• Column space / Image: $C(A)$, since it corresponds to $C(A) \equiv \{ y : \exists x, y = Ax \}$
• Null space $N(A) \equiv \{ k : Ak = 0 \}$.
• Row space: row spans the row space, so it's all linear combinations of the rows of $A$. This is the same as all combinations of the columns of $A^T$. Row space is denoted by $C(A^T)$.
• Null space of $A^T$: $N(A^T)$, also called as the "left null-space of $A$".

Let $A$ be $m \times n$. The Null space of $A$ is in $\mathbb R^n$. The column space is in $\mathbb R^m$. The rows of $A$ are in $\mathbb R^n$. The nullspace of $A^T$ is in $\mathbb R^m$.

We want a basis for each of those spaces, and what are their dimensions?

• The dimension of the column space is the rank $r$.
• The dimension of the row space is also the rank $r$.
• The dimension of the nullspace is $n - r$.
• Similarly, the left nullspace must be $m - r$.

§ Basis for the column space

The basis is the pivot columns, and the rank is $r$.

§ Basis for the row space

$C(R) \neq C(A)$. Row operations do not preserve the column space, though they have the same row space. The basis for the row space of $A$ and $R$ since they both have the space row space, we just read off the first $r$ rows of $R$.

§ Basis for null space

The basis will be the special solutions. Lives in $\mathbb R^n$

§ Basis for left null space

It has vectors $y$ such that $A^T y = 0$. We can equally write this as $y^T A = 0$. Can we infer what the basis for the left null space is from the process that took us from $A$ to $R$? If we perform gauss-jordan, so we compute the reduced row echelon form of $[A_{m\times n} I_{m \times m}]$, we're going to get $[R E]$ where $E$ is whatever the identity matrix became.

Since the row reduction steps is equivalent to multiplying by some matrix $M$, we must have that:

So the matrix that takes $A$ to $R$ is $E$! We can find the basis for the left nullspace by lookinag at $E$, because $E$ gives us $EA = R$.

§ Reference

• Gilbert strang, lecture 10: four fundamental subspaces