Not sure how to elaborate on the question more than the title, so I'll just reiterate:
Is A Subspace of a 3D Vector Space a 2D Plane?
I'm just trying to visualize/understand this concept.
Yes a plane (is there any other sort than 2D?) is a proper subspace of a 3D Euclidian space.
So is a line.
The 3D space is spanned by 3 non parallel vectors, eg the orthogonal basis vectors i, j, k, but they need not be orthogonal.
Any two intersecting vectors will span a plane, so any vector in the plane can be expressed as C = aV + bW, and any such vector C is in the plane.
One single vectors spans a line.
I hope that you rewrote that question. No one teaching you should be that careless.
The space $3D$ is $\{(x,y,z):\{x,y,z\}\subset\mathbb{R}\}$ whereas $2D$ is $\{(x,y):\{x,y\}\subset\mathbb{R}\}$
It is clear no subset of $3D$ is a subset of $2D$. So it is hard to know exactly what is being asked.
The set $\{(x,y,0):x+y=0\}$ is a subspace of $3D$ Can you prove that? Do you know how to prove that a subset is a subspace of a space?
Now it is true that $\{(x,y,0):x+y=0\}$ can be embedded in $2D$ as a subspace $\{(x,y):x+y=0\}$ of $2D$. So you need to re-read the original question for it true meaning.
1. This was my own inquiry, not a question that was given to me.
2. That is why I said "subspace" and not "subset". Also, had this been a few years ago when I last took a Lin Alg class I would have been able to, but I was just reviewing the concepts and making sure that I understood what they actually meant.
3. Useful info from both of you, thanks guys!
Suppose that $\mathcal{V}$ is a vector space and $\mathcal{A}\subset\mathcal{V}$ now that is simply a set.
To prove that $\mathcal{A}$ is a subspace of $\mathcal{V}$, two properties are necessary:
1) $\bf{0}\in\mathcal{A}$ and
2) if $\alpha$ is a scalar and $\{u,v\}\subset\mathcal{A}$ then $(\alpha u+v)\in\mathcal{A}$.