Dimension of the rowspace of a matrix based on x and b

Question

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

Answer 1

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, $\operatorname{rank}(A)=1$.

Answer 2

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 $\text{rank}(A)=1$.

Answer 3

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.)