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# 16b The fundamental subspaces: Nullspace and columnspace.
Recall in [[1 teaching/smc-spring-2024-math-13/linear-algebra-notes/16-solution-sets-of-a-linear-system-Ax-b|notes 16]], we mentioned two sets of vectors associated with an $n\times k$ matrix $A$ -- its **nullspace** and its **columnspace**. Recall we write for an $n\times k$ matrix $A$, we have $$
\text{nullspace}(A) = \{\vec x\in \mathbb{R}^{k}: A\vec x = \vec 0\}
$$and $$
\text{columnspace}(A)=\operatorname{span} (\vec a_{1},\ldots,\vec a_{k})\subset\mathbb{R}^{n},
$$where $\vec a_{1},\ldots,\vec a_{k}$ are the columns of $A=[\ \vec a_{1} \ |\cdots |\ \vec a_{k }\ ]$.
**Remark.** Notice that the nullspace and the columnspace of the same matrix may consist of vectors that live in different "dimensions"! For instance, a $5\times 8$ matrix $A$ has $\text{nullspace}(A)$ consisting vectors that live in $\mathbb{R}^{8}$, while $\text{columnspace}(A)$ consists of vectors that live in $\mathbb{R}^{5}$. Make good note of this!!
These are very important "subspaces" associated to a matrix $A$. We will define the word "subspace" later. For now just think of them as collections of vectors.
Recall also from [[1 teaching/smc-spring-2024-math-13/linear-algebra-notes/16-solution-sets-of-a-linear-system-Ax-b|notes 16]], we can also write $\text{nullspace}(A)$ as a span of some number of vectors (the number being the number of free variables of in an echelon form of $A$).
Let us practice finding these spaces by writing them out as span of a set of vectors.
**Example.** Let $A =\begin{bmatrix}0 & 1 & 3 \\ 5 & 3 & 1\end{bmatrix}$. Write $\text{nullspace}(A)$ and $\text{columnspace}(A)$ as spans of some number of vectors.
$\blacktriangleright$ First, columnspace is always "easier", it is just the span of the columns. So in this case it is $$
\text{columnspace}(A) = \text{span}\{\begin{bmatrix}0\\5\end{bmatrix},\begin{bmatrix}1\\3\end{bmatrix},\begin{bmatrix}3\\1\end{bmatrix}\},
$$done! Well, technically this is correct. But we think geometrically (and trust our intuition), we can see that this is just the whole of "2-dimensional plane" $\mathbb{R}^{2}$. We will see how to "get rid of redundancy" later. For now, this is fine.
Now for the nullspace, we need to solve for the vectors $\vec x$ such that $A\vec x = \vec 0$. So we row reduce the augmented matrix $$
\begin{bmatrix}
0 & 1 & 3 & \vdots & 0 \\
5 & 3 & 1 & \vdots & 0
\end{bmatrix} \stackrel{\text{row}}\sim \begin{bmatrix}
\color{blue} 5 & 3 & 1 & \vdots & 0 \\
0 & \color{blue} 1 & 3 & \vdots & 0
\end{bmatrix}
$$which we can make $x_{3}$ the free variable, and $x_{2}=-3x_{3}$ and $x_{1} = \frac{1}{5}(-3x_{2}-x_{3})=\frac{8}{5}x_{3}$. So solutions to $A\vec x=\vec 0$ are of the form $$
\vec x = \begin{bmatrix}
x_{1} \\ x_{2} \\ x_{3}
\end{bmatrix}= x_{3 }\begin{bmatrix}
8 / 5 \\
-3 \\
1
\end{bmatrix}\quad \text{where \(x_3\) is free.}
$$Hence we have $$
\text{nullspace}(A) = \operatorname{span}\{ \begin{bmatrix}
8 / 5 \\
-3 \\
1
\end{bmatrix}\},
$$a "1-dimensional line" in $\mathbb{R}^{3}$. $\blacklozenge$
A couple observations.
(1) The nullspace and columnspace may consist of vectors of **different** kinds! In this example of a $2\times 3$ matrix $A$, the columnspace are vectors from $\mathbb{R}^{2}$, while the nullspace are vectors from $\mathbb{R}^{3}$. Make a note of where these numbers come from.
(2) An observation of import: The matrix $A =\begin{bmatrix}0 & 1 & 3 \\ 5 & 3 & 1\end{bmatrix}$ has an echelon form with 2 pivots and 1 free variable. The $\text{columnspace(A)}$ is a "2-dimensional plane", and the $\text{nullspace(A)}$ is a "1-dimensional line". And, $2+1=3$, where $A$ has $3$ columns. This is the manifestation of **a very important theorem**, that you may have seen in a homework assignment. Can you think about which one?
We will speak more of nullspace and columnspace later. This is just a quick look to get the terminology down, as well as these observations.