Eigenvectors and eigenvalues Brian Krummel March 30, 2020 We will now look at eigenvalues and eigenvectors of a square matrix. When solving a problem in linear algebra, it is often convenient to choose our basis, or coordinate system. For a square matrix A, it is often convenient to work with a basis of eigenvectors, which are nonzero vectors for which multiplication by A is the same by scaling by a number λ. λ is called an eigenvalue. Definition 1. Let A be an n × n matrix. We call a scalar λ an eigenvalue if there is a nonzero vector X 2 Rn such that AX = λX: We call the nonzero vector X an eigenvector. Example 1. Let 2 0 A = : 0 3 T Then E1 = 1 0 satisfies AE1 = 2E1, so E1 is an eigenvector with corresponding eigenvalue T 2. E1 = 0 1 satisfies AE2 = 3E2, so E2 is an eigenvector with corresponding eigenvalue 3. Notice that this tells us something very important about A: A acts on E1 and E2 by scaling: x2 AE2 E2 E1 AE1 x1 We of course have A0 = λ0 for every scalar λ, but this does not tell us anything insightful about A the way that 2; 3 being eigenvalues does. x2 AB2 Example 2. Show that B1 is an eigenvector of A and find the corresponding eigenvalue, where AB1 5 −1 1 A = ;B = : −1 5 1 1 B2 B1 Answer. We compute x1 5 −1 1 4 1 AB = = = 4 = 4B : 1 −1 5 1 4 1 1 So B1 is an eigenvector with corresponding eigenvalue 4. 1 x2 Ae2 e2 The main reason we care about eigenvalues and eigenvectors is that some matrices are \diag- onalizable" in terms of their eigenvalues and eigenvectors.e1 Ae1 Forx1 instance, the other eigenvector of A is 1 B = 2 −1 with eigenvalue 6. A scales by 4 in the B1-direction and A scales by 6 in the B2-direction. x2 AB2 AB1 B2 B1 x1 Let's discuss how we can find eigenvalues and eigenvectors. Let A be an n × n matrix. Notice that AX = λX for a scalar λ and nonzero vector X 2 Rn if and only if (A − λI) X = 0 has a nontrivial solution X. Hence λ is an eigenvalue if and only if the matrix A − λI is a singular matrix. In particular, λ must satisfy the characteristic equation det(A − λI) = 0: We call det(A − λI) the characteristic polynomial. det(A − λI) is a degree n polynomial of λ. Solving the characteristic equation for an n × n matrix A will give us n eigenvalues counting real and complex roots and repeated roots. Example 3. Let 1 3 A = : 2 2 Find all the eigenvalues for A and a basis for the corresponding eigenspaces. Answer. Finding eigenvalues of A. Solving the characteristic equation 1 − λ 3 2 det(A − λI) = = (1 − λ)(2 − λ) − 6 = λ − 3λ − 4 = (λ − 4)(λ + 1) = 0: 2 2 − λ Therefore the eigenvalues of A are λ = 4; −1. Having found an eigenvalue λ of A, we recall that a corresponding eigenvector X is a solution to the homogeneous linear system (A − λI) X = 0: 2 The eigenspace of A corresponding to the eigenvalue λ is the solution set to (A − λI) X = 0. Equivalently, the eigenspace is the null space Nul(A − λI). Clearly the eigenspace is a subspace of Rn. The eigenspace consists of the zero vector 0 and all non-zero eigenvalues of A corresponding to the eigenvalue λ. Since the eigenvectors corresponding to λ form a subspace (together with 0), there are infinitely many eigenvectors corresponding to λ. For instance, if X is an eigenvector such that AX = λX, then cX is also an eigenvector such that A(cX) = λ(cX) for every nonzero scalar λ. More generally, any linear combinations of eigenvectors corresponding to the same eigenvalue λ is again an eigenvector corresponding to λ. Consequently, to “find the eigenvalues of A corresponding to λ", we tend to want to find a basis for the eigenspace corresponding to λ. We do this by solving the homogeneous linear system (A − λI) X = 0 and letting the resulting spanning vectors be the basis for the eigenspace. Note that each eigenspace has many bases. In particular, how one approaches the algebra to solving (A−λI) X = 0 might yield different bases for the eigenspace. Example 3 continued. Finding eigenvector corresponding to 4. We solve (A − 4I)x = 0. −3 3 1 1 A − 4I = −! : 2 −2 0 0 x2 is a free variable and x1 is a pivot variable with x1 = −x2. Since there is only one free variable, the eigenspace corresponding to 4 has dimension 1. In general, the dimension of the eigenspace is equal to the number of free variables of (A − λI) X = 0. Since the eigenspace corresponding to 4 is one-dimensional, it suffices to find one eigenvector which spans this eigenspace. In particular, recalling that x1 = −x2, an eigenvector of A corresponding to λ = 4 is −1 : 1 Finding eigenvector corresponding to −1. We solve (A + I)x = 0. 2 3 1 3=2 A + I = −! : 2 3 0 0 x2 is a free variable and x1 is a pivot variable with x1 = (−3=2) x2, so a basis for the eigenspace of A corresponding to λ = −1 is (by setting x2 = 2) −3 : 2 Note that we set x2 = 2 to clear fractions. Any other choice of x2 would have simply yielded another eigenvector spanning the eigenspace. Example 4. Find the eigenvalues of the rotation matrix 0 −1 A = : 1 0 3 Answer. Solving the characteristic polynomial gives us −λ −1 2 det(A − λI) = = λ + 1 = 0 1 −λ p so λ = ±i, where i = −1. Thus the eigenvalues of A are complex numbers. We will eventually discuss complex eigenvalues and eigenvectors. Example 5. Find the eigenvalues and eigenvectors of 2 3 4 4 3 A = 4 0 5 2 5 : 0 0 3 Answer. Finding eigenvalues of A. Finding the characteristic polynomial by computing the determinant of an upper triangular matrix 3 − λ 4 4 2 det(A − λI) = 0 5 − λ 2 = (3 − λ) (5 − λ): 0 0 3 − λ Therefore the eigenvalues of A are the diagonal entries λ = 3; 5. Note that λ = 3 is an eigenvalue with multiplicity two since it comes from the factor (3−λ)2 with exponent 3. λ = 5 is an eigenvalue with multiplicity one since it comes from the factor (5 − λ) with exponent 1. In general, every n × n matrix has exactly n eigenvalues counting multiplicity. Example 5 illustrates the following important theorem. Theorem 1. Let A be an n × n upper triangular matrix. Then the eigenvalues of A are the diagonal entries of A. Example 5 continued. Finding eigenvector corresponding to 5. Using Gaussian elimination to solve (A − 5I)X = 0: 2 −2 4 4 3 2 −2 4 0 3 2 1 −2 0 3 (−1=2) R1 7! R1 A − 5I = 4 0 0 2 5 −! 4 0 0 1 5 −−−−−−−−−! 4 0 0 1 5 0 0 −2 0 0 0 0 0 0 Thus x2 is a free variables and x1; x2 are pivot variables with x1 = 2x2 and x3 = 0. That is, the eigenspace corresponding to λ = 5 is one-dimensional and is spanned by the eigenvector 2 2 3 4 1 5 : 0 Finding eigenvector corresponding to 2. Using Gaussian elimination to solve (A − 3I)X = 0: 2 0 4 4 3 2 0 1 1 3 A − 3I = 4 0 2 2 5 −! 4 0 0 0 5 : 0 0 0 0 0 0 4 Thus x1; x3 are free variables and x2 is a pivot variable with x2 = −x3. That is, the basis for the eigenspace corresponding to λ = 3 is 82 3 2 39 < 1 0 = 4 0 5 ; 4 −1 5 : : 0 1 ; Notice that the dimension of the eigenspace corresponding to λ = 3 is equal to the number of free variables, which is two. Together, all three eigenvectors 82 3 2 3 2 39 < 2 1 0 = 4 1 5 ; 4 0 5 ; 4 −1 5 : 0 0 1 ; form a basis of R3. Warning! Let A be a square matrix and suppose we use row reduction to find an echelon form U of A. The eigenvalues of A are not typically the diagonal entries / eigenvalues of U. For instance, suppose we had 2 0 (1=2) R1 7! R1 1 0 A = −−−−−−−−! = U 0 3 (1=3) R2 7! R2 0 1 so that by scaling the rows we find that the reduced echelon form of U of A is the identity matrix. Then A has eigenvalues λ = 2; 3, but U has eigenvalue λ = 1 with multiplicity two. That is, A and U have different eigenvalues as a result of scaling the rows. Note that we may compute the determinant det(A − λI) using row reduction, but this involves applying row reduction to A − λI. However, it was important that we subtracted λ from the diagonal entries before applying row reduction. This is not the same as row reducing A to find U { with no λ in sight { and then computing the eigenvalues of U. Note that not only does scaling rows change eigenvalues. Any elementary row operation { including interchanging rows and adding a multiple of one row to another row { can change the eigenvalues of a matrix. Example 6. Find the eigenvalues and eigenvectors of 2 2 1 4 3 3 6 0 1 −4 1 7 A = 6 7 : 4 0 1 6 0 5 0 0 0 2 Answer.