> [!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 <iframe width="560" height="315" src="https://www.youtube.com/embed/vyYrvhbDhW4?si=9p72z9REUj1u0iuC" title="YouTube video player" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" referrerpolicy="strict-origin-when-cross-origin" allowfullscreen></iframe>