Find the dimension of the row space of $A$ this was an exam question and I want to see if I argued it correctly.
$A$ is a $5\times 3$ matrix with nullspace
$$N(A) = \operatorname{span}\{ [2, 1, 0], [1, 3, 0], [1, 1, 0] \},$$
and I have to find the dimension of the row space.
Using rank nullity theorem, I argued that the nullity of $A = 3$ based on $N(A)$, and $q$ being no. of columns, which is $3$, I reached $\operatorname{rank}(A) = 0$.
My understanding is that the transpose $\operatorname{rank}(A^T)$ gives the same value, but for the dimension of row space, which makes sense.
So my eventual answer was $0$.
Is this correct? Thank you!
 A: $N(A)$ has dimension $2$, so $\mathrm{rank}(A)$ has dimension $3-2=1$, so $\mathrm{rank}(A^T)=1$, i.e. the dimension of row space of $A$ is $1$.
A: Sadly it's not quite correct. Just because the number of vectors in the system of generators is $3$ it doesn't mean that the dimension of the null space is $3$ as well.
In fact, those three vectors can only generate at max the subspace $\mathbb{R}^2 \times \{0\}$ because all have $0$ in its third component. Therefore $\dim N(A) \leq 2$. As two of them are pairwise linearly independent we also have $\dim N(A) \geq 2$.
Therefore we get $\text{rank} (A) = 3 -2 =1$.
The rank of the transpose is the same as the rank of the matrix, because dimension of column and dimension of row space are equal, thus $\text{rank} (A^T) = 1$ as well.
Edit: You could calculate $\dim N(A)=2$ with other techniques as well:
We can see that the dimension is $<3$ because we have a representation of $0$ with non-zero coefficients:
$$
2*(2,1,0) + (1,3,0) - 5*(1,1,0) = 0
$$
But we can generate the space $\mathbb{R}^2 \times \{0\}$ because we can create the generators of this space:
\begin{align}
(1,0,0) &= (2,1,0) - (1,1,0) \\
(0,1,0) &= \frac{1}{2} (1,3,0) - \frac 12 (1,1,0)
\end{align}
Therefore the dimension is at least $2$ and at most $2 \Rightarrow$ it is $2$.
