Can a vector be in two subspaces at once? In all vector spaces two operations are possible :


*

*Addition of two vectors. 

*Multiplication of vectors by scalar quantities. 


A subspace of vector is a non empty subset  that satisfies the requirements for a vector space.Linear combinations stay in the subspace.
 A: What book are you using? There are several good books that motivate and describe vector spaces.
A vector space, is essentially an object that has been defined to contain a group of vectors and that space, all of them, follow a certain group of axioms.
You mention subspaces, a subspace must contain the zero vector, so as another commenter has said you could imagine two intersecting planes, that intersect the origin, i.e., contain the zero vector. A vector can be in two subspaces at once. The zero vector as the most trivial example is in every subspace.
A: I have two bags with me. Let me name the two bags, ok? Let us say one of the bags is called $V$ and the other is called $\Bbb K$. Also, you do not know what I have in the bags. But that does not keep me from naming those things. Let me say that the things that are inside the $V$ bag are called vectors and that the things inside the $\Bbb K$ bag are called scalars. Let me also tell you this (let $v, u, w \in V; k, q \in \Bbb K$):


*

*If you take two objects from the $V$ bag, you can add them together and get, as a result, some other vector that you can also find in your bag. The order in which you add them does not matter. The addition of those objects also has the associativity property; ($v + u \in V, v + u = u + v, (v + u) + w = v + (u + w)$)

*If you take one object of each bag, you can multiply them and get some vector that is in the $V$ bag; ($kv \in V$)

*There is one object in the $V$ bag, let me call it 0, that when added to any other vector, yields the other vector unchanged; ($0 + v = v$)

*For every vector $v$ in the $V$ bag, there is another one you can find (just gotta look for it) that, when adding both together, you get the 0 vector. Let me call it $-v$; ($\forall v \in V \exists (-v)\in V: v + (-v) = 0$)

*There is some scalar, let me call it 1, that when multiplied by any vector, leaves that vector unchanged; ($1v = v$)

*If you take two scalars from the $\Bbb K$ bag, you can also multiply them together. Plus, if you take some vector from the $V$ bag, you can multiply it by some scalar first, then by the other, or just by the product of the two at once; ($kq \in \Bbb K; k(qv) = (kq)v$)

*Adding vectors and multiplying by scalars is distributive: the product of the sum is the sum of the products, $k(v + u) = kv + ku$, and $(k + q)v = kv + qv$.
Therefore, if you have some two bags that you can name as being like the $V$ and $\Bbb K$ I have, you have found yourself a vector space $V$ over the scalars $\Bbb K$. The scalars are usually just $\Bbb R$ or $\Bbb C$ so I wouldn't worry too much about them. They work as the "numbers".
Any bag you find with those properties is a vector space! What bags can you use? Let me just give some examples:


*

*The bag $V$ with all real numbers, using the real numbers as scalars;

*The bag $V$ with all vectors (in the geometric sense) that have a starting point in the origin, with $\Bbb K = \Bbb R$;

*The bag $V$ with all matrices of some dimensions $m\times n$, for example $2 \times 3$, with $\Bbb K = \Bbb C$;

*The bag $V$ with all continuous functions and $\Bbb K = \Bbb R$;

*The bag $V$ with all polynomials with degree, at most, 4 and $\Bbb K = \Bbb R$. Actually, any bag with all polynomials with degree at most $n$, for some $n \in \Bbb N$ you pick, and $\Bbb K = \Bbb R$;

*The bag $V$ with all bounded sequences and $\Bbb K = \Bbb R$;

*$\cdots$
A vector space is just a bag with things you know how to add, multiply by some "numbers" (scalars) and that satisfy some properties.
Answering your first question, you can have some vector in different vector spaces. For example, consider the identity matrix that is $2\times2$. That matrix is both in the vector space of all matrices that are $2\times2$ and of all matrices that are $2\times2$ AND diagonal. (Let us assume I am talking about $2\times2$ matrices) As a matter of fact, all diagonal matrices are also matrices, and therefore the vector space of all diagonal matrices is a subspace of the space of all matrices and therefore you can also have a vector belonging to different subspaces at the same time. (The vector space of all matrices is a subspace of itself)
A: Let $V=\Bbb R^2$ and let $v=\begin{pmatrix} 1 \\ 0\end{pmatrix}$. Then $v\in \{(x,0)^T: x\in\Bbb R\}$ and $v\in V$ and clearly these are different spaces.
