Computing a characteristic polynomial I need help computing this characteristic polynomial. I tried myself to understand even the answer and I didn't manage.
Here's the characterstic polynomial:
$$
        \begin{matrix}
        t & -1 & 0 & \cdots &&&&& 0\\
        0 & t & 0 & \cdots &&&&& 0\\
        \vdots & & \ddots \\
        &&&&&&&&  0 \\
        0 & 0 & 0 & \cdots &&&&&  -1 \\
       -1 & 0 & 0 & \cdots  & &&&&t\\
        \end{matrix}
$$
Here's a picture:
Picture (Before the charcteristic polynomial)
The answer is : $\Lambda^n +(-1)^{n+1}(-1)^{n-1}(-1) = \Lambda^n + (-1)^{2n+1} = \Lambda^n - 1$
Can anyone explain me how  ? I did try with using $C_1$ for the determinant. but it won't get the same result as I wrote up there.
 A: To compute the determinant, consider, e.g., using the Laplace expansion along the first column. With
$$
A=\begin{bmatrix}
t & -1 &        & \\
  & t  & -1 & \\
  &    & \ddots & \ddots \\
  &    &        & t & -1 \\
-1  &    &        & & t
\end{bmatrix}\in\mathbb{C}^{n\times n},
$$
$$
\det(A)=t\det(A_{11})+(-1)(-1)^{n+1}\det(A_{n1}),
$$
where $A_{ij}$ is obtained from $A$ by removing the row $i$ and column $j$.
Also note, that the $(n-1)\times(n-1)$ matrices $A_{11}$ and $A_{n1}$ are triangular so that their determinants are products of the diagonal entries. The matrices $A_{11}$ and $A_{n1}$ contain $t$ and $-1$ on the diagonal, respectively, so
$$
\det(A_{11})=t^{n-1}, \quad \det(A_{n1})=(-1)^{n-1}.
$$
Putting this together,
$$
\det(A)=t\cdot t^{n-1}+(-1)(-1)^{n+1}(-1)^{n-1}=t^n-1.
$$
A: I think the given matrix is a Companion Matrix.
(Please follow the above link and observe that your matrix has the transpose form,which is also mentioned in the said page)
Now it is easy to find that characteristic polynomial is $t^n+(-1)=t^n-1$ 
(Assuming that it is a $n\times n$ matrix)
Please rectify me if I am wrong.
