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

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.

• What is the definition of a cyclic vector? (What is the definition of a cyclic space?) Commented Jul 31 at 7:53
• @Gerry Myerson, I have attached the definition of Cyclic Space and Cyclic Vector. Commented Jul 31 at 8:22

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.