Prove that $C$ is diagonalizable if and only if $$
C=\begin{bmatrix}
0 & 1 & 0 &\cdots & 0\\
0 & 0 & 1 &\cdots & 0 \\
\vdots&\vdots &\vdots&\ddots&\vdots\\
0 & 0 & 0 &\cdots &1 \\
-\alpha_0 &-\alpha_1 &-\alpha_2 &\cdots&-\alpha_{n-1}
\end{bmatrix}
$$ 
Prove that $C$ is diagonalizable if and only if the polynomial of $C$ has $n$ distinct roots
 A: In general please try to let us know what you've tried and what is confusing you. 
If the characteristic polynomial has $n$ distinct roots, then $C$ must be diagonalizable. Why? Because then we will have $n$-distinct eigenvectors which will span the space. 
Conversely, suppose the matrix is diagonalizable. Let's write $\text{char}(C)$. We have
$$f(x)=-x^n-\alpha_{n-1}x^{n-1}-\dots-\alpha_2x^2-\alpha_1x-\alpha_0$$
Now what happens if they're not all distinct? 
A: For a companion matrix, the characteristic polynomial and the minimal polynomial agree. That is, each eigenvalue occurs in a single Jordan block. With $n$ distinct eigenvalues, these blocks are just 1 by 1 on the diagonal. If there are fewer than $n$ distinct eigenvalues, the pigeonhole principle says that some eigenvalue is in a Jordan block that is $k$ by $k$ with $k \geq 2.$ In particular, the Jordan form is not diagonal then. 
A: You can prove by recursive relation.  I assume that you are working in the field $K$ over which the polynomial $$p(x):=x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_1x+a_0\in K[x]$$ splits into linear factors.  
We say $p(x)$ is the characteristic polynomial of a sequence $\left(t_k\right)_{k\in\mathbb{Z}_{\geq  0}}$ of elements in the field $K$ if
$$t_{k}+a_{n-1}t_{k-1}+a_{n-2}t_{k-2}+\ldots+a_1t_{k-n+1}+a_0t_{k-n}=0$$
for all integers $k\geq n$.  If $p(x)$ has a root $\lambda$ with multiplicity $m>1$, then show that, for each $j=0,1,2,\ldots,m-1$, $p(x)$ is the characteristic polynomial of $\left(t_k\right)_{k\in\mathbb{Z}_{\geq 0}}$ with $$t_k=\binom{k}{j}\,\lambda^{k-j}\text{ for all integers }k\geq 0\,.$$
Here, we use the convention that $\binom{k}{j}\,\lambda^{k-j}=0$ for all nonnegative integers $k<j$, and $\binom{k}{j}\,\lambda^{k-j}=1$ when $k=j$, regardless of whether $\lambda=0$. (Hint: The polynomial $x^{k-n}\,p(x)\in K[x]$ has $\lambda$ as a repeated root of multiplicity $m$ for each integer $k\geq n$, so that the coefficient of $(x-\lambda)^j$, for each $j=0,1,2,\ldots,m-1$, is zero in the expansion of $x^{k-n}\,p(x)$ as a polynomial in $x-\lambda$.)
Now, write $$v_j:=\begin{bmatrix}\binom{0}{j}\,\lambda^{0-j}&\binom{1}{j}\,\lambda^{1-j}&\cdots&\binom{n-2}{j}\,\lambda^{n-2-j}&\binom{n-1}{j}\,\lambda^{n-1-j}\end{bmatrix}^\top$$ for every $j=0,1,2,\ldots,m-1$.  Show that $v_0,v_1,v_2,\ldots,v_{m-1}$ are $m$ linearly independent over $K$, $v_0$ is an eigenvector of $C$ corresponding to the eigenvalue $\lambda$, and $v_1,v_2,\ldots,v_{m-1}$ are not an eigenvector of $C$, but these $m-1$ vectors are generalized eigenvector of $C$ corresponding to the eigenvalue $\lambda$.  
In fact,
$$(C-\lambda\,I)^s\,v_j=v_{j-s}$$
for each $s=0,1,2,\ldots,j$ and $j=0,1,2,\ldots,m-1$.  Here, $I$ denots the $n$-by-$n$ identity matrix.  Thus, the generalized eigenvalue $\lambda$ corresponds to the (indecomposable) Jordan block of $C$ of dimension $m\times m$, where $m$ is the multiplicity of the root $\lambda$ of $p(x)$.  Hence, $C$ is a diagonagonalizable over $K$ if $p(x)$ splits over $K$ and every root of $p(x)$ is simple.
