# Reducing Span
When we make a span of discrete vectors, sometimes one or more of those vectors can be scalar multiples of another.
> Take, for example, an example from An Introduction To Linear Algebra For Science and Engineering by Norman, D., & Wolczuk, D. The solution is adapted and is my original interpretation of the steps.
If we have a span of the below:
$
\text{Span} \left\{
\begin{bmatrix}
3 \\ 1 \\ -3
\end{bmatrix},
\begin{bmatrix}
0 \\ -1 \\ 1
\end{bmatrix},
\begin{bmatrix}
3 \\ 0 \\ -2
\end{bmatrix}
\right\}
$
Notice that by definition, our [[Vector Equations|vector equation]] is the following:
$\vec{x} = c_1\begin{bmatrix}
3 \\ 1 \\ -3
\end{bmatrix} + c_2\begin{bmatrix}
0 \\ -1 \\ 1
\end{bmatrix} + c_3\begin{bmatrix}
3 \\ 0 \\ -2
\end{bmatrix}, \quad c_1,c_2,c_2 \in \mathbb{R} $
Notice that the first two vector equations added together give us the last vector equation, so we can rewrite the [[Vector Equations|vector equation]] as the following (assume $c_1,c_2,c_2 \in \mathbb{R}$ is always true)
$\begin{array}{c}
\begin{bmatrix}
3 \\ 1 \\ -3
\end{bmatrix} + \begin{bmatrix}
0 \\ -1 \\ 1
\end{bmatrix} = \begin{bmatrix}
3 \\ 0 \\ -2
\end{bmatrix} \\ \\
\vec{x} = c_1\begin{bmatrix}
3 \\ 1 \\ -3
\end{bmatrix} + c_2\begin{bmatrix}
0 \\ -1 \\ 1
\end{bmatrix} + c_3\begin{bmatrix}
3 \\ 0 \\ -2
\end{bmatrix} \\
\vec{x} = c_1\begin{bmatrix}
3 \\ 1 \\ -3
\end{bmatrix} + c_2\begin{bmatrix}
0 \\ -1 \\ 1
\end{bmatrix} + c_3 \left\{ \begin{bmatrix}
3 \\ 1 \\ -3
\end{bmatrix} + \begin{bmatrix}
0 \\ -1 \\ 1
\end{bmatrix}\right\}
\\\\
\vec{x} = (c_1 + c_3 )\begin{bmatrix}
3 \\ 1 \\ -3
\end{bmatrix} + (c_2 +c_3)\begin{bmatrix}
0 \\ -1 \\ 1
\end{bmatrix}
\\
\text{Cant simplify futher than this so:} \\
\text{Let $c_1 + c_3 = s$} \\
\text{Let $c_2 + c_3 = t$} \\ \\
\vec{x} = s\begin{bmatrix}
3 \\ 1 \\ -3
\end{bmatrix} + t\begin{bmatrix}
0 \\ -1 \\ 1
\end{bmatrix}, \quad s,t \in \mathbb{R}
\end{array}$
## Generalization
From the example above, notice that our vector equation before we simplified it is technically the same as the vector equation after simplifying, because one vector was a scalar multiple of the other.
$\begin{array}{c}
\text{Some vector $\vec{v_i}$ where $1 \leq i \leq k$ can be written as a linear combination of } \\
\vec{v_1}, \dots, \vec{v}_{i-1}, \vec{v}_{i+1},\dots , \vec{v} _ k\\\\
\text{If and only if:}
\\
Span\{\vec{v_1}, \dots, \vec{v_k}\} = Span\{\vec{v_1}, \dots, \vec{v}_{i-1}, \vec{v}_{i+1},\dots , \vec{v} _ k\}
\end{array}$
# Examples
---
> 🧠 Enjoy this walkthrough? [Support Math & Matter](https://github.com/rajeevphysics/Obsidan-MathMatter) with a star and help others learn more easily.
---