Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Puzzled about characteristic polynomial output from my calc

  1. Dec 18, 2012 #1
    Hello everyone, first time poster, long time reader here!

    I'm an ex-math major and while I'm no longer pursuing a degree anymore in mathematics, I still continue onwards in my spare time trying to learn as much as I can about it because it's always been something I've enjoyed partaking in and sharing the knowledge with others. I did manage to pull off three years of classes at my university while as a math major. I have a pretty firm grip with calc/odes/(beginning) pdes, intro linear algebra, basic proofs and logic, intro group theory, intro complex analysis, and some very basic topology.

    I decided to treat myself with an early X-Mas present of a new calculator (ti-nspire cx cas); been trying to follow along with the examples given in the manual. Anyway, I'm just puzzled about the output on my calc regarding calculating char polynomials. It currently has the operation defined as [itex]p_{A}(\lambda) = det(\lambda * I - A)[/itex] with I as the nxn ident matrix, A is any nxn square matrix. When I use the charPoly function on the calculator with the matrix [itex]A = \left(\begin{array}{ccc} 1 & 3 & 0\\ 2 & -1 & 0\\ -2 & 2 & 5 \end{array}\right)[/itex], the calc outputs [itex]-\lambda^3+5\lambda^2+7\lambda-35[/itex]

    When I try to do it by hand, my answer is off by a factor of -1. Hopefully someone can point out to me where my mistake is:

    So I get [itex]\lambda*I - A = \left(\begin{array}{ccc} \lambda-1 & -3 & 0\\ -2 & \lambda+1 & 0\\ 2 & -2 & \lambda-5 \end{array}\right)[/itex]

    [itex]det(\lambda*I - A) = (\lambda - 1) \left|\begin{array}{cc} \lambda+1 & 0\\ -2 & \lambda-5 \end{array}\right| + 3 \left|\begin{array}{cc} -2 & 0\\ 2 & \lambda-5 \end{array}\right| [/itex]

    [itex]= (\lambda-1)(\lambda+1)(\lambda-5)+3[-2(\lambda-5)][/itex]

    [itex] = \lambda^3 -5\lambda^2 - 7\lambda + 35[/itex]

    Not really that big of a deal per se, just trying to find where I've made a mistake; any insight would be greatly appreciated.

    P.S. Hands down, PF is one of the BEST sites out there! :!!)
  2. jcsd
  3. Dec 18, 2012 #2
    Your calculation seems correct.

    Somehow I think that your calculator defines the characteristic polynomial as [itex]\det(A-\lambda I)[/itex]. And not as [itex]\det(\lambda I -A)[/itex] like the manual claims.

    Can you test this by finding the characteristic polynomial of the [itex]3\times 3[/itex] zero matrix?? Under your definition, it should be [itex]\lambda^3[/itex].
  4. Dec 18, 2012 #3
    Hmm this is quite funky...
    [itex]\det(A-\lambda I)[/itex] yields [itex]-\lambda^3[/itex], the same as if I were to output it with the charPoly() function. However, when I type in [itex]\det(\lambda I -A)[/itex], it spits out [itex]\lambda^3[/itex]

    Even after clearing out any stored variables/history and resetting it, nothing changes. Wonder if it's a kink with the software?
  5. Dec 19, 2012 #4


    User Avatar
    Science Advisor
    Homework Helper

    It looks like a kink in the documentation, not the software.

    The characteristic polymonial should be ##\det(A - \lambda I)##, as Micromass said.
  6. Dec 19, 2012 #5


    User Avatar
    Science Advisor

    Some texts define the characteristic polynomial as ##det(A-λI)##, others as ##det(λI-A)##. So both you and your software are correct, or rather, neither is more right than the other.
    It has little practical significance, since these two only differ with a factor ##-1##, if ##n## (the size of the matrix) is odd.
  7. Dec 21, 2012 #6
    It's the same thing. You can just multiply by -1. The polynomial is equal to 0, so yeah.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook