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Scalars z such that A-zI is singular

  1. Mar 8, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider the (2x2) symmetric matrix A = [a b; b d]

    Show that there are always real scalars z such that A-zI is singular [Hint: Use quadratic formula for the roots from the previous exercise ] t^2 - (a + d)t + (ad - bc) =0

    2. Relevant equations



    3. The attempt at a solution

    I just want to see if I did the problem correctly..

    First, i did the whole cross multiplication, and simplified to the required quadratic formula, given in the hint

    z2 - z(a + d) + (ad - b2) = 0

    Then I found roots to the above equation, but thatss where I want for you guys to double check my logic..

    The roots for z1/2 = ((a + b) +/- sqrt[(a - d)2+4b2])/2 (sorry for the messy equation :-S)
    Since (a - d)2+4b2 is always positive, then the root for the above equation does exist, therefore there are real scalars z such that A-zI is singular...

    Is that good, or should I add something else?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 8, 2010 #2

    vela

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    Re: Eigenvalues

    That's fine, though you should have (a+d) instead of (a+b) in your expression for the roots.
     
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