1. Not finding help here? Sign up for a free 30min 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!

Question about a Linear Algebra Proof?

  1. Apr 4, 2012 #1
    If we have a matrix with n distinct eigenvalues I understand why we have n distinct eigenvectors.

    My question is why is it if we have x[itex]_{1}[/itex]....x[itex]_{n}[/itex] eigenvectors of A with the largest eigenvalue equal to 1 and the rest of the eigenvalues are less than or equal to one, why can we assume that for any y[itex]_{0}[/itex] = c[itex]_{1}[/itex] x[itex]_{1}[/itex] + . . .+c[itex]_{n}[/itex] x[itex]_{n}[/itex]

    for some constants c[itex]_{1}[/itex] ..c[itex]_{n}[/itex] with c[itex]_{1}[/itex] [itex]\neq[/itex]0.

    Why does c[itex]_{1}[/itex] [itex]\neq[/itex]0.?

    I am assuming that we can get any vector because the eigenvectors form a basis for the space we are in.
     
  2. jcsd
  3. Apr 4, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    What are you actually trying to prove? Just because 1 is the largest eigenvalue there's no reason why the eigenvectors must be a basis. They might not be. And if you are given that x1..xn are a basis there is no reason why c1 must be nonzero. Suppose y0=x2?
     
    Last edited: Apr 4, 2012
  4. Apr 4, 2012 #3
    I wasnt trying to prove anything I was trying to uderstand why the statement is written the way it is. The eigen vectors are linearly independent so since there is n of them doesnt that mean they form a basis?
     
  5. Apr 5, 2012 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    If an n by n matrix has n independent eigenvectors, then you can take the eigenvectors to be a basis (you can take any n independent vectors to be a basis for an n dimensional space).

    But I don't understand what you mean by "for any y 0 = c 1 x 1 + . . .+c n x n".
    What does the "for any" mean here? If you just mean that any vector in the space can be written as a linear combination of the basis vectors, of course that is true. But it is NOT true that c1 cannot be 0. For example, if y0 is any vector you can take y0= xn. In that case, cn= 1 while all other coefficients, including c1, are 0.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Question about a Linear Algebra Proof?
Loading...