> [!summary]
The image describes a set of vectors in the domain mapped to the codomain. The image is a subspace of the codomain.
>
**Key Equations:**
$\mathrm{Im}(S) = \left\{ L (\vec{s})| \vec{s}\in S \right\} \subseteq W$
>[!info]+ Read Time
**⏱ 1 min**
# Definition
If a [[Vector Spaces|vector space]] ($V$) is [[Linear Mapping (Transformations)|linearly mapped]] to another [[Vector Spaces|vector space]] $W$, via [[Linear Mapping (Transformations)|transformation]] $L: V\to W$. Then any [[Subspace|subspace]], $S \subseteq V$ will be mapped to a [[Subsets|subset]] of $W$. The [[Sets|set]] of all vectors in $S$ under $L$ is called the image of $S$ under $L$. The image itself is a [[Subspace|subspace]] of $W$ formally defined below.
$
\begin{array}{c}
\text{Image of S under L} = \mathrm{Im}(S)= L(S) = \left\{ L(\vec{s})| \vec{s}\in S \right\} \subseteq W
\end{array}
$
![[im_1.png]]
## Proof of Subspace
> [!warning] Assumptions
To prove that the $\mathrm{Im}(S)$ is a [[Subspace|subspace]] of $W$ through the [[Subspace|subspace test]], assume the following:
> - Zero vector is defined as $\vec{0}_{V}\in V$ and $\vec{0}_{W}\in W$
> - $\exists u,v\in V$ such that $y_1=S(u)$ and $y_2=s(v)$
> - $\alpha \in \mathbb{F}$
**Zero Vector**
$
S(0_V) = S(0 \cdot v) = 0 \cdot S(v) = 0_W , \quad 0_W \in \operatorname{Im}(S)
$
**Closed under vector addition**
$
y_1 + y_2 = S(u) + S(v) = S(u + v) \in \operatorname{Im}(S) .
$
**Closed under scalar multiplication:**
$
\alpha y = \alpha S(v) = S(\alpha v) \in \operatorname{Im}(S)
$
# Resources
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