this is a question that appeared on webassign for my linear algebra course. I believe it's a bad question, and want to know what you guys think.
Question: Suppose that $A$ is an $n \times m$ matrix with $rank(A)=4$, $nullity(A)=3$ and that $col(A)$ is a subspace of $\mathbb{R}^5$. What are the dimensions for $A$?
By the rank nullity theorem, we know that $rank(A) + nullity(A) = \text{# of columns of A}$. Therefore $A$ has $7$ columns.
I believe this is a bad question because they want us to deduce that $n=5$ from the fact that $col(A)$ is a subspace of $\mathbb{R}^5$. The following matrix has $n=6$ and $col(A)$ is still a subspace of $\mathbb{R}^5$.
$$ A = \left( \begin{matrix} 2 & 3 & 2 & 2 & 1 & 3 & 7 \\ 4 & 3 & 2 & 4 & 1 & 4 & 3 \\ 1 & 2 & 3 & 3 & 1 & 1 & 2 \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 6 & 5 & 4 & 3 & 0 & 1 & 9 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} \right) $$
Here, $col(A)$ will be a subspace of $\mathbb{R}^6$, as the corresponding linear transformation $T(x)=Ax$ has codomain $\mathbb{R}^6$. However, $col(A)$ is also a subspace of $\mathbb{R}^5$ as the last row of the matrix is all zero.
Can somebody verify that my thoughts about this being a bad question are valid, or if not, point out the flaw in my thinking? Thank you.
In your example, the columnspace is not a subspace of $\Bbb{R}^5$, but rather $\Bbb{R}^5 \times \{0\}$ (a subspace of $\Bbb{R}^6$). While this subspace of $\Bbb{R}^6$ is isomorphic to $\Bbb{R}^5$, it is not the same thing.
The columnspace of an $n \times m$ real matrix is always a subset of $\Bbb{R}^n$. If $a \neq b$, then $\Bbb{R}^a \cap \Bbb{R}^b = \emptyset$ ($a$-tuples of real numbers are only $b$-tuples if $a = b$). So, if the columnspace contains even a single element of $\Bbb{R}^5$, then the matrix must have $5$ rows.