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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...

BCLC
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Noble.
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  • For what sort of application are you looking to use the characteristic polynomial? You've tagged linear algebra, but is there something more specific? – Ian Coley Mar 23 '13 at 06:09
  • @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? – Noble. Mar 23 '13 at 06:11
  • It would probably be easier to use cofactor expansion. – EuYu Mar 23 '13 at 06:59
  • @EuYu Yeah, I thought about trying that as well. Were you thinking expansion along the first column, then second etc. until it becomes a 2x2 matrix? – Noble. Mar 23 '13 at 07:01
  • I am not sure if you know what type of matrix this is, but it is very well known. Notice that if you expand along the first row, you end up with two cofactors. The cofactor for the first column has the same form as your original matrix, just smaller. The cofactor for the last column is easily evaluated. I would suggest induction at this point. You can play around with $2\times 2$ and $3\times 3$ matrices to guess a formula for the general form. I recommend doing all of this first, but if you get stuck then googling "companion matrix" will offer some spoilers. – EuYu Mar 23 '13 at 07:06
  • @EuYu Thanks for the suggestions, I've noticed now expansion along the first row does seem to be the right way to go about this problem. I'll update my main post or submit an answer myself once I've tried working it through. Thanks. – Noble. Mar 23 '13 at 07:15
  • Sure thing. Feel free to ping me if you have any questions. – EuYu Mar 23 '13 at 07:20
  • @EuYu I've posted my work as an answer, I'd appreciate it if you could confirm it's correct (I think it is) or if there's any errors. Thanks a lot. – Noble. Mar 23 '13 at 08:05
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    See this answer for several proofs. – Marc van Leeuwen Mar 23 '13 at 08:12
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    @MarcvanLeeuwen Thank you Marc for the link, however, as a first-year undergrad some of the discussion goes over my head. – Noble. Mar 23 '13 at 08:26
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    possible duplicate of The characteristic and minimal polynomial of a companion matrix – Marc van Leeuwen Mar 23 '13 at 08:30
  • @Noble I can sympathise with that. Note however that the answer I gave there gives one computation by elementary row operations, and one by development by the last column. I thought I had added at some point in time a third computation by development by the first row, the proof you found, to the answer; however apparently I didn't (maybe it was at some other similar question). – Marc van Leeuwen Mar 23 '13 at 08:37
  • Possible duplicate of The characteristic and minimal polynomial of a companion matrix – José Carlos Santos Nov 12 '17 at 13:46

1 Answers1

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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.

Noble.
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