I have a confusion about Subspaces and i hope that can be cleared.. Say $E,F,G$ are subspaces of the same vector space. Say we want to show that $(E\cap F)+(E\cap G)$ is a subspace of $E\cap(F+G)$.

My first question is, Does being a subspace necessarily mean being a subset? And if so, to show the mentioned above, do i have to, first, show that $(E\cap F)+(E\cap G) \subset E\cap(F+G)$ ?, i know three conditions have to be satisfied for a set to be a subspace of some vector space, and the first one among them is having the zero vector in it, How do i do that for the one above? I mean i don't know what is the zero vector for this vector space, Is it the same as the zero vector of the vector space of $E,F,G$ ? I was able to show that it is closed under addition and multiplication so that is not my question..

Thanks everyone!

  • $\begingroup$ First, how does your text/notes define the set $S + T$? And yes, we require $S \subset X$ to say $S$ is a subspace of $X$. $\endgroup$ Commented Mar 12, 2018 at 18:28
  • $\begingroup$ $S+T =\lbrace X=x+y \vert x\in S$ et $y\in T \rbrace$ @SeanRoberson $\endgroup$ Commented Mar 12, 2018 at 18:52

1 Answer 1


Subset, yes, it is necessary: For $x+y$, $x\in E\cap F$, $y\in E\cap G$, then as $E$ is a vector space and $x,y\in E$, then $x+y\in E$. Also, $x\in F$ and $y\in G$, so $x+y\in F+G$, so we conclude that $x+y\in E\cap(F+G)$.

  • $\begingroup$ So , if i'm asked to show whether B is a subspace of A, if $\exists a \in B$ but not in $A$, it's enough to conclude that $B$ isn't a subspace of $A$?. $\endgroup$ Commented Mar 12, 2018 at 19:03
  • $\begingroup$ Yes, you are right. $\endgroup$
    – user284331
    Commented Mar 12, 2018 at 19:08
  • $\begingroup$ What about the zero vector? $\endgroup$ Commented Mar 12, 2018 at 19:12
  • $\begingroup$ There is only one zero vector, and that zero vector belongs to all $E,F,G$, and the ambient space, so it is apparently that the zero vector belongs to $(E\cap F)+(E\cap G)$. $\endgroup$
    – user284331
    Commented Mar 12, 2018 at 19:14
  • $\begingroup$ So it's safe to assume that the zero vector is in there since every subspace has it.. Thanks for your time. $\endgroup$ Commented Mar 12, 2018 at 19:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .