# Jordan basis of $A$ when $A$ is the companion matrix?

The actual question: when $A$ is the companion matrix, why the general form of $M_i$ (the group of columns of the Jordan matrix $M$ that belongs to the block associated to $\lambda_i$) is:

$$M_i^{h,j} = \binom {h-1}{j-1} \lambda_i^{h-j}$$

$A$ has order $n$, $h$ is the row index (thus $1 \le h \le n$), $j$ is the column index ($1 \le j \le \mu_i$ with $\mu_i$ the algebraic multiplicity of $\lambda_i$).

Explanation for the first column of $M_i$. Given the characteristic polynomial:

$$p(z) = \frac 1a_n(a_nz^n + \dots + a_1z + a_0)$$

The corresponding companion matrix $A$ is:

$$\begin{bmatrix} 0 & 1 &&& \\ & \ddots & 1 &&\\ &&\ddots&\ddots &\\ &&& 0 & 1\\ - \frac{a_0}{a_n} & - \frac{a_1}{a_n} & \cdots & -\frac{a_{n-2}}{a_n} & -\frac{a_{n-1}}{a_n} \end{bmatrix}$$

Because $A$ is the companion matrix, we know that there are $d$ distinct eigenvalues, each having a geometric multiplicity 1.

The Jordan matrix is made of $d$ blocks: $J = diag(J(\lambda_1), \dots, J(\lambda_d))$ .Each block $J(\lambda_i)$ associated with the eigenvalue $\lambda_i$ has the form:

$$J(\lambda_i) = \begin{bmatrix} \lambda_i & 1 & & \\ & \lambda_i & \ddots &\\ & & \ddots & 1\\ & & & \lambda_i\\ \end{bmatrix}$$

The Jordan matrix $M$ can be partitioned in $d$ groups (a generic group would be $M_i$). The columns of $M_i$ forms the Jordan chain:

$$M_i = \begin{bmatrix}v_{i,1} & v_{i, 2} & \dots & v_{i, \mu_i}\end{bmatrix}$$

The first vector $v_{i,1}$ is the eigenvector, others are the generalized eigenvectors. Last index is $\mu_i$, the algebraic multiplicity of the eigenvalue $\lambda_i$.

Let's calculate the first column of $M_i$. We know that $AM_i = M_iJ(\lambda_i)$. Computing the product on column at time and we get (for the first column):

$$Av_{i,1} = \lambda_iv_{i,1}$$

The above is the exact definition of eigenvector. Solving $(A-\lambda_iI)v_{i,1} = 0$ is easy, because of the structure of $A$:

$$A-\lambda_iI = \begin{bmatrix} -\lambda_i & 1 &&& \\ & \ddots & 1 &&\\ &&\ddots&\ddots &\\ &&& -\lambda_i & 1\\ - \frac{a_0}{a_n} & - \frac{a_1}{a_n} & \cdots & -\frac{a_{n-2}}{a_n} & -\frac{a_{n-1}} {a_n} -\lambda_i \end{bmatrix}$$

First $n-1$ are linearly independent, $A-\lambda_iI$ is singular so the last row can be ignored. First $n-1$ gives us (note that $v_{i,1}^{1} \dots v_{i,1}^{n}$ are the coordinates of the vector):

\begin{align} v_{i,1}^{2} &= \lambda_i v_{i,1}^{1}\\ v_{i,1}^{3} &= \lambda_i v_{i,1}^{2} = \lambda_i^2 v_{i,1}^{1} \\ v_{i,1}^{4} &= \lambda_i v_{i,1}^{3} = \lambda_i^3 v_{i,1}^{1} \\ \vdots\\ v_{i,1}^{n} &= \lambda_i^{n-1} v_{i,1}^{1} \end{align}

Done! The general form of the first row of $M_i$ is then:

$$M_{i,1} = \begin{bmatrix} 1 \\ \lambda_i \\ \lambda_i^2 \\ \vdots \\ \lambda_i^{n-1} \end{bmatrix}$$

I cannot decipher your notation, notably of your actual question. However, you seem to be asking why for a companion matrix eigenvalues cannot have geometric multiplicity larger than $1$, or equivalently why there is only a single unbroken Jordan block for each eigenvalue $\lambda$ (if there would be several Jordan blocks for $\lambda$, each one would give an independent eigenvector for $\lambda$).
The clue to this is that companion matrices $A$ of size $n$ cannot by annihilated by a nonzero polynomial $P$ of degree less than $n$. To see that apply $P(A)$ on the right to the row vector $(1~0~0~\ldots~0)$, and see that it doesn't vanish.
Now assume that eigenvalue $\lambda$ has multiplicity $m$ as root of the characterisitic polynomial, but there is no Jordan block for $\lambda$ of size $m$ (there are multiple blocks, each of size less than $m$). That means that the corresponding block of $(A-\lambda I)^{m-1}$ already vanishes. It follows that the polynomial obtained by dividing the characterisitic polynomial by $X-\lambda$ already annihilates $A$. But we have seen that it cannot.