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Characteristic polynomial has degree n and leading coefficent (-1)^n

  1. Jul 22, 2013 #1
    I've been looking for proof of the fact that the characteristic polynomial of an n by n matrix has degree n with leading coefficient ## (-1)^{n} ##.

    I first tried proving it myself but my method is a bit strange (it does use induction though) and I am doubting the rigor, so could perhaps someone please post a link to the proof? I've been looking around without much success. Thanks!

  2. jcsd
  3. Jul 22, 2013 #2


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    If you want to do this by induction, then my suggestion would be to use the cofactor expansion of the determinant. There are other ways of doing this though that avoid induction.
  4. Jul 22, 2013 #3


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    http://www.math.uri.edu/~eaton/NotesCh5.pdf , see Theorem 5.3.
  5. Jul 22, 2013 #4
    Well, it seems fairly straightforward. You have a matrix ##\mathbf{A}=\begin{bmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\ a_{2,1} & a_{2,2} & \cdots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \cdots & a_{n,n}\end{bmatrix}##.

    To find the characteristic polynomial, we consider ##\det(\mathbf{A}-\lambda\mathbf{I})=\det\begin{bmatrix} a_{1,1}-\lambda & a_{1,2} & \cdots & a_{1,n} \\ a_{2,1} & a_{2,2}-\lambda & \cdots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \cdots & a_{n,n}-\lambda\end{bmatrix}=0##. The proof should then become fairly evident when considering the definition of the determinant. I'm tired and lazy right now, though, so I'm going to leave it at that.

    As an aside, some authors choose to define the characteristic polynomial as ##\det(\lambda\mathbf{I}-\mathbf{A})=0##, in which case the factor of ##(-1)^n## is unnecessary.

    I'd like to see your proof by induction, though. That might be a fun read. :wink:
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