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Math 511 Fall 2009 Final Exam Corrected: Linear Operators and Eigenvalues - Prof. Paul How, Exams of Linear Algebra

The corrected version of the final exam for math 511: linear algebra, fall 2009. The exam covers topics such as eigenvalues, eigenvectors, diagonalizability, and characteristic polynomials. Students are required to answer true/false questions, short answers, and proof-based questions.

Typology: Exams

2009/2010

Uploaded on 02/24/2010

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Math 511 Fall 2009 Final Exam - Corrected (twice)
(The change is to add the word “not” in question 3 (b).
Name
Student Number
1. Answer Tfor true or Ffor false. three points each
Assume that all vector spaces are finite dimensional and that the field in R.
(a) A linear operator on an ndimensional vector space must have nor fewer eigenvalues.
(b) A linear operator on an ndimensional vector space with at least one eigenvector will have infinitely
many eigenvectors.
(c) If vis an eigenvector of the linear operator Tcorresponding to eigenvalue 3 the T(v)3v= 0.
(d) If Tis a linear operator on a finite dimension vector space Vand αand βare two bases for V, then
[T]αand [T]βhave the same eigenvectors.
(e) If Tis a linear operator on a finite dimension vector space Vand αand βare two bases for V, then
[T]αand [T]βhave the same eigenvalues.
(f) If Tis a linear operator on a finite dimensional vector space Vand if v1and v2are eigenvectors
corresponding to the eigenvalues λ1and λ2respectively then v1+v2is an eigenvector corresponding to
the eigenvalue λ1+λ2of T.
(g) If a linear operator Ton a vector space Vof dimension nhas at least n1 distinct eigenvalues then
Tis diagonalizable.
(h) If λis an eigenvalue of the linear operator Tand v1and v2are eigenvectors corresponding to the
eigenvalue λ, then v1=cv2for some scalar c.
(i) If λ1is an eigenvalue of a linear operator Tthen λis also an eigenvalue of T.
(j) If fT(t) (the characteristic polynomial of the linear operator T) doesn’t split, then Tis not diago-
nalizable.
2. Short Answer, five points each
(a) For the matrix A=
021
131
241
i. Find the eigenvalues.
ii. Find a basis βfor R3consisting of eigenvectors of A.
iii. What is [LA]β?
(b) Why is
1 2 2
012
0 2 3
not diagonalizable. Note that λ= 1 is an eigenvalue of the matrix.
(c) Find a polynomial p(t) such that p(A) = 0 where Ais the matrix from the previous problem.
(d) Give an example of a 3 ×3 non-zero matrix Band a polynomial p(x) of degree 2 such that P(B) = 0.
(The right hand side of this equation is the zero 3 ×3 matrix.)
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Math 511 Fall 2009 Final Exam - Corrected (twice)

(The change is to add the word “not” in question 3 (b). Name Student Number

  1. Answer T for true or F for false. three points each

Assume that all vector spaces are finite dimensional and that the field in R.

(a) A linear operator on an n dimensional vector space must have n or fewer eigenvalues.

(b) A linear operator on an n dimensional vector space with at least one eigenvector will have infinitely many eigenvectors.

(c) If v is an eigenvector of the linear operator T corresponding to eigenvalue 3 the T (v) − 3 v = 0.

(d) If T is a linear operator on a finite dimension vector space V and α and β are two bases for V , then [T ]α and [T ]β have the same eigenvectors.

(e) If T is a linear operator on a finite dimension vector space V and α and β are two bases for V , then [T ]α and [T ]β have the same eigenvalues.

(f) If T is a linear operator on a finite dimensional vector space V and if v 1 and v 2 are eigenvectors corresponding to the eigenvalues λ 1 and λ 2 respectively then v 1 + v 2 is an eigenvector corresponding to the eigenvalue λ 1 + λ 2 of T.

(g) If a linear operator T on a vector space V of dimension n has at least n − 1 distinct eigenvalues then T is diagonalizable.

(h) If λ is an eigenvalue of the linear operator T and v 1 and v 2 are eigenvectors corresponding to the eigenvalue λ, then v 1 = cv 2 for some scalar c.

(i) If λ 1 is an eigenvalue of a linear operator T then −λ is also an eigenvalue of T.

(j) If fT (t) (the characteristic polynomial of the linear operator T ) doesn’t split, then T is not diago- nalizable.

  1. Short Answer, five points each

(a) For the matrix A =

i. Find the eigenvalues. ii. Find a basis β for R^3 consisting of eigenvectors of A. iii. What is [LA]β?

(b) Why is

 (^) not diagonalizable. Note that λ = 1 is an eigenvalue of the matrix.

(c) Find a polynomial p(t) such that p(A) = 0 where A is the matrix from the previous problem. (d) Give an example of a 3 × 3 non-zero matrix B and a polynomial p(x) of degree 2 such that P (B) = 0. (The right hand side of this equation is the zero 3 × 3 matrix.)

  1. Proof, ten points each

(a) Assume that T : V → V is a linear operator on the finite dimensional vector space V. Show that there are bases α and β for V such that [T ]βα is a diagonal matrix. (b) Prove: If V is a finite dimensional vector space and T is a linear operator on V then T is one to one if and only if zero is not an eigenvalue of T. (c) Prove: If T is a linear operator on the vector space V and T has an inverse and U is any other linear operator on V , then T U and U T have the same characteristic polynomial. (This will be like the proof that two similar matrices have the same characteristic polynomial. You can us the fact that if β is any basis for V , then [T ]β has an inverse and that for any two matrices, A and B, det(A) det(B) = det(AB).) (d) Prove that for any linear operator S on a vector space V and for every eigenvalue λ of S, −λ is an eigenvalue of −S.

  1. Extra credit, 10 points

Assume that T and U are linear operators on R^3 , that U T = −T U and that U T has three distinct eigenvalues. Prove that one of the eigenvalues must be zero. (Hint: Use the results of the previous three problems and do the proof by contradiction: Assume the hypotheses and assume that none of the eigenvalues are zero. By problem 3b, U T is one to one and therefore T is one to one.)