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!

Homework Help: Eigenvalue(s) for a matrix

  1. Aug 1, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the eigenvalues of A, when A =[3,1,1; 0,5,0; -2,0,1]

    2. Relevant equations

    3. The attempt at a solution

    I took the det(A-[itex]\lambda[/itex]I)=0 and found the characteristic polynomial to be:


    I couldn't figure out whole roots to the equation and wolframalpha gave me
    [itex]\lambda[/itex]=5, 5-√2, 5+√2

    Does this seem like the right answer? I have checked and double checked the determinant. Would I still be able to find a basis for the eigenspace?
  2. jcsd
  3. Aug 1, 2012 #2


    User Avatar
    Gold Member

    For matrix:
    $$A= \begin{bmatrix}3 & 1 & 1 \\ 0 & 5 & 0 \\ -2 & 0 & 1\end{bmatrix}$$
    Here is the characteristic equation that i get:
    Last edited: Aug 1, 2012
  4. Aug 1, 2012 #3
    Because the second column is [0, 5, 0], the polynomial should look like (5 - x)Q(x), where Q(x) is a second-degree polynomial. So it should be very easy to get its roots. Do it.
  5. Aug 1, 2012 #4
    Sorry the matrix was $$A= \begin{bmatrix}3 & 1 & 1 \\ 0 & 5 & 0 \\ -2 & 0 & 7\end{bmatrix}$$
    not 2,0,1 in the last row. That was my mistake.
  6. Aug 1, 2012 #5
    Yes, those are the eigenvalues.
  7. Aug 1, 2012 #6


    Staff: Mentor

    The second row is [0, 5, 0].
  8. Aug 1, 2012 #7
    This is not clear from the original notation. Anyway, that does not matter for determinants so my argument holds.
  9. Aug 2, 2012 #8


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Expand the determinant along the second row.
  10. Aug 2, 2012 #9


    User Avatar
    Gold Member

    Well, the original notation used is actually very common and can be recognized as a standard, since it is also used in popular software like MATLAB.
  11. Aug 2, 2012 #10
    Let's put it this way: not clear for me, I always seem to forget the conventions for rows vs columns in a linear transcription. This may stem from the fact that even when working with paper and pencil, I am about equally likely to write down a matrix column after column and row after row. Let's hope this thread will leave a dent in my state of confusion :)
  12. Aug 2, 2012 #11


    User Avatar
    Gold Member

    Your characteristic equation from post #1 is correct. Now, you just need to solve it. Usually, you would start by trial and error to find one factor of the polynomial (you already know the easy one is (λ-5)) and then do long division to obtain a quadratic in the quotient, which you would then solve using the quadratic formula.
  13. Aug 2, 2012 #12
    The point I made in #3 is that none of this is required. By tackling the determinant along [0, 5, 0], one gets the polynomial directly in the factored form.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook