Proving a specific solution vector of the characteristic polynomial is a cyclic vector in linear algebra

Question

I am working on a problem in linear algebra involving the characteristic polynomial of a matrix. The problem I encountered is as follows:

Let $A$ be an $n \times n$ matrix over a field $K$, with the characteristic polynomial $f(\lambda) = P_1(\lambda)P_2(\lambda) \cdots P_k(\lambda)$, where each $P_i(\lambda)$ $(1 \leq i \leq k)$ is a distinct monic irreducible polynomial over $K$.

Suppose $\alpha_i$ is a non-zero solution to the equation $P_i(A)x = 0$. Prove that $\alpha = \alpha_1 + \cdots + \alpha_k$ is a cyclic vector for $A$ in the cyclic space $K^n$.

My current understanding is as follows:

• I know how to prove $K^n$ is a cyclic space.
• The matrix $A$ has $n$ different eigenvalues over $ℂ$, and I can figure out this case rather than over $K$(i.e. $f(\lambda) = (\lambda-\lambda_1)(\lambda-\lambda_2)\cdots(\lambda-\lambda_n)$).
• The factorization of the characteristic polynomial might indicate different eigenvalues of the matrix $A$ and their corresponding eigenspaces.
• Since $\alpha_i$ is a solution to $P_i(A)x = 0$, $\alpha_i$ may lie in the eigenspace corresponding to $P_i$.
• The basic idea might be to prove $\alpha, A\alpha, A^2\alpha, \cdots,A^{n-1}\alpha$ are linearly independent.

However, I am unsure how to combine these observations to prove that $\alpha$ is a cyclic vector.

My approach so far has been to directly analyze the definition of a cyclic vector, but I have not yet arrived at a clear proof.

I would appreciate any guidance on how to prove this or any hints or theorems that might be useful.

Note:

Cyclic Subspace and Cyclic Vector

For a given vector $v$ and matrix $A$, the cyclic subspace is the smallest invariant subspace generated by $v$, which is formally expressed as

$$ \text{span}\{v, Av, A^2v, \dots\} $$

The cyclic subspace consists of all linear combinations of the vectors generated by the action of the matrix $A$ on the vector $v$.And $v$ is called cyclic vector over the cyclic subspace.

Cyclic Space

Given an $n \times n$ matrix $A$ and a vector $v$ in the vector space $K^n$ (where $K$ is a field), if the subspace generated by the vector $v$,

$$ \text{span}\{v, Av, A^2v, \dots, A^{n-1}v\} $$

equals the entire vector space $K^n$, then the vector space $k^n$ is called the cyclic space.

In other words, a cyclic space is defined as the entire vector space generated through the action of the matrix $A$ on a cyclic vector.

Answer 1

The problem can be approached using a straightforward strategy. As mentioned above, the goal is to prove that the set $$\{\alpha, A\alpha, \ldots, A^{n-1}\alpha\}$$ is linearly independent.

Let$ \sum_{i=1}^{n} l_i A^{i-1}\alpha = 0$, and we need to show that $l_i = 0$ for all $i = 1, 2, \ldots, n$.

Step 1: Proving $\{\alpha_i\}_{i=1}^k$ is linearly independent

First, consider the set $\{\alpha_i\}_{i=1}^k$, where $P_i(A)\alpha_i = 0$ for $i = 1, 2, \ldots, k$.

Since the characteristic polynomial $f(\lambda)$ factors as $f(\lambda) = \prod_{i=1}^k P_i(\lambda)$, where the elements of the set $\{P_i(\lambda)\}_{i=1}^k$ are coprime, the vector space $V$ can be decomposed as: $$ V = \ker P_1(A) \oplus \ker P_2(A) \oplus \cdots \oplus \ker P_k(A), $$ where $\alpha_i \in \ker P_i(A)$. Therefore, the set $\{\alpha_i\}_{i=1}^k$ is linearly independent.

Step 2: Proving $\{\alpha, A\alpha, \ldots, A^{n-1}\alpha\}$ is linearly independent

Next, we prove that $\{\alpha, A\alpha, \ldots, A^{n-1}\alpha\}$ is linearly independent:

The linear combination of $\{A^{j-1}\alpha\}_{j=1}^n$ can be rewritten as a combination of $\{\alpha_i\}_{i=1}^k$: $$ \sum_{j=1}^n l_j (A^{j-1}\alpha) = \sum_{j=1}^n l_j \sum_{i=1}^k A^{j-1}\alpha_i = \sum_{j=1}^n \sum_{i=1}^k l_j A^{j-1} \alpha_i = \sum_{i=1}^k \left(\sum_{j=1}^n l_j A^{j-1}\right) \alpha_i. $$

Since $\{\alpha_i\}_{i=1}^k$ is linearly independent, if $\sum_{i=1}^{n} l_i A^{i-1}\alpha = 0$, then $\sum_{j=1}^n l_j A^{j-1} = 0$.

Step 3: Utilizing the diagonalizability of $A$

It is also known that $A$ is diagonalizable over the complex field (this can be verified independently). Thus, $A$ can be decomposed as: $$ A = P \Lambda P^{-1}, $$ where $\Lambda$ is a diagonal matrix. Substituting this into the equation, we get: $$ \sum_{j=1}^n l_j \Lambda^{j-1} = 0. $$

Since $A$ is not a null matrix, neither is $\Lambda$. Therefore, it must be that $l_i = 0$ for all $i = 1, 2, \ldots, n$.

Conclusion

Thus, we have proved the linear independence of the set $\{\alpha, A\alpha, \ldots, A^{n-1}\alpha\}$.

Q.E.D.