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Let $U, X$ be vector spaces and $f: U \rightarrow X$ be a linear map. Let $X= X_1 \oplus \cdots \oplus X_r$, where $X_i$ is subspace of $X$. Let $\pi_i : X \rightarrow X_i$ be the projection onto $X_i$. How strong a condition is to assume that $\pi_i \circ f (U) \subseteq f(U)$, i.e. that the image of $f$ is $\pi_i$-invariant for all $i=1,...,r$?

I did some thinking in the case where $U=\mathbb{R}, X=\mathbb{R}^2$ and the condition seems quite strong. On the other hand for $U=X=\mathbb{R}$ the condition is trivial.

Any insights?

To the more experienced of our community: how interesting would be a theorem that uses this assumption on the map $f$?

Thanks :-)

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What is B? B=f? –  AlexE Jan 3 '12 at 18:08
    
@Alex: yes, i will correct it! Thanks. –  Manos Jan 3 '12 at 18:09

1 Answer 1

up vote 2 down vote accepted

Do you have a concrete case where such an f is given or are you asking just out of curiosity?

Nevertheless, under this condition the image of f splits as a direct sum of its projections, i.e. $f(U) = \bigoplus \pi_i(f(U)).$ And this is in fact equivalent to $\pi_i \circ f(U) \subset f(U)$. Defining $\pi_i \circ f =: f_i$, we see that every information we want to know about f is stored in the $f_i$ (under the assumption that we don't have any further structure on X or any further properties of f which are not compatible with this decomposition), i.e. we can restrict our investigations to the $f_i$. So your condition just breaks the f down into smaller pieces which might be easier to study.

But ... if the decomposition of X and the map f have nothing to do with each other, then I think your condition won't hold that often.

The other way round, given the map f, we can always find such a decomposition of X, e.g. $X = \operatorname{im}(f) \oplus W$, where $W$ is some arbitrary complement of $\operatorname{im}(f)$.

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I am studying a particular problem, which i want to decompose into subproblems, and it seems that this is a necessary condition for this to happen. However, as you noted, $f$ and the decomposition of $X$ have nothing to do with each other. –  Manos Jan 3 '12 at 18:48

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