = λ3 − λ2 − 17λ − 15: Thus the characteristic equation of A is λ3 − λ2 − 17λ − 15 = 0: 2. The roots of the characteristic equation above are the desired eigenvalues; fortunately we already know from the calculations above that the polynomial factors as λ3 − λ2 − 17λ − 15 = (λ + 1)(λ − 5)(λ + 3): So the roots of the equation (and the eigenvalues of A) are λ1 = −1; λ2 = −3; and λ3 = 5: 3. The desired eigenvectors x1, x2, and x3 must satisfy the equations Ax1 = −x1;Ax2 = −3x2; and Ax3 = 5x3; 6 Section 5.1 respectively. For a general vector x in R3, Ax is given by 0 1 0 1 5 −2 0 x1 @ A @ A Ax = 0 −3 0 x2 1 4 −1 x3 0 1 5x1 − 2x2 @ A = −3x2 : x1 + 4x2 − x3 Starting with λ1 = −1, we want 0 1 0 1 5x1 − 2x2 −x1 @ A @ A −3x2 = −x2 ; x1 + 4x2 − x3 −x3 so we see that 6x1 − 2x2 = 0 −2x2 = 0 x1 + 4x2 = 0: Now we must choose x1 = x2 = 0, but x3 does not show up in the system, so we are free to choose a value for it as we wish, say x3 = 1. Thus 0 1 0 @ A x1 = 0 1 is an eigenvector for A associated with the eigenvalue λ1 = −1. Next, let's look for an eigenvector associated with the eigenvalue λ2 = −3: we want 0 1 0 1 5x1 − 2x2 −3x1 @ A @ A −3x2 = −3x2 ; x1 + 4x2 − x3 −3x3 so we have 8x1 − 2x2 = 0 0 = 0 x1 + 4x2 + 2x3 = 0: Since we have two equations in three unknowns, we are free to choose one of the variables as we wish; we can easily satisfy the first equation by setting x2 = 4, so that x1 = 1.
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