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Let $\ m_T(x)=\prod p_i(x)^{m_i}$ be the minimal polynomial of an operator $T$ and $\ p_T(x)=\prod p_i(x)^{h_i}$ the characteristic polynomial.

Let $ V_i= \operatorname{Ker}{(p_i(x))^{h_i}} $

I don't understand why $\dim V_i = \deg\,(p_i)*h_i$

Thanks.

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Are we assuming $p_i$ are irreducible over our ground field? –  Arturo Magidin Jun 28 '11 at 19:19
    
@Arturo in the name of the question he sais "irred factors". So I guess the answer to your question is yes. –  Listing Jun 28 '11 at 19:22
    
@Listing: Yes, but posts should be self-contained. I'm not usually required to look in the spine of the book (or the title page of a book) for the hypothesis of a theorem stated in the book; are you? –  Arturo Magidin Jun 28 '11 at 20:08
    
If you see it that way, you are right :) –  Listing Jun 29 '11 at 5:15

2 Answers 2

Suppose that there are $k$ distinct irreducible factors in $m_T$. I'll use $\chi$ to denote characteristic polynomials, and reserve $p$ for those irreducible factors.

$V_i$ is an $T$-invariant subspace of $V$, so we can consider the restriction $T|_{V_i}:V_i\to V_i$. Let us write it $T_i$, for simplicity.

It is clear that the minimal polynomial of $T_i$ divides $p_i^{h_i}$, so it is in fact a power of $p_i$. It follows that the characteristic polynomial $\chi_{T_i}$ of $T_i$ is also a power of $p_i$.

Now it is easy to see that $V$ is the direct sum of $V_1$, $\dots$, $V_k$, each of which is invariant, so that $\chi_T=\chi_{T_1}\cdots \chi_{T_k}$. It follows immediately that $\chi_{f_i}$ has to be equal to $\chi_i^{h_i}$ and then, since the degree of $\chi_{f_i}$ is equal to $\dim V_i$, you get what you want.

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I'm assuming that the $p_i$ are pairwise distinct, monic, and irreducible (over the ground field).

Claim 1. If $i\neq j$, then $V_i\cap V_j=\{\mathbf{0}\}$.

Proof of Claim 1. Suppose $\mathbf{v}\in V_i\cap V_j$; since $p_i$ and $p_j$ are distinct and irreducible, they are coprime, so there exist polynomials $f(x)$ and $g(x)$ such that $f(x)p_i^{h_i}(x) + g(x)p_j^{h_j}(x) = 1$. Since $T$ commutes with itself and with all scalars, evaluation at $T$ is compatible with the polynomial operations, so $$\mathbf{v} = 1\mathbf{v} = (f(T)p_i^{h_i}(T) + g(T)p_j^{h_j}(T))(\mathbf{v}) = f(T)p_i^{h_i}(T)(\mathbf{v}) + g(T)p_j^{h_j}(T)(\mathbf{v}) = \mathbf{0}+\mathbf{0} = \mathbf{0}.$$ So the claim holds. QED

Claim 2. $V_i$ is $T$ invariant for all $i$. That is, $T(V_i)\subseteq V_i$ for all $i$. In particular, $p(T)(V_i)\subseteq V_i$ for all polynomials $p(x)$.

Proof of Claim 2. Let $\mathbf{v}\in V_i$; then $$p_i(T)^{h_i}(T(\mathbf{v})) = (p_i^{h_i}(T)T)(\mathbf{v}) = T(p_i^{h_i}(T))(\mathbf{v}) = T(\mathbf{0}) = \mathbf{0},$$ so $T(\mathbf{v})\in V_i$. (We are using the fact that $p_i^{h_i}(x)x =xp_i^{h_i}(x)$, and evaluation at $T$ respects products). QED

Claim 3. If $i\neq j$, then $p_i^{h_i}(T)\Bigm|_{V_j}$ is one-to-one and onto $V_j$ to $V_j$.

Proof of Claim 3. The kernel of $p_i^{h_i}(T)$ is $V_i$, which intersects trivially with $V_j$ by Claim 1; the conclusion now follows from Claim 2. QED

Claim 4. If $p_T(x)$ is a power of a single irreducible factor, $p_T(x) = (p(x))^h$, then $\dim(V) = h\deg(p)$.

Proof of Claim 4. By the Cayley-Hamilton Theorem, we know that $p_T(T)$ is the zero linear transformation, so $V\subseteq \mathrm{Ker}((p(x))^h) \subseteq V$. Therefore, $\dim(\mathrm{Ker}((p(x))^h) = \dim(V)$. Since $\dim(V)$ is equal to the degree of the characteristic polynomial, it follows that $\deg(p_T(x)) = \dim (V)$. But $\deg(p_T(x)) = \deg((p(x))^h) = h\deg(p)$, proving the claim. QED

We can now prove the result. Let $m$ be the number of distinct irreducible factors of $p_T(x)$; we proceed by induction on $m$.

If $m=1$, then the result follows by Claim 4. Assume the result holds for linear transformations whose characteristic polynomial has strictly fewer than $m$ distinct irreducible factors, and let $$p_T(x) = \prod_{i=1}^m p_i(x)^{h_i}.$$ Let $\mathbf{W}=\mathrm{Im}(p_1(T)^{h_1})$ be the range of $p_1(T)^{h_1}$, and let $U$ be the restriction of $T$ to $W$. Let $q(x)$ be the minimal polynomial of $U$. Then $q(x)$ divides $m_T(x)$, since $\mathbf{W}$ is a $T$-invariant subspace of $V$, and $p_1(x)$ does not divide $q(x)$, because $$p_2(T)^{h_2}\cdots p_m(T)^{h_m}(\mathbf{W})=\{\mathbf{0}\}$$ by the Cayley-Hamilton Theorem.

Therefore, $q(x)$ has strictly fewer than $m$ distinct irreducible factors, so the characteristic polynomial of $U$ has strictly fewer than $m$ distinct irreducible factors and we can apply the induction hypothesis to $U$. Moreover, the characteristic polynomial of $U$ divides $p_2(x)^{h_2}\cdots p_m(x)^{h_m}$. Let $$p_U(x) = p_2(x)^{k_2}\cdots p_m(x)^{k_m}$$ be the characteristic polynomial of $U$, with $0\leq k_j\leq h_j$. By the induction hypothesis, we have that $\dim(\mathbf{W}) = k_2\deg(p_2)+\cdots+k_m\deg(p_m)$.

Now, $V$ can be decomposed as $\mathbf{W}\oplus \mathrm{Ker}(p_1(T)^{h_1})$. If we restrict $T$ to $\mathrm{Ker}(p_1(T)^{h_1})$, then the characteristic polynomial of $T$ is of the form $p_1(T)^{k_1}$ for some $k_1$, $1\leq k_1\leq h_1$, so by Claim 4 we know that $\dim(\mathrm{Ker}(p_1(T)^{h_1}) = k_1\deg(p_1)$. We therefore have: $$\begin{align*} \dim V &= \mathrm{rank}(p_1(T)^{h_1}) + \mathrm{nullity}(p_1(T)^{h_1})\\ &= \dim(\mathbf{W}) + \dim(\mathrm{Ker}(p_1(T)^{h_1}))\\ &= k_2\deg(p_2)+\cdots + k_m\deg(p_m) + k_1\deg(p_1)\\ &\leq h_2\deg(p_2) + \cdots + h_m\deg(p_m) + h_1\deg(p_1)\\ &= \deg(p_T(x))\\ &= \dim V. \end{align*}$$ Therefore, all inequalities are equalities, so $k_i=h_i$ for each $i$. Therefore, $\dim(\mathrm{Ker}(p_i(T)^{h_i})) = \dim(V_i) = h_i\deg(p_i)$, as claimed. QED

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