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$$ C=\begin{bmatrix} 0 & 1 & 0 &\cdots & 0\\ 0 & 0 & 1 &\cdots & 0 \\ \vdots&\vdots &\vdots&\ddots&\vdots\\ 0 & 0 & 0 &\cdots &1 \\ -\alpha_0 &-\alpha_1 &-\alpha_2 &\cdots&-\alpha_{n-1} \end{bmatrix} $$

Prove that $C$ is diagonalizable if and only if the polynomial of $C$ has $n$ distinct roots

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  • $\begingroup$ It's unclear what you are asking. What is the polynomial of $C$? Whats your approach to the solution? Check "Companion Matrix" on Wikipedia. $\endgroup$ – Flowrian Oct 21 '18 at 13:39
  • $\begingroup$ If characteristic polynomial has $n$ distict roots ,then vandermonde matrix makes your $C$ diagonalizable. See this in Wikipedia page $\endgroup$ – Chinnapparaj R Oct 21 '18 at 14:14
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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?

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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.

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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.

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