I need some help with these proofs.

1. Prove that the intersection of any collection of subspaces of V is a subspace of V.

2. Suppose that U is a subspace of V. What is U + U ?

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- Jan 22nd 2007, 02:12 PMtaypezLinear Algebra II
I need some help with these proofs.

1. Prove that the intersection of any collection of subspaces of V is a subspace of V.

2. Suppose that U is a subspace of V. What is U + U ? - Jan 22nd 2007, 02:23 PMThePerfectHacker
The subspace must contain $\displaystyle \bold{0}$ thus their intersection is non-empty.

We need to show vector addition closure and scalar multiplication closure.

Let $\displaystyle \b{u},\b{v} \in W_1\cap W_2$

Then,

$\displaystyle \b{u}+\b{v} \in W_1$ because it is a subspace since $\displaystyle \b{u},\b{v}\in W_1$.

And,

$\displaystyle \b{u}+\b{v} \in W_2$ because it is a subspace since $\displaystyle \b{u},\b{v}\in W_2$.

Thus,

$\displaystyle \b{u}+\b{v}\in W_1\cap W_2$.

Next, let $\displaystyle k$ be any scalar.

Then,

$\displaystyle k\b{u}\in W_1$ because it is a subspace and $\displaystyle \b{u}\in W_1$.

And,

$\displaystyle k\b{u}\in W_2$ because it is a subspace and $\displaystyle \b{u}\in W_2$.

Thus,

$\displaystyle k\b{u}\in W_1\cap W_2$.

I presume that means the direct product (or direct sum).

Meaning,

$\displaystyle U+U=\{(x,y)|x,y\in U\}$.

I seems that,

$\displaystyle U+U$ will be a subspace $\displaystyle V+V$. - Jan 22nd 2007, 02:31 PMtopsquark
:eek: If I may just say that has to be the most interesting reason for deleting a message I've ever seen...

-Dan - Jan 22nd 2007, 02:45 PMtaypez
2) Makes sense, but this problem comes before Direct Sums. Would there be another proof perhaps using just sums?

- Feb 17th 2009, 05:59 PMcookiesyum