Theorem: Let $\mathcal{A}$ be a linear operator on a finite dimensional vector space $V$, there exists a basis such that $V$ can be represented by the direct sum of some $\mathcal{A}$-cyclic subspace of $V$.
I want to proof this theorem in an "linear algebraic" way, but my proof is stuck somewhere and can't get the final result.
I proof this theorem by induction on dimension of $V$.
When $d=1$, the theorem is correct obviously.
Suppose the throrem is correct when $d\le n-1$. When $d=n$, choose any $v\in V,v\ne 0$, then we can generate $v$ to be a $\mathcal{A}$ cyclic subspace(denote by $U$). Note that U is $\mathcal{A}$ invariant, we can consider the quotient space $V/U$ which dimension is less than $n$.
But by induction, we can only know that under this basis, $\mathcal{A}$ can be represented by an upper triangular matrix, not a diagonal matrix. I don't know what conditions I overlooked.
In summary, I want a "linear algebraic" proof of this theorem and the uniqueness of Frobenius normal form.
Thanks.