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Let $T$ be a linear operator on a finite dimensional space $V$, and let $p=p_{1}^{r_{1}} \cdots p_{k}^{r_{k}} $ be the minimal polynomial for $T$, and let $V= W_{1} \oplus\cdots\oplus W_{k}$ be the primary decomposition for $T$, i.e., $W_{j}$ is the null space of $p_{j}(T)^{r_{j}}$. Let $W$ be any subspace of $V$ which is invariant under $T$. Prove that $W= (W \cap W_{1})\oplus (W \cap W_{2})\oplus \cdots \oplus (W \cap W_{k})$.

Zev Chonoles
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user71433
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    Welcome to MSE! Can you share your thoughts on the problem and what you have tried? It helps us to understand where the issues are and shows that you are trying. Regards – Amzoti Apr 08 '13 at 04:00

2 Answers2

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To answer the new question in the OP (thus completing EuYu’s answer) : Notice first that the nullspace $W_i={\sf Ker}(p_i(T)^{r_i})$ can also be written as a image space $W_i={\sf Im}(q_i(T))$ where $q_i=\prod_{j\neq i}p_j(T)^{r_j}$.

Indeed, the inclusion ${\sf Im}(q_i(T)) \subseteq W_i$ follows from the Cayley-Hamilton theorem. Conversely, let $w\in W_i$. The polynomial $p_i^{r_i}$ is coprime to $q_i$. We then have a Bezout identity

$$ (p_i^{r_i})A_i+q_iB_i=1 \tag{1} $$

for some polynomials $A_i,B_i$. It follows that $w=q_i(T)(B_i(T)w)$, showing $w\in {\sf Im}(q_i(T))$. In fact, this very argument shows that $E_iw=q_i(T)(B_i(T)w)$ for any $w\in V$ at all (not just $w\in W_i$), so that $E_i=q_i(T)B_i(T)$ is a “polynomial” in $T$. So any $T$-invariant subspace is automatically $E_i$-invariant also.

Ewan Delanoy
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  • Thank you for completing my argument Ewan. When I first answered this question, I remembered it being an exercise in the book "Linear Algebra" by Hoffman and Kunze. In the same section, the fact that the projections are polynomials in $T$ is given as a proposition. I guess it felt obvious to me at the time, even though it's not as direct as I would've liked in hindsight. – EuYu Jan 08 '14 at 21:46
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Let $E_1,\ \cdots,\ E_k$ be the projections associated with the decomposition. Let $\mathbf{w}\in W$ have the unique representation in terms of the direct sum as $$\mathbf{w} = \mathbf{w}_1 + \cdots + \mathbf{w}_k$$ We wish to show that $\mathbf{w}_i \in W$. Without loss of generality we work with $\mathbf{w}_1$. Then $$E_1\mathbf{w} = E_1(\mathbf{w}_1 + \cdots + \mathbf{w}_k) = E_1\mathbf{w}_1 = \mathbf{w}_1$$ Since $W$ is $T$-invariant, it follows that $W$ is also $E_1$-invariant. Therefore $\mathbf{w}_1 = E_1\mathbf{w}\in W$.

Now for each $i$, it follows that $\mathbf{w}_i \in W$ and by assumption, we also have $\mathbf{w}_i\in W_i$. Therefore $\mathbf{w}_i\in W\cap W_i$. We then have $$W\subseteq (W\cap W_1)\oplus \cdots \oplus (W\cap W_k)$$ On the other hand, we also have $$(W\cap W_1)\oplus \cdots \oplus (W\cap W_k)\subseteq W$$ This is trivially true by closure of addition. Therefore $$W = (W\cap W_1)\oplus \cdots \oplus (W\cap W_k)$$

EuYu
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