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The question I'm currently working on has boiled down to

$\chi_A(t) = \det \begin{bmatrix} t & 0 & 0 & \cdots & 0 & -a_0 \\ -1 & t & 0 & \cdots & 0 & -a_1 \\ 0 & -1 & t & \cdots & 0 & -a_2 \\ 0 & 0 & -1 & \cdots & 0 & -a_3 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & t & -a_{n-1} \\ 0 & 0 & 0 & \cdots & -1 & t-a_n \end{bmatrix}$

I can see I want to use EROs (and taking note of the changes, if any, the EROs make to the determinant) to end up with a lower/upper triangular matrix, so then the determinant is just the product of the entries along the main diagonal, but I'm not sure how to get it.

Swapping the first column and the last column would multiply the determinant by $-1$,but would leave us with only a $t$ and $-1$ in the last column to get rid of, giving a lower triangular matrix - but I can't see how I'd do this.

Many thanks in advance!

EDIT:

Swapping the first and last column multiplies the determinant by $-1$, so

$\chi_A(t) = -\det \begin{bmatrix} -a_0 & 0 & 0 & \cdots & 0 & t \\ -a_1 & t & 0 & \cdots & 0 & -1 \\ -a_2 & -1 & t & \cdots & 0 & 0 \\ -a_3 & 0 & -1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ -a_{n-1} & 0 & 0 & \cdots & t & 0 \\ t-a_n & 0 & 0 & \cdots & -1 & 0 \end{bmatrix}$

Adding $t$ times row two to row one doesn't change the determinant and so

$\chi_A(t) = -\det \begin{bmatrix} -a_0 - ta_1 & t^2 & 0 & \cdots & 0 & 0 \\ -a_1 & t & 0 & \cdots & 0 & -1 \\ -a_2 & -1 & t & \cdots & 0 & 0 \\ -a_3 & 0 & -1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ -a_{n-1} & 0 & 0 & \cdots & t & 0 \\ t-a_n & 0 & 0 & \cdots & -1 & 0 \end{bmatrix}$

This doesn't seem to have helped...

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  • $\begingroup$ For what sort of application are you looking to use the characteristic polynomial? You've tagged linear algebra, but is there something more specific? $\endgroup$ – Ian Coley Mar 23 '13 at 6:09
  • $\begingroup$ @FrankMcGovern Hi Frank. No, the question was originally about finding the matrix with respect to a basis, and the last step is just to find the characteristic polynomial of the linear operator - so it really is just solving that equation and nothing more (so I wasn't sure of the tags). Sorry if I've caused confusion. I could post the full question if that helps? $\endgroup$ – Noble. Mar 23 '13 at 6:11
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    $\begingroup$ See [this answer][1] for several proofs. [1]: math.stackexchange.com/a/78937/18880 $\endgroup$ – Marc van Leeuwen Mar 23 '13 at 8:12
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    $\begingroup$ @MarcvanLeeuwen Thank you Marc for the link, however, as a first-year undergrad some of the discussion goes over my head. $\endgroup$ – Noble. Mar 23 '13 at 8:26
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    $\begingroup$ possible duplicate of The characteristic and minimal polynomial of a companion matrix $\endgroup$ – Marc van Leeuwen Mar 23 '13 at 8:30
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Looking at the $2 \times 2$ and $3 \times 3$ forms of this matrix, we see that:

$\det \begin{bmatrix} t & -a_0 \\ -1 & t-a_1 \end{bmatrix} = t(t-a_1) - a_0 = t^2 - a_1t - a_0$

and, by expansion along the first row:

$\det \begin{bmatrix} t & 0 & -a_0 \\ -1 & t & -a_1 \\ 0 & -1 & t-a_2 \end{bmatrix} = t \times\det \begin{bmatrix} t & -a_1 \\ -1 & t-a_2 \end{bmatrix} + (-a_0) \det\begin{bmatrix} -1 & t \\ 0 & -1 \end{bmatrix}$

$= t[t(t-a_2) - a_1] - a_0 = t^3 - a_2t^2 - a_1t - a_0 $

So it looks like:

$\det \begin{bmatrix} t & 0 & 0 & \cdots & 0 & -a_0 \\ -1 & t & 0 & \cdots & 0 & -a_1 \\ 0 & -1 & t & \cdots & 0 & -a_2 \\ 0 & 0 & -1 & \cdots & 0 & -a_3 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & t & -a_{n-1} \\ 0 & 0 & 0 & \cdots & -1 & t-a_n \end{bmatrix} = t^{n+1} - a_nt^n - a_{n-1}t^{n-1} - ... - a_2t^2 - a_1t - a_0$

Which we can prove by induction.

Assume that:

$\det \begin{bmatrix} t & 0 & 0 & \cdots & 0 & -a_0 \\ -1 & t & 0 & \cdots & 0 & -a_1 \\ 0 & -1 & t & \cdots & 0 & -a_2 \\ 0 & 0 & -1 & \cdots & 0 & -a_3 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & t & -a_{n-2} \\ 0 & 0 & 0 & \cdots & -1 & t-a_{n-1} \end{bmatrix} = t^{n} - a_{n-1}t^{n-1} - a_{n-2}t^{n-2} - ... - a_2t^2 - a_1t - a_0$

Then, by expansion along the first row:

$\det \begin{bmatrix} t & 0 & 0 & \cdots & 0 & -a_0 \\ -1 & t & 0 & \cdots & 0 & -a_1 \\ 0 & -1 & t & \cdots & 0 & -a_2 \\ 0 & 0 & -1 & \cdots & 0 & -a_3 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & t & -a_{n-1} \\ 0 & 0 & 0 & \cdots & -1 & t-a_n \end{bmatrix} = t \det \begin{bmatrix} t & 0 & 0 & \cdots & 0 & -a_1 \\ -1 & t & 0 & \cdots & 0 & -a_2 \\ 0 & -1 & t & \cdots & 0 & -a_3 \\ 0 & 0 & -1 & \cdots & 0 & -a_4 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & t & -a_{n-1} \\ 0 & 0 & 0 & \cdots & -1 & t-a_n \end{bmatrix} $

$+ (-1)^{n+1} \times (-a_0)(-1)^n $

$ = t[t^{n} - a_{n}t^{n-1} - a_{n-1}t^{n-2} - ... - a_3t^2 - a_2t - a_1] + (-1)^{2n+1} a_0$

$ = t^{n+1} - a_nt^n - a_{n-1}t^{n-1} - ... - a_2t^2 - a_1t - a_0$

Proof complete.

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  • $\begingroup$ Nice explanation +1 $\endgroup$ – bubba Mar 23 '13 at 9:56
  • $\begingroup$ Looks good to me. $\endgroup$ – EuYu Mar 23 '13 at 20:06

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