Dimension of the rowspace of a matrix based on x and b In lecture 13 of Gibert Strang's lecture videos from MIT 18.06 Linear Algebra Spring 2005, he reviews the following problem:
What is the dimension of the rowspace of A where
$$
    Ax=\begin{bmatrix}
    2 \\
    4\\
    2 \\
    \end{bmatrix}
$$
and
$$
x=\begin{bmatrix}
2\\
0\\
0\\
\end{bmatrix}
+c\begin{bmatrix}
1\\
1\\
0\\
\end{bmatrix}
+d\begin{bmatrix}
0\\
0\\
1\\
\end{bmatrix}
$$
According to him, the answer is 1 as the nullspace of A has dimension 2. How did he conclude this?
Here is the link to the lecture when he starts describing the problem
 A: This is the rank-nullity theorem. 
The matrix $A$ is $3\times3$, so $\operatorname{rank}(A)=3-\operatorname{dim}\operatorname{Ker}(A)$. 
You are given the solutions $x$ which parametrise a 2-dimensional affine subspace of $\mathbb{R}^3$, hence the kernel of $A$ is 2-dimensional - namely, $\operatorname{Ker}(A)=\operatorname{span}\left(\begin{bmatrix}
1\\
1\\
0\\
\end{bmatrix}
,\begin{bmatrix}
0\\
0\\
1\\
\end{bmatrix}\right)$.
Hence, $\operatorname{rank}(A)=1$.
A: For ease of notation, I'll rewrite this as
$$A\vec x=\vec b$$
$$\vec x=\vec u_1+c\,\vec u_2+d\,\vec u_3$$
This expands to
$$A\vec u_1+c\big(A\vec u_2\big)+d\big(A\vec u_3\big)=\vec b$$
The left side has variables $c$ and $d$, but the right side is a constant $\vec b$. Therefore,
$$A\vec u_2=A\vec u_3=\vec 0$$
and the equation simplifies to $A\vec u_1=\vec b\neq\vec 0$. This shows that $\vec u_2$ and $\vec u_3$ are in the nullspace of $A$, while $\vec u_1$ is not. The three $\vec u$'s are a basis for $\mathbb R^3$, so the nullspace of $A$ is 2-dimensional.
(I assume you were asking "How did he conclude that the nullspace of $A$ has dimension 2?". The other answers interpreted the question differently.)
A: First note that $A$ must be $3$ by $3$. So $Ax=0$ represents $3$ systems in $3$ variables. Second, since $\text{rank}(A)=\text{rank}(A^T)$ it’s enough to find the rank of $A$. Since the null space of $A$ is two dimensional, that means that when finding the reduced row echelon form $A$ we encountered two free variables. That means, $3-2=1$ variables are not free. Which means $A$ has $1$ pivot. Which means $\text{rank}(A)=1$.
