Suppose we have an $n \times n$ matrix such that $M^k = I_n$ (see details) Suppose we have a matrix $M$ of the following form: 
$$
\pmatrix{
a&1&\cdots&&0&0\\
&a&1&\cdots&0&0\\
&&\ddots &\ddots &\vdots&\vdots\\
&&&a&1&0\\
&&&&a&1\\
&&&&&a
}
$$
That is, a constant multiple of the identity matirx with ones on the super diagonal.  Show that if there exists a natural number $k$ such that $M^k = I_n$ then $n = 1$.  
The proof is easy in the $2 \times 2$ case, where we can find a general form for the $k$th power of such a matrix, and see that the diagonal entries are $a^k$ (which must equal one) while the super-diagonal entry is $ka^{k-1}$ (which must equal zero, and this is not possible if $k$ and $a$ are nonzero).  
I am having a hard time doing it for the $n \times n$ case.  I know my diagonal entries will be $a^k$ but I have much more to consider as far as the super-diagonal entries are concerned.  Can somebody help me with this?  
 A: The only eigenvalue of $M$ is $a$. The polynomial with simple roots $x^k-1$ annihilates $M$ so $M$ is diagonalizable and so it's similar to $aI_n$. Hence $M=aI_n$ which isn't possible for $n\ge2$.
A: Denote $M$ as $M_a$, so we can talk of $M_0$, which is a nilpotent matrix: $M_0^n=0$. Obviously $a\ne0$.
Since $M^k=I$ implies $M^{hk}=I$ for every $h$, we can assume that $k>n$. Now $M_a=aI+M_0$ and
$$
I=M_a^k=\sum_{i=0}^k\binom{k}{i}a^{k-i}M_0^i=
\sum_{i=0}^{n-1}\binom{k}{i}a^{k-i}M_0^i
$$
This is a contradiction unless $n=1$: just check the entry on place $(1,2)$, using the fact that $M_0^i$ has zero at that place, for $i>1$.
A: You can check that if $M$ has that form then $(M - aI_n)^n = 0$ which we can write as $f(M) = 0$ where $f(x) = (x - a)^n$. If in addition, $g(M) =0$ where $g(x) = x^k - 1$ then $\gcd(f,g)(M) = 0$. But note,


*

*the roots of $f$ are $a, \dots, a$ ($n$ times)

*the roots of $g$ are $\exp(2\pi i/n), \dots, \exp(2\pi i k/n), \dots, \exp(2\pi i n/n) = 1$

*any root of $\gcd(f,g)$ is a root of $f$ and a root of $g$


So either $\gcd(f,g) = 1$ (they are coprime) which is a contradiction since then $I_n = \gcd(f,g)(M) = 0$. Or, $\gcd(f,g) = (x - a)$ and $a = \exp(2 \pi i k/n)$ for some $1 \le k \le n$. Thus $\gcd(f,g)(M) = M - aI_n$ so $M = aI_n$. And $aI_n$ only has that superdiagonal form for $n = 1$.
A: The characteristic polynomial of $M$ is clearly $p_M(x)=(-1)^n(x-a)^n$, while
the polynomial $q(x)=x^k-1$ annihilates $M$, and hence its minimal polynomial divides $q(x)$. 
But $q$ has simple roots, and hence $M$ is diagonalisable, and since $p_M(x)=(-1)^n(x-a)^n$, has only one root $(x=a)$, then its minimal polynomial should be $m(x)=x-a$. But 
$$
m(M)=M-aI=0,
$$
only if $n=1$.
