1. Find the eigenvalues and eigenvectors of the following matrices. 5 7 a) A = −3 −5 λ is an eigenvalue of A iff det(A − λI) = 0. 5 − λ 7 det(A − λI) = det = (5 − λ)(−5 − λ) − 7 · (−3) = λ2 − 4. −3 −5 − λ Thus, det(A − λI) is zero for λ1 = 2 and λ2 = −2 only, so these are the eigenvalues of A. In order to find the eigenvector that belongs to λ1 the system of linear equations (A−λ1E)x1 = 0 is solved by Gaussian elimination. 5 − λ 7 0 3 7 0 1 3=7 0 1 = ∼ ∼ 1 3=7 0 −3 −5 − λ1 0 −3 −7 0 0 0 0 −3=7y That is the form of eigenvector x is: x = , where y 2 arbitrary. 1 1 y R −3=7t The subspace spanned by eigenvectors belonging to λ is: : t 2 . 1 t R Similarly, to calculate the eigenvectors that belong to λ2 = −2 the system (A − λ2E)x2 = 0 is solved using Gaussian elimination. 5 − λ 7 0 7 7 0 1 1 0 2 = ∼ ∼ 1 1 0 −3 −5 − λ2 0 −3 −3 0 0 0 0 −y That is the form of eigenvector x is: x = , where y 2 arbitrary. 2 2 y R −t The subspace spanned by eigenvectors belonging to λ is: : t 2 . 2 t R 00 0 −21 b) A = @3 −2 −3A 6 −6 1 λ is an eigenvalue of A iff det(A − λI) = 0. 0−λ 0 −2 1 det(A − λI) = det @ 3 −2 − λ −3 A = −λ((−2 − λ)(1 − λ) − (−6) · (−3)) − 2(3 · (−6) − 6 −6 1 − λ 6(−2 − λ)) = −λ3 − λ2 + 8λ + 12. Easy to see that −2 is a root of the cubic polynomial above, hence by long division we obtain: det(A − λI) = −λ3 − λ2 + 8λ + 12 = (λ + 2)(−λ2 + λ + 6) = (λ + 2)(−λ − 2)(λ − 3). Thus the eigenvalues are: λ1 = −2 s λ2 = 3. In order to find the eigenvectors that belong to λ1 the nonzero solutions of the system (A − λ1E)x1 = 0 are determined using Gaussian elimination. 0 1 0 1 0 1 −λ1 0 −2 0 2 0 −2 0 1 0 −1 0 @ 3 −2 − λ1 −3 0 A = @ 3 0 −3 0 A ∼ @ 0 0 0 0 A ∼ 6 −6 1 − λ1 0 6 −6 3 0 0 −6 9 0 1 0 −1 0 ∼ 0 1 −3=2 0 0 z 1 So the form of eigenvector x1 is: x1 = @ 3=2z A, where z 2 R n f0g arbitrary. z 80 1 9 < t = Thus the eigenspace of λ1 is: @ 3=2t A : t 2 R . : t ; Similarly, the calculation of eigenvectors that belong to λ2 = −3: 0 1 0 1 0 1 −λ2 0 −2 0 −3 0 −2 0 1 0 2=3 0 @ 3 −2 − λ2 −3 0 A = @ 3 −5 −3 0 A ∼ @ 0 −5 −5 0 A ∼ 6 −6 1 − λ2 0 6 −6 −2 0 0 −6 −6 0 1 0 2=3 0 ∼ 0 1 1 0 0 −2=3z 1 Thus the form of x2 is: x2 = @ −z A, where z 2 R n f0g arbitrary. z 80 1 9 < −2=3t = The eigenspace that belongs to λ2 is: @ −t A : t 2 R . : t ; 05 0 01 c) A = @0 1 2A 0 1 3 λ is an eigenvalue of A iff det(A − λI) = 0. 05 − λ 0 0 1 det(A−λI) = det @ 0 1 − λ 2 A = (5−λ)((1−λ)(3−λ)−1·2) = (5−λ)(λ2 −4λ+1). 0 1 3 − λ p Thus the eigenvalues are λ = 5, and from the quadratic formula λ = 4± 12 , that is λ = p p 1 2;3 2 2 2 + 3 s λ3 = 2 − 3. Calculation of eigenvectors that belong to λ1: 0 5 − λ 0 0 0 1 0 0 0 0 0 1 1 0 1 −1=2 0 0 1 − λ 2 0 = 0 −4 2 0 ∼ ∼ @ 1 A @ A 0 0 −3=2 0 0 1 3 − λ1 0 0 1 −2 0 0 1 0 0 ∼ 0 0 −3=2 0 0 x 1 Thus x1 is: x1 = @ 0 A, where x 2 R n f0g arbitrary. 0 80 1 9 < t = The eigenspace of λ1 is: @ 0 A : t 2 R . : 0 ; p Calculation of eigenvectors that belong to λ2 = 2 + 3: we find the nonzero solutions of (A − λ2E)x2 = 0 by Gaussian elimination. 0 1 0 p 1 5 − λ2 0 0 0 3 − 3 0p 0 0 @ 0 1 − λ2 2 0 A = @ 0 −1 − 3 2p 0 A ∼ 0 1 3 − λ2 0 0 1 1 − 3 0 0 1 0 1 1 0 0p 0 1 0 0p 0 ∼ @ 0 1 1 p− 3 p 0 A ∼ @ 0 1 1 − 3 0 A 0 0 2 + (1 + 3)(1 − 3) 0 0 0 0 0 0 1 p 0 Thus x2 is: x2 = @ ( 3 − 1)z A, ahol z 2 R n f0g. z 80 1 9 < p 0 = The eigenspace of λ2 is: @ ( 3 − 1)t A : t 2 R . : t ; p Similarly, calculation of eigenvectors that belong to λ2 = 2 − 3: 0 1 0 p 1 5 − λ3 0 0 0 3 + 3 0p 0 0 @ 0 1 − λ3 2 0 A = @ 0 −1 + 3 2p 0 A ∼ 0 1 3 − λ3 0 0 1 1 + 3 0 0 1 0 1 1 0 0p 0 1 0 0p 0 ∼ @ 0 1 1p + 3 p 0 A ∼ @ 0 1 1 + 3 0 A 0 0 2 + (1 − 3)(1 + 3) 0 0 0 0 0 0 1 p 0 Thus, x3 is: x3 = @ (− 3 − 1)z A, wher z 2 R n f0g arbitrary. z 80 1 9 < p 0 = The eigenspace of λ3: @ (− 3 − 1)t A : t 2 R . : t ; 01 1 01 d) A = @0 1 0A 0 1 1 λ is an eigenvalue of A iff det(A − λI) = 0. 01 − λ 1 0 1 det(A − λI) = det @ 0 1 − λ 0 A = (1 − λ)((1 − λ)(1 − λ) − 0) = (1 − λ)3. 0 1 1 − λ Thus the unique eigenvalue of matrix A is λ = 1. Calculation of eigenvectors: 0 1 0 1 1 − λ1 1 0 0 0 1 0 0 @ 0 1 − λ1 0 0 A = @ 0 0 0 0 A ∼ 0 1 0 0 0 1 1 − λ1 0 0 1 0 0 0 x 1 Thus the form of the eigenvector is: x = @ 0 A, where x; z 2 R, furthermore at least one of x z and z is nonzero. 80 1 9 < t1 = Thus the eigenspace of λ = 1 is: @ 0 A : t1; t2 2 R . : t2 ; 2. LetV be the vectorspace of real polynomials of degree at most 6. Determine the eigenvalues and eigenvectors of the following linear transformations. a) f(x) ! 0 The matrix of this transformation is the 6 × 6 all-zero matrix (in arbitrary basis). This transformation assigns to any polynomial the constant 0 polynomial, that is to f(x) the polynomial 0 · f(x). This means the only eigenvalue is 0, and every nonzero plynomial is an eigenvector, so the eigenspace of eigenvalue 0 is the whole space V . b) f ! f 0 The matrix of the transformation in the usual basis fx6; x5; : : : ; x; 1g is: 00 0 0 0 0 0 01 B6 0 0 0 0 0 0C B C B0 5 0 0 0 0 0C B C B0 0 4 0 0 0 0C B C B0 0 0 3 0 0 0C B C @0 0 0 0 2 0 0A 0 0 0 0 0 1 0 This transformation assigns every polynomial its derivative. Since differentiation decreases the degree of a polynomial (except for degree 0), f 0(x) = λf(x) can only hold if f(x) is of degree 0, that is a constant polynomial. The derivative of those is the constant 0 polynomial. Thus, the only eigenvalue is 0, and every constant, but nonzero polynomial is an eigenvector. The eigenspace belonging to eigenvalue 0 is ff(x) = cjc 2 Rg. c) f ! xf 0 6 5 The given transformation A assigns to polynomial f(x) = a6x + a5x + ··· + a1x + a0 the polynomial 6 5 A(f(x)) = 6 · a6x + 5 · a5x + ··· + 1 · a1x; (1) thus the matrix of A in the usual basis fx6; x5; : : : ; x; 1g is: 06 0 0 0 0 0 01 B0 5 0 0 0 0 0C B C B0 0 4 0 0 0 0C B C A = B0 0 0 3 0 0 0C B C B0 0 0 0 2 0 0C B C @0 0 0 0 0 1 0A 0 0 0 0 0 0 0 In order to find the eigenvalues, the roots of det(A − λI) are determined: 06 − λ 0 0 0 0 0 0 1 B 0 5 − λ 0 0 0 0 0 C B C B 0 0 4 − λ 0 0 0 0 C B C det(A − λI) = det B 0 0 0 3 − λ 0 0 0 C = B C B 0 0 0 0 2 − λ 0 0 C B C @ 0 0 0 0 0 1 − λ 0 A 0 0 0 0 0 0 −λ = (6 − λ)(5 − λ) ··· (1 − λ)(−λ) Thus the eigenvalues of A are 6; 5; 4; 3; 2; 1and 0. In order to find the eigenvectors, one has to 6 5 consider that by (1) the image of polynomial f(x) = a6x + a5x + ··· + a1x + a0 is 6-times itself exactly if a5 = a4 = ··· = a0 = 0.