Algebraic properites of the direct sum

[Bipolarity]

Is the following statement true? I am trying to see if I can use it as a lemma for a larger proof:

Let $V$ be a vector space and let $W, W_{1},W_{2}...W_{k}$ be subspaces of $V$.
Suppose that $W_{1} \bigoplus W_{2} \bigoplus ... \bigoplus W_{k} = W$
Then is it always the case that:
$(W_{1} \cap W) \bigoplus (W_{2} \cap W) \bigoplus ... \bigoplus (W_{k} \cap W) = W$

In essence, this is asking whether there is a distributive law compatible with the set intersection operation and the direct sum operation. I am only asking for a determination of whether the statement is true for false. I will work out the proof/counterexample for myself.

Thanks!

BiP

 

[lavinia]

Ask yourself whether the Wi are subspaces of W

 

[Bipolarity]

Ask yourself whether the Wi are subspaces of W

It is obvious if they are subspaces of $W$, but what if they aren't?

BiP

 

[lavinia]

It is obvious if they are subspaces of $W$, but what if they aren't?

BiP

If W is the direct sum of the Wi then show me an element of one of the Wi's that is not in W

 

[Erland]

Is the following statement true? I am trying to see if I can use it as a lemma for a larger proof:

Let $V$ be a vector space and let $W, W_{1},W_{2}...W_{k}$ be subspaces of $V$.
Suppose that $W_{1} \bigoplus W_{2} \bigoplus ... \bigoplus W_{k} = W$
Then is it always the case that:
$(W_{1} \cap W) \bigoplus (W_{2} \cap W) \bigoplus ... \bigoplus (W_{k} \cap W) = W$

In essence, this is asking whether there is a distributive law compatible with the set intersection operation and the direct sum operation. I am only asking for a determination of whether the statement is true for false. I will work out the proof/counterexample for myself.

Thanks!

BiP

As it stands, the answer is obviously yes, since each $W_i$ is a subspace of $W$, and hence $W_ i\cap W=W_i$.

But I assume that there is a typo and that you meant $W_{1} \bigoplus W_{2} \bigoplus ... \bigoplus W_{k} = V$. Then, the answer is no. It is almost trivial to find a counterexample in R².