On Hoffman and Kunze's Linear Algebra Section 9.6, page 355, Theorem 20 states without proof that:
Let $T$ be a normal linear operator on a finite dimensional inner product space $V$ (dim $V \geq 1$). Then there exist $r$ nonzero vectors $\alpha_1,…,\alpha_r$ in $V$ with respective $T$-annihilators $e_1,…,e_r$ such that
(i)$$V = Z(\alpha_1 ; T) \oplus … \oplus Z(\alpha_r;T);$$
(ii)if $1 \leq k \leq r-1$ then $e_{k+1}$ divides $e_k$;
(iii) $Z(\alpha_j;T)$ is orthogonal to $Z(\alpha_k;T)$ when $j \neq k$. Furthermore, the integer $r$ and the annihilators $e_1,…e_k$ are uniquely determined by conditions (i) and (ii) and the fact that no $\alpha_k$ is $0$.
The below link also mentions this question:
https://mathoverflow.net/questions/306759/error-in-hoffman-kunze-normal-operators-on-finite-dimensional-inner-product-spa
I am not sure how to overcome the difficulty of the $Z(\alpha_j;T)$ being orthogonals to each other. I certainly get the proof of the cyclic decomposition theorem,but this theorem is the "strengthened" cyclic decomposition theorem for a normal matrix. The author says that the key to proving this theorem is to know that for a normal operator any $T$-invariant subspace is also $T^*$-invariant. If I can show all but the second part of (iii), the uniqueness of the "original" cyclic decomposition theorem garantees the second part of (iii), so I may disregard the second part of (iii).
My attempt: First choose $\alpha_1$ so that the degree of $e_1$ is maximal. Now choose $\alpha_2 \neq 0$ among the orthogonal compliment of $Z(\alpha_1;T)$ so that $e_2$ divides $e_1$ and the degree of $e_2$ is maximal. Now choose $\alpha_3 \neq 0$ among the orthogonal compliment of $Z(\alpha_1;T) \oplus Z(\alpha_2;T)$ so that $e_3$ divides $e_2$ and the degree of $e_3$ is maximal. As the degree of subspace increases in each step, this process will end in a finite number of steps. The resulting final subspace $W = Z(\alpha_1 ; T) \oplus … \oplus Z(\alpha_r;T)$ satisfies all three conditions possibly except the condition that $V=W$. Now suppose $W$ is a proper subspace. I can find a vector $\beta$ that is orthogonal to $W$ and show (by the fact that the components of $W$ must all be $T^*$-invariant) that $Z(\beta;T)$ is orthogonal to each $Z(\alpha_k;T)$, but then the problem arises in satisfying the condition (ii). Certainly the degree of the $T$-annihilator $g$ of $Z(\beta;T)$ is not greater than $e_r$ but I have to show that $g$ divides $e_r$.
Best Answer
First, your idea actually works but it requires the following observation: If you choose $\alpha_1$ such that the degree of $T$-annihilator $e_1$ of $\alpha_1$ is of maximal degree, then in fact the $T$-annihilator of $\alpha_1$ is the minimal polynomial of $T$. Then, if you consider the $T$-invariant decomposition $$ V = Z(\alpha_1;T) \oplus Z(\alpha_1;T)^{\perp}, $$ the $T$-annihilator of any non-zero $\alpha \in Z(\alpha_1;T)^{\perp}$ will necessarily divide $e_1$ and then you can continue this process and you are guranteed to arrive at the end to an orthogonal $T$-invariant decomposition $$ V = Z(\alpha_1;T) \oplus Z(\alpha_2;T) \oplus \dots \oplus Z(\alpha_r;T) $$ with $e_r | e_{r-1} | \dots | e_1$.
Instead of justifying the observation above (which follows immediately from the existence of the cyclic decomposition or can be proved independently), I suggest you choose $\alpha_1$ such that the $T$-annihilator of $\alpha_1$ is the minimal polynomial of $T$. This is possible by a corollary of the cyclic decomposition theorem which appears on page 237. Then choose $\alpha_2$ such that the $T$-conductor of $\alpha_2$ is the minimal polynomial of $T|_{Z(\alpha_1;T)^{\perp}}$ and continue inductively to obtain the required orthogonal cyclic decomposition.