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I am trying to prove to myself something that my linear algebra book is telling in the section talking about dimension:

Show that the subspaces of $\mathbb{R}^{3}$ are precisely $\mathbb{R}^{3}$, {0}, all lines through the origin and all planes through the origin.

I can clearly see that any subpsace of $\mathbb{R}^{3}$ with dimension 3 is going to be equal to $\mathbb{R}^{3}$.

Intuitively, I can see all of the lines and planes formed by the axes being made up of independent vectors, the span of which would have dimensions of 2.

But what about the rest? The ones that only share the origin, but don't lie on any axis? I'm having trouble wrapping my head around proving those.

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    Those are not vector spaces because they do not have the $0$ vector (i.e. the origin). – 2017-02-12
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    They are *affine subspaces*, i.e. translated from a subspace by a vector. – 2017-02-12
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    @AdamHughes If the span of a set describes a line or a plane that intersects the origin, how could it not contain a zero vector? I think you misunderstood the question, so I edited it to be more clear. – 2017-02-12
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    To your edited question: if all you want to do is visualize them you need only rotate any other line or plane. – 2017-02-12

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Tl;dr: the $2$-dimensional subspaces are possiby tilted planes, like this one.

Here's an example of a $2$-dimensional subspace which is not generated by the axes in any way:

Consider the vectors $$u=(1, 2, 3), \quad v=(1, 4, 9).$$ It's not hard to show that $\{u, v\}$ is linearly independent, so that set generates a $2$-dimensional subspace. Now what does that subspace look like? Well, imagine drawing the vectors $u$ and $v$ in $\mathbb{R}^3$; they describe a plane! In general, any two vectors which don't lie in the same line generate a plane. This plane is tilted - while it goes through the origin, it does not contain any of the usual axes.

Conversely, if $V$ is a $2$-dimensional subspace, then $V$ is the span of some set $\{a, b\}$ of vectors. Well, $a$ and $b$ - as above - describe a (possibly tilted) plane containing the origin.

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    So, if I was to state what you said formally, would I be wrong to say: "Assume $U \subset \mathbb{R}^{3}$ and $DimU=2$, then we can find a basis $B=(u,v)$ the span of which describes a plane. As all planes intersecting the origin of $\mathbb{R}^{3}$ can be described this way, All such planes are subspaces of $\mathbb{R}^{3}$." How do I prove that those planes NOT intersecting $\mathbb{R}^{2}$ are not subspaces of $\mathbb{R}^{3}$? – 2017-02-12
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    @wesleyNeill Do you mean "not intersecting **the origin**"? If so, what's one vector that *any* subspace has to contain, by definition? – 2017-02-12
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    Yessir, that is what I meant, and you I just smacked my forehead in exasperation at the answer. Thanks. – 2017-02-12