1. ## 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

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

3. Hello Opalg,
hi everyone

Originally Posted by Opalg
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.

Originally Posted by Opalg
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

4. Originally Posted by Rapha
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.

5. Okay then.
I was able to unterstand the solution completely, so thank you.

All the best,
Rapha