# Linear Algebra II

• Jan 22nd 2007, 02:12 PM
taypez
Linear 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 PM
ThePerfectHacker
Quote:

Originally Posted by taypez
I need some help with these proofs.

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

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

Quote:

Originally Posted by taypez
2. Suppose that U is a subspace of V. What is U + U ?

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 PM
topsquark
: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 PM
taypez
2) Makes sense, but this problem comes before Direct Sums. Would there be another proof perhaps using just sums?
• Feb 17th 2009, 05:59 PM
Quote:

Originally Posted by ThePerfectHacker
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 $\displaystyle V+V$.

What would U + V be then? It is a space, but is it a subspace of V?