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: Linear algebra

  1. Nov 16, 2006 #1
    I have two questions

    1. If I have a 2x2 matrix A with entries a, b, c, d where a is the upper left corner, b upper right corner, c lower left, and d lower right. I have eigenvalues L1 and L2. I need to show that L1^2 + L2^2 <= a^2 + b^2 + c^2 + d^2. So far I've done this: I know det(A)=L1*L2 and that tr(A) = L1 + L2. So L1 + L2 = a + b square each side and get L1^2 + L2^2 +2*L1*L2 = a^2 + d^2 +2ad subtract 2*L1*L2 from each side and get L1^2 + L2^2 = a^2+d^2+2ad-2*L1*L2 where since L1*L2 is det(A) I have a^2+d^2 + 2ad-2ad+2bc thus L1^2 + L2^2 = a^2 + d^2 + 2bc. Now all I would really need to show is that 2bc <= b^2+c^2. I also need to know when this inequality is equal. Am I on the right track or does anyone have any advice?

    2.If A is an nxn matrix where the sum of each row is 1 and all entires are positive. if v is an eigenvector of A with positive componentsShow the the associated eigenvalue is less than are equal to 1. Also, if we drop the requirement that the components of the eigenvector v be positive, is it still true that the associated eigenvalue is less than or equal to 1 in absolute value terms? Justify your answer. For this problem I was thinking I could consider the largest entry of v and take the corresponding entry in A. the largest that this A value could be is 1. If the value in v is less than 1 then the eigenvalue is then the most the eigenvalue could be is 1. If the value in v is 1 then the eigen value could be at most whatever the corresponding term in A is. If the value in v is greater than 1 then the eigenvalue is 1. I think I might be on the right track but I'm not really sure if this is a real clear/valid argument or if it's complete or how to put it together so it sounds lilke a clear proof. For the secone part I'm not really sure how to do it. Does anyone have any advice/help.

  2. jcsd
  3. Nov 16, 2006 #2


    User Avatar
    Science Advisor

    1. Very good so far. For any b, c, [itex](b-c)^2= b^2- 2bc+ c^2\ge 0[/itex] so [itex]b^2+ c^2\ge 2bc[/itex]. Of course, equality holds when b= c.
  4. Nov 16, 2006 #3
    Thanks, I think I got the first one now. Any advice on the second one? Am I on the right track?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook