# 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

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

## 3 Answers

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?

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.

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 $$k, and $$\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.