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

Question

Why does $U+U = U?$ I want to think of this geometrically, say in $\mathbb{R^{3}}$ we have some random plane in space that intersects the origin. How is this subspace, when added to itself, equal to itself?

I am not sure what this notion of 'addition' means. How does one add one vector space to another? — Kevin Driscoll, Sep 13 '13 at 03:43
I would assume it is ${ x+y \ | \ x \in U, \ y \in U }$ which is clearly $U$ again here. — James S. Cook, Sep 13 '13 at 03:45
@JamesS.Cook This makes sense, but then the result follows directly from the definition of a subspace. — Kevin Driscoll, Sep 13 '13 at 03:47
@Kevin Driscoll to be clear, trivial in the sense of linear algebra. Perhaps not trivial in terms of geometry, the beauty of linear algebra is manifest in the wealth of cases this proposition covers (see answer by Clive Newstead and discussion by MJD). — James S. Cook, Sep 13 '13 at 13:56

score 13 · Accepted Answer · answered Sep 13 '13 at 03:44

13

$U+U$ is the subset of $V$ consisting of everything that you can possibly get when you add two elements of $U$ to each other. Since $U$ is a subspace, adding any two of its elements lands you back in the subspace, so you get $U+U \subseteq U$. Conversely, $u=u+0$ for all $u \in U$, and so $U \subseteq U+U$.

In general, given subspaces $U, W \le V$, the sum $U+W$ is what you can possibly obtain by adding a vector in $U$ and a vector in $W$. Geometrically thinking, you can imagine dragging $U$ along $W$, then $U+W$ is the space it fills up.

For instance, if $\ell_1$ and $\ell_2$ are two different lines through the origin in $\mathbb{R}^3$ then $\ell_1 + \ell_2$ is the (unique) plane containing both lines. Why? Because if you drag one line along another line, the space it sweeps out is a plane.

answered Sep 13 '13 at 03:44

Clive Newstead

64,925

This answer is excellent, thank you – St Vincent Sep 13 '13 at 03:48
3

I think this is an excellent answer, but that you missed the opportunity to point out here that if $\ell_1 = \ell_2$, then $\ell_1 + \ell_2$ is what you get if you drag $\ell_1$ along itself, and this is exactly $\ell_1$ again. – MJD Sep 13 '13 at 04:21
Thanks for that addition MJD – St Vincent Sep 13 '13 at 04:36

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

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