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Math Help - Determinant

  1. #1
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    Determinant

    Hi!
    I need some help on the following problem.

    Let A = \begin{pmatrix} -a_{n-1} & -a_{n-2} & ... & -a_0 \\ 1 & 0 & & \\  & \ddots & \ddots & \\ & &1 & 0 \end{pmatrix} \in \mathbb{R}^{n \times n}, n \in \mathbb{N}.

    Show that \forall t \in \mathbb{C}

    det(t*I - A) = \sum^{n-1}_{j=0} a_j t^j + t^n, where

    I \in \mathbb{R}^{n \times n } is the identity matrix.


    Actually I tried this:

    det(tI-A) = det \begin{pmatrix} t+a_{n-1} & a_{n-2} & ... & a_0 \\ 1 & t & & \\ & \ddots & \ddots & \\ & &1 & t \end{pmatrix}

    = \pm 1*det \begin{pmatrix} t+a_{n-1} & a_{n-2} & ... & a_2 & a_0 \\ 1 & t & & & \\ & \ddots & \ddots & \\ & & & & t \end{pmatrix}\pm t* det \begin{pmatrix} t+a_{n-1} & a_{n-2} & ... & a_2 & a_1 \\ 1 & t & & & \\ & \ddots & \ddots & \\ & & & & 1 \end{pmatrix}

    I guess that doesn't work that way. Furthermore I am confused about + or -, it depends on the line and colomn, whatever it is not that important, because my try leads to nothing.

    Thanks for any comment, help would be much appreciated.

    All the best,
    Rapha
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  2. #2
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    Quote Originally Posted by Rapha View Post
    det(tI-A) = det \begin{pmatrix} t+a_{n-1} & a_{n-2} & ... & a_0 \\ 1 & t & & \\ & \ddots & \ddots & \\ & &1 & t \end{pmatrix}
    In fact, it should be \det(tI-A) = \begin{vmatrix} t+a_{n-1} & a_{n-2} & ... & a_0 \\ -1 & t & & \\ & \ddots & \ddots & \\ & &-1 & t \end{vmatrix} (with –1s on the subdiagonal).

    Using row and column operations, add t times the first column to the second column, then t times the (new) second column to the third column, and so on until you add t times the (n-1)th column to the n'th column. That kills off all the t's on the main diagonal (except for the first one), and you are left with \begin{vmatrix}\scriptstyle t+a_{n-1} &\scriptstyle t^2+ta_{n-1}+a_{n-2} & ... &\scriptstyle t^n + \sum_{j=0}^{n-1}a_jt^j \\ -1 & 0 & & 0\\ 0& \ddots & \ddots &0 \\ 0& \ldots&-1 & 0 \end{vmatrix}. Now expand down the last column.
    Last edited by Opalg; January 20th 2009 at 08:30 AM.
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  3. #3
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    Hello Opalg,
    hi everyone

    Quote Originally Posted by Opalg View Post
    In fact, it should be \det(tI-A) = \begin{vmatrix} t+a_{n-1} & a_{n-2} & ... & a_0 \\ -1 & t & & \\ & \ddots & \ddots & \\ & &-1 & t \end{vmatrix} (with 1s on the subdiagonal).

    Using row and column operations, add t times the first column to the second column, then t times the (new) second column to the third column, and so on until you add t times the (n-1)th column to the n'th column. That kills off all the t's on the main diagonal (except for the first one), and you are left with \begin{vmatrix}\scriptstyle t+a_{n-1} &\scriptstyle t^2+ta_{n-1}+a_{n-2} & ... &\scriptstyle \sum_{j=0}^{n-1}a_jt^j \\ -1 & 0 & & 0\\ 0& \ddots & \ddots &0 \\ 0& \ldots&-1 & 0 \end{vmatrix}.
    Wow, what a great idea. Thank you. I was able to understand your steps, very good explanation.

    Quote Originally Posted by Opalg View Post
    Now expand down the last column.
    \begin{vmatrix}\scriptstyle t+a_{n-1} &\scriptstyle t^2+ta_{n-1}+a_{n-2} & ...&\scriptstyle \sum_{j=0}^{n-2}a_jt^j  &\scriptstyle \sum_{j=0}^{n-1}a_jt^j \\ -1 & 0 & &0& 0\\ 0& \ddots & \ddots &0 \\ 0&0& \ldots&-1 & 0 \end{vmatrix}

    =\pm (-1)* \begin{vmatrix} t+a_{n-1} & t^2+ta_{n-1}+a_{n-2} & ... &\sum^{n-3}_{j=0}a_jt^j & \sum^{n-1}_{j=0}a_jt^j \\ -1 & 0 & & & \\ 0 & \ddots & \ddots & \ddots & 0 \\ 0 & 0 & ... &-1 &0 \end{vmatrix}

    If I repeat it (n-1 times) I have

    \pm (-1)^{n-1} * \sum^{n-1}_{j=0}a_jt^j

    Again I have a problem with - or +.
    I still don't get it, where does the +t^n come from [according to det(t*I - A) = \sum^{n-1}_{j=0} a_j t^j + t^n ]?

    Kind regards,
    Rapha
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  4. #4
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    Quote Originally Posted by Rapha View Post
    I still don't get it, where does the +t^n come from [according to det(t*I - A) = \sum^{n-1}_{j=0} a_j t^j + t^n ]?
    Oops, my mistake. The element in the top right-hand corner of the matrix should be t^n + \sum_{j=1}^na_jt^j (I have corrected it in my previous comment). Then when you expand down the last column you get (-1)^{n-1}\Bigl(t^n + \sum_{j+1}^na_jt^j\Bigr)(-1)^{n-1}. The first (-1)^{n-1} is the sign attached to the element in the (1,n)-position in the matrix, and the second one comes from multiplying together the 1s on the subdiagonal. So altogether we have an even number of 1s, and their product is +1.
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  5. #5
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    Okay then.
    I was able to unterstand the solution completely, so thank you.

    All the best,
    Rapha
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