Linear Algebra basis. What does it mean? Could someone check if my understanding is correct for:
(c) find a basis from vectors that span a subspace,
(d) find a basis for a column space of a matrix,
(e) find a basis for the row space of a matrix,
(f) find a basis for a nullspace of a matrix,
My attempt at understanding:
c) is basically like column space? Let A be a m x n matrix, Let H be the RREF of A so A ~ H. According to the pivots of H the basis would be {Ac1,...,Acn} (Depending on where the pivots are on H) (c means column)
d) Same as c?
e) Let A be a m x n matrix, Let H be the RREF of A so A ~ H. According to where there are nonzero values in the rows of H we can just let the basis be {Hr1, Hr2,...,Hrn} (r means row)
f) Is this like the homogeneous space? If the matrix has a Ax = 0 solution what the values of coefficients would also be the null set? 
Another question: How do I check my basis for rowspace / column space because the answers arent unique?
 A: Much of what you say is correct.
(c) You can think about a subspace formed from a set of vectors without worrying about them being the image of a linear map. Given a set of vectors $V = \{v_1,v_2,\ldots,v_p\}$, a basis for the subspace spanned by them is a collection of linearly independent vectors in $V$. For example if the first $q$ are linearly independent, then the subspace generated by $V$ would have a basis $\{v_1,v_2,\ldots,v_q\}$. You can find the linearly independent vectors of $V$ by forming the matrix $A = [v_1 | v_2|\ldots |v_p]$ with the column vectors of $V$ and row-reducing. The nonzero pivots give the linearly independent vectors. 
(d and e) - I think your understanding is good. It is possible to consider it more abstractly if you want but what you wrote down is correct.
(f) The nullspace of a matrix $A$ is the space $N = \{x \in \mathbb{R}^n | Ax = 0\}$. This is the set of vectors which map to $0$ under the action of $A$. This can be found by row-reducing the augmented matrix $[A|0]$. Depending on whether or not you want to know what it is or how to compute it, the following question may prove helpful: How to solve this to find the Null Space.
You can check your bases by seeing if they do what you claim they do. For the column space, can every column vector in your matrix be written as a linear combination of the vectors you wrote down as the basis? Is this also true for the rows of your matrix and the basis for the row space? 
A: (C) Put your vectors as the rows of a matrix.  Use Gauss-Jordan to get it into RREF.  Then the nonzero rows will be a basis.
(D) Find the RREF.  Take note of the columns with pivots in them.  Then the columns of your original matrix corresponding to those same pivot-columns in your RREF will be a basis.
(E) Find the RREF.  The nonzero rows will be a basis.
(F) Let's look at an example so that it'll hopefully be more clear.  Take this RREF matrix $$\begin{bmatrix} 1 & 0 & 0 & 1 & -2 \\ 0 & 0 & 1 & 0 & 5 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$


*

*First just count the columns without pivots.  That's the number of vectors you should have in your basis.  In this case that's three.

*Create one vector for each of the columns of your RREF without a pivot in it (that'll be the 2nd, 4th, and 5th columns in this case).  It should have the same number of rows as your matrix has columns.  Fill it with all 0's: $\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0\end{bmatrix}$, $\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0\end{bmatrix}$, $\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0\end{bmatrix}$.

*Add a 1 in to each vector in the row corresponding to the column number of the nonpivot column: $\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0\end{bmatrix}$, $\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0\end{bmatrix}$, $\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1\end{bmatrix}$.

*For any nonzero number in the nonpivot columns of your RREF, add the negative of that number to the row with the corresponding pivot column number.  So for instance, that $5$ in the RREF becomes $-5$ and goes in the vector we're constructing that's associated with the last column.  And it'll go in the 3rd row because the row that $5$ is in is the one with the pivot in the 3rd column (does that make sense?).  Doing the same for the $1$ and $-2$, we get: $\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0\end{bmatrix}$, $\begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \\ 0\end{bmatrix}$, $\begin{bmatrix} 2 \\ 0 \\ -5 \\ 0 \\ 1\end{bmatrix}$.


Those are then the basis vectors.
