dimension of column space and null space I am trying to answer a series of questions given the dimension of a matrix. 
Suppose that $A$ is a $6 \times 12$ matrix. 
The column space is a subspace of $\mathbb{R}^n$. What is n?
$n = 6$ because there can only be 6 pivot columns. 
The null space is a subspace of $\mathbb{R}^m$. What is m?
$m = 12$? Not so sure about this question. 
Is it possible to have rank = 4, dimension of null space = 8? 
$rank \leq min(m,n)$ for $m \times n$ matrix,
rank + nullity = number of columns. 
It is possible. 
Is it possible to have rank = 8, dimension of null space = 4? 
rank + nullity = number of columns 
but $rank \nleq min(m,n)$.
It is not possible.
Are my answers valid for the three questions that I answered?
I am a bit confused with the second question. 
Any help would be great.
Thank you for reading.
 A: 
The column space is a subspace of $\mathbb{R}^n$. What is n?
$n = 6$ because there can only be 6 pivot columns.

Your answer is technically correct, but misleading.  I would say the following: the column-space is a subspace that contains the columns of $A$.  Because the columns of $A$ each have $6$ entries, an element of the column space has $6$ entries which is to say that the column space is a subspace of $\Bbb R^6$.

The null space is a subspace of $\mathbb{R}^m$. What is m?
$m = 12$? Not so sure about this question.

Your answer is correct; here's a reason.  The nullspace of $A$ is the set of column-vectors ($k \times 1$ vectors for some $k$) $x$ satisfying $Ax = 0$.  However, in order for $Ax$ to make sense, the "inner dimensions" of $m \times n, k \times 1$ need to match, which is to say that $k = n = 12$.  So indeed, the nullspace is a subspace of $\Bbb R^{12}$.

Is it possible to have rank = 4, dimension of null space = 8?
$rank \leq min(m,n)$ for $m \times n$ matrix,
rank + nullity = number of columns.
It is possible.
Is it possible to have rank = 8, dimension of null space = 4?
rank + nullity = number of columns
but $rank \nleq min(m,n)$.
It is not possible.

Both of these are correct; I have nothing to add.
