minimal polynomial of $n$ by $n$ matrix I want to know if $x^{n-1}-1$ can be a minimal polynomial of n by n matrix with real entries. 
If yes, find the matrix. 
If no, why so? 
I am thinking that when we subtract something from diagonal element and expand the matrix to get characteristic polynomial, the eigenvalue must be in the expression. So the minimal polynomial in the question gives roots of unity as eigenvalue, which are complex. 
So no such matrix. Am I correct? 
 A: Any monic polynomial can be the minimal polynomial of a matrix. If $$p(x) = a_0 +a_1x + \cdots + a_{n-1}x^{n-1} + x^n.$$
then $p$ is the minimal and characteristic polynomial of the $n \times n$ matrix $$C(p) = \begin{pmatrix}
0 & 0 & \dots & 0 & -a_0\\
1 & 0 & \dots & 0 & -a_1\\
0 & 1 & \dots & 0 & -a_1\\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \dots & 1 & -a_{n-1}\\
\end{pmatrix}_.$$
$C(p)$ is called the "Companion Matrix" of $p(x)$.
I just realized that your question asked about $x^{n-1}-1$ as the minimal polynomial of a matrix of size $n$. In that case you can take a block diagonal matrix:
$$\begin{pmatrix}
1 & 0 \\
0 & C(x^{n-1}-1)\\
\end{pmatrix}$$
To answer your question of a matrix "satisfying" a polynomial, I will answer your question with the specific polynomial in your question.
To say that a matrix $A$ satisfies the polynomial $x^{n-1}-1$ means that $A^{n-1}-I = \textbf{0}$. Where $I$ denotes the identity matrix and $\textbf{0}$ denotes the zero matrix.
$\textbf{Example:}$ Let $p(x) = x^3 + 2x^2 - x + 4$. Then $$C(p) = \begin{pmatrix}
0 & 0 & -4\\
1 & 0 & 1 \\
0 & 1 & -2\\
\end{pmatrix}_.$$
To see that $C(p)$ satisfies $p(x)$,
\begin{align*}
C(p)^3 &= \begin{pmatrix}
-4 & 8 & -20\\
1 & -6 & 13\\
-2 & 5 & -16\\
\end{pmatrix}\\
C(p)^2 &= \begin{pmatrix}
0 & -4 & 8 \\
0 & 1 & -6 \\
1 & -2 & 5\\
\end{pmatrix}
\end{align*}
So 
\begin{align*}
p(C(p)) &= C(p)^3 + 2 C(p)^2 - C(p) + 4I\\
&= \begin{pmatrix}
-4 & 8 & -20\\
1 & -6 & 13\\
-2 & 5 & -16\\
\end{pmatrix} + 2 \begin{pmatrix}
0 & -4 & 8 \\
0 & 1 & -6 \\
1 & -2 & 5\\
\end{pmatrix} - \begin{pmatrix}
0 & 0 & -4\\
1 & 0 & 1 \\
0 & 1 & -2\\
\end{pmatrix} + \begin{pmatrix}
4 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & 4 \\
\end{pmatrix} \\
&= \begin{pmatrix}
-4 & 8 & -20\\
1 & -6 & 13\\
-2 & 5 & -16\\
\end{pmatrix} + \begin{pmatrix}
0 & -8 & 16 \\
0 & 2 & -12 \\
2 & -4 & 10\\
\end{pmatrix} + \begin{pmatrix}
0 & 0 & 4\\
-1 & 0 & -1 \\
0 & -1 & 2\\
\end{pmatrix} + \begin{pmatrix}
4 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & 4 \\
\end{pmatrix}\\
&= \begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}_.
\end{align*}
A: You want an $n$ by $n$ matrix with minimal polynomial $x^{n-1} -1$. It is sufficient (and necessary if $n$ is even) that the matrix has the invariant factors: $x-1, x^{n-1} - 1$. This means that all such matrices must have the rational canonical form: $C(x -1) \oplus C(x^{n-1} - 1)$ (which means that we can take this matrix as an example), where $C(p(x))$ is the Fröbenius Companion Matrix of a monic polynomial. 
For instance, if $n=5$, you get the following:
$$\begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{bmatrix}$$
