Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 :-)

share|cite|improve this question
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
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 $\operatorname{im}(f)$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.