B is a basis for W Let V be a vector space, B be a basis for V and W be a subspace of V. Then is B is a basis for W?
It is tough for me to deal with the bases. I do understand what is a subspace, but not concepts of basis.
Thank you
 A: If $B$ is a basis for $V$ it means that the elements in $B$ are linearly independent and that $V=sp(B)$.
If $B$ is a basis for $W$ it means that the elements in $B$ are linearly independent and that $W=sp(B)$
So $B$ is a basis for $W$ if and only if $W=V$.
A: Take $V = \Bbb R^2$, $B = \{(1,0),(0,1)\}$, and $W = \langle (1,0) \rangle$.
Then, $(0,1) \in B$ but $(0,1)$ is not even in $W$.
A: A dictionary definition for “basis” will say something along the lines of “underlying support or foundation.” 
Giving the same title to a linearly independent spanning set for a vector space turns out to be rather intuitive. Bases are the “building blocks” of a vector space.
For example, consider $\mathbb{R}^3$ and $B = \{ (1,0,0),(0,1,0),(0,0,1) \}$. Vector spaces behave well under addition and scalar multiplication by design so what is a “nice” way to form operations between vectors? 
Well, that is where the notion of a linear combination comes in. If we add scalar multiples of vectors in a vector space then we end up with a vector in that vector space. So, the reason that we want our basis to be a spanning set is because we want to build any vector in our space using a linear combination of vectors from our basis. 
Why do we want linear independence? Well, this is to rid ourselves of redundancies from our linear combinations. It’d be nice to form our vectors in a unique way.
Now, is it possible to cleverly find vectors in our space that can “form” any other vector in our vector space? Yes! A set of such vectors is called a basis.
For example, if we consider any $v = (x, y, z) \in \mathbb{R}^3$ then we may use $B$ to form $v$. How?
Well, $ v = (x, y, z) = x(1,0,0)+y(0,1,0)+z(0,0,1).$
That wasn’t so bad! You can find infinitely other bases for this vector space but the important thing is that they will have cardinality $3$. 
Now, what if we consider a subspace? For example, consider $W = \{(x, y, 0)| x, y \in \mathbb{R} \}.$ $W$ is the $xy$-plane sitting flat in $3$ dimensions with a vertical component of $0$ height. So, we take linear combinations of vectors of $B$ then it is true that we can form any vector of $W$ but we have a lot of extra vectors that can be formed outside of $W$ so $B$ is not a basis for $W$. What would be a basis for $W$?
Consider $A = \{(1,0,0),(0,1,0) \}.$ Then any vector $u = (x,y,0) \in W$ can be formed from linear combinations of the vectors in $A$ since $u = (x,y,0) = x(1,0,0)+y(0,1,0)$ and so we have found a set of vectors that can “form” any vector in $W$. Hence we have a basis for $W$. 
Note that our basis for $W$ differs from our basis for $V$. The subspace $W$ is a vector space in its own right but it properly sits within another vector space, namely $\mathbb{R}^3$ and so we don’t expect these spaces to share a basis.
This might not have been exactly what you were looking for but I hope it helps.
