1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Show that given n distinct eigenvalues, eigenvectors are independent

  1. Apr 3, 2017 #1
    1. The problem statement, all variables and given/known data
    Let ##T## be a linear operator on a vector space ##V##, and let ##\lambda_1,\lambda_2, \dots, \lambda_n## be distinct eigenvalues of ##T##. If ##v_1, v_2, \dots , v_n## are eigenvectors of ##T## such that ##\lambda_i## corresponds to ##v_i \ (1 \le i \le k)##, then ##\{ v_1, v_2, \dots , v_n \}## is linearly independent.

    2. Relevant equations


    3. The attempt at a solution
    The proof is by induction.
    First, suppose that ##n=1##. Then ##v_1 \ne 0## since it is an eigenvector and hence ## \{ v_1 \}## is linearly independent.
    Now assume that the theorem holds of ##n=k##, and that we have ##k+1## eigenvectors corresponding to distinct eigenvalues. We wish to show that these ##k+1## eigenvectors are linearly independent.
    Suppose that ##a_1, a_2 , \dots, a_n## are scalars such that ##a_1 v_1 + a_2 v_2 + \cdots + a_{k+1} v_{k+1} = 0## (1). Apply ##T## to both sides to get ##a_1 \lambda_1 v_1 + a_2 \lambda_2 v_2 + \cdots + a_{k+1} \lambda_{k+1} v_{k+1} = 0## (2).

    Now, my plan to continue the proof is to multiply (1) by ##\lambda_{k+1}## and then subtract (2) from it to get a linear relation only in terms of the vectors ##v_1, v_2 , \dots , v_k##. However, I am concerned about assuming what ##\lambda_{k+1}## is. Do I need to split the proof into two cases, one where ##\lambda_{k+1} = 0## and one where ##\lambda_{k+1} \ne 0##?
     
  2. jcsd
  3. Apr 3, 2017 #2

    fresh_42

    Staff: Mentor

    I think so. If you formulate the induction hypothesis as for all ##n \leq k##, then in the case ##\lambda_{k+1}=0## you can apply the argument on ##\lambda_{k}\neq 0## and then consider (1) once again before the multiplication by ##\lambda_{k}##.

    You could as well start with ##\lambda_1=0## and anchor the induction here to both cases.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted