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What does it mean for the characteristic polynomial to equal 0?

  1. Feb 12, 2010 #1
    1. The problem statement, all variables and given/known data

    I'm working on a problem which is show that the matrix A = ([tex]^{a}_{c}[/tex] [tex]^{b}_{d}[/tex]) is diagonalisable iff (a-d)2 + 4bc > 0 or a-d = b = c = 0

    2. Relevant equations



    3. The attempt at a solution

    I've shown if (a-d)2 + 4bc > 0 then the characteristic polynomial has 2 distinct roots so 2 distinct eigenvalues so is diagonalisable.

    However, if (a-d)2 + 4bc [tex]\geq[/tex] 0 then it has a repeated root, so I have split it into cases:

    If a-d = b = c then A is already diagonal

    However, if a-d = b = 0 (c non-zero), the characteristic polynomial of A is already equal to 0. Does this mean A has no eigenvalues? Infinitely many? I will encounter a similar problem if a-d = c = 0 (b non-zero).

    I am also unsure how to tackle (a-d)2 = 4bc.
     
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  3. Feb 12, 2010 #2

    vela

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    Does the problem statement have a greater-than sign or a greater-than-or-equal-to sign in it? As you have written it, you don't need to worry about the cases you're talking about since you've already shown in A is already diagonal or it has two distinct roots. I'm guessing it should have been the greater-than-or-equal-to sign in the original statement.

    If a-d=0 and b=0 and c not zero, you have a matrix of the form

    [tex]\begin{pmatrix}a & 0 \\ c & a\end{pmatrix}[/tex]

    which will have a characteristic polynomial of the form [itex](a-\lambda)^2=0[/itex].
     
  4. Feb 12, 2010 #3
    The question goes show the matrix [tex]\begin{pmatrix}a & b \\ c & d\end{pmatrix}[/tex] is diagonal if and only if a-d = b = c = 0 or (a-d)2+4bc > 0.

    So I have covered the case for (a-d)2+4bc > 0

    I have also shown if a-d=b=c=0 then it is diagonal.

    However I need to show that if a-d=b = 0 but c non-zero, a-d=c=0 but b non-zero, and (a-d)2 = -4bc (so b or c would have to be negative) do not lead to diagonal matrices.
     
  5. Feb 13, 2010 #4

    HallsofIvy

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    You mean the case (a-d)2+ 4bc= 0 since you have already handled (a-d)2+ 4bc> 0.

    You are only talking about 2 by 2 matrices so there cannot be more than 2 eigenvalues. In this particular case, 0 is a double eigenvalue.
    The real problem is that you are trying to prove something that is not true.
    The matrix
    [tex]\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}[/tex]
    satisifies the condition that [itex](a-d)^2+ 4bc= 0^2+ 0= 0[/itex] but is NOT diagonalizable.

    Your original statement said [itex](a-d)^2+ 4bc> 0[/itex] and you have already proved that. Why did you change to [itex](a-d)^2+ 4bc\ge 0[/itex]?

     
  6. Feb 13, 2010 #5
    The way I've tried to answer this question is to find the characteristic polynomial of the matrix which is x2 - (a+d)x + ad-bc

    This has 2 distinct real roots if (a-d)2 +4bc > 0 (so is definitely diagonalisable)

    It has a repeated real root if (a-d)2 + 4bc = 0, in which case don't know if it is diagonalosable. This is the part I'm trying to split into cases. I think I have managed the first 3, but I don't know how to show if (a-d)2 = -4bc (could happen if b or c was negative) then the matrix is not diagonalisable?
     
  7. Feb 13, 2010 #6

    vela

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    You've shown the RHS of the iff implies the LHS. What you want to show now is the LHS implies the RHS. It was a little confusing why you were still messing around with cases on the RHS, but your approach seems to be to prove the contrapositive, ~RHS -> ~LHS, which is logically fine. It might be easier, however, to prove LHS->RHS directly, depending on what you know about diagonalizable matrices. (I don't know how the proof goes; I'm just throwing the idea out there for you to ponder.)
     
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