

Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Problem set 8 for the linear algebra course (math 110) at the university level. The problem set includes various problems related to eigenvalues, eigenvectors, determinants, and characteristic polynomials. Students are required to prove various mathematical statements and find the rational canonical form of a linear transformation. The document also introduces putzer's method for finding the eigenpolynomials and eigenvectors of a matrix. Each problem is worth a certain number of points.
Typology: Assignments
1 / 2
This page cannot be seen from the preview
Don't miss anything!


Problem Set 8 (due Friday October 29) MATH 110: Linear Algebra
Each problem is worth 10 points. PART 1
Problem 1(20) a) Prove that if A is an n × n matrix with all its eigenvalues equal to λ, then
exA^ = eλx
n∑− 1
k=
xk k!
(A − λI)k.
b) A 3 × 3 matrix A has all its eigenvalues equal to λ. Show that
exA^ =
eλx((λ^2 x^2 − 2 λx + 2)I + (− 2 λx^2 + 2x)A + x^2 A^2 ).
Problem 2(15) Curtis p. 226 2,3. Problem 3 (10) Let A and B be n × n matrices with detA = detB and trA = trB. Prove that A and B have the same characteristic polynomial if n = 2, but that this need not be the case if n > 2. Problem 4 (10) Suppose that the minimal polynomial of a linear transformation T : V → V satisfies m(x) = (x − α 1 )e 1 · · · (x − αs)es
as in the Triangular Form Theorem. Find the rational canonical form of T. PART 3 - Optional Problem (Putzer’s Method) Let λ 1 ,... , λn be the eigenvalues of an n × n matrix A, and define a sequence of polynomials in A as follows: P 0 (A) = I, and for k = 1,... , n:
Pk(A) =
∏^ k
m=
(A − λmI).
Show that exA^ =
n∑− 1
k=
rk+1(x)Pk(A)
where the function r 1 (x),... , rn(x) are defined recursively by the linear dif- ferential equations: r′ 1 (x) = λ 1 r 1 (x), r 1 (0) = 1
and for k = 1,... , n − 1:
r k′+1(x) = λk+1rk+1(x) + rk(x), rk+1(0) = 0.