# Is the unit disk in $\Bbb R^2$ a subspace?

This is the original Spanish version of the exercise:

My understanding is that if we make a circle which has a center in [0,0] and a radius of 1 and we take all the points of that circle.... is this set of points a vector space?

How do they define addition of 2 members of a vector space? As a scalar product? If so, than the set is not a vector space since it is possible to do the scalar product of 2 such vectors, that the product will be out of the circle, so the set can't be a vector space...

I assume they meant the points of the circle are represented by vectors in the vector space, right?

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What point must a subspace of $\mathbb R^2$ contain? When can a subspace of $\mathbb R^2$ be bounded? –  John Douma Jul 12 '13 at 16:02
Each of the point of $\Bbb R^2$ can be identified as the arrowtip of a vector which begins at $(0,0)$, yes. These are sometimes called "position vectors." –  rschwieb Jul 12 '13 at 16:27

The operations would be those inherited from $\Bbb R^2$, that is, $(a,b)+(c,d)=(a+c,b+d)$ and $\lambda(a,b)=(\lambda a,\lambda b)$. That's what it means to be a subspace, that it is a nonempty set that is a vector space with operations "fitting into/matching" the containing space.

"Multiplication" is a word usually reserved for operations other than addition. In vector spaces, the binary operation involved is usually referred to as "addition."

Hint: what happens to a vector in $\Bbb R^2$ when you multiply it by a large positive scalar?

Concretely: take a look at $5(1/2,1/2)$.

Intuitively, one dimensional subspaces of $\Bbb R^2$ are lines. If it has more than one dimension, it would have to fill up $\Bbb R^2$ completely!

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The exact translation of your exercise is the following:

"Let $C$ be the set of points which are inside the unitary circle in the plane [viewed as a vector space of dimension $2$ over $\mathbb R$ with the standard operations]. Is this $C$ a vector subspace of the plane?"

The answer is: "No, it is not". The reason for this answer is that for any non-zero element $x$ of $C$, there exists $\lambda\in\mathbb R$ such that $\lambda\cdot x\notin C$.

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