Image of Subset

63. Image of Subset

For a linear transformation $𝑇 : ℝ^{𝑛} \to ℝ^{𝑚}$ and a subset $𝑉 \subseteq ℝ^{𝑛}$ , the image of $𝑉$ under $𝑇$ is:

𝑇 (𝑉) = {𝑇 (\vec{𝑥}) | \vec{𝑥} \in 𝑉}

— the set of all outputs you get by applying $𝑇$ to every input in $𝑉$ .

Example

{\vec{𝑥}}_{0} = [\begin{matrix} - 2 \\ - 2 \end{matrix}] {\vec{𝑥}}_{1} = [\begin{matrix} - 2 \\ 2 \end{matrix}] {\vec{𝑥}}_{2} = [\begin{matrix} 2 \\ - 2 \end{matrix}]

\begin{matrix} 𝐿_{0} = {{\vec{𝑥}}_{0} + 𝑡 ({\vec{𝑥}}_{1} - {\vec{𝑥}}_{0}) | 0 \leq 𝑡 \leq 1} \\ 𝐿_{1} = {{\vec{𝑥}}_{1} + 𝑡 ({\vec{𝑥}}_{2} - {\vec{𝑥}}_{1}) | 0 \leq 𝑡 \leq 1} \\ 𝐿_{2} = {{\vec{𝑥}}_{2} + 𝑡 ({\vec{𝑥}}_{0} - {\vec{𝑥}}_{2}) | 0 \leq 𝑡 \leq 1} \end{matrix}

Triangle: $𝑆 = {𝐿_{0}, 𝐿_{1}, 𝐿_{2}}$ .

Apply transformation $𝑇 (\vec{𝑥}) = 𝐴 \vec{𝑥}$ with $𝐴 = [\begin{matrix} 1 & - 1 \\ 2 & 0 \end{matrix}]$ :

𝑇 ({\vec{𝑥}}_{0}) = [\begin{matrix} 0 \\ - 4 \end{matrix}] 𝑇 ({\vec{𝑥}}_{1}) = [\begin{matrix} - 4 \\ - 4 \end{matrix}] 𝑇 ({\vec{𝑥}}_{2}) = [\begin{matrix} 4 \\ 4 \end{matrix}]

By linearity, line segments map to line segments: $𝑇 (𝐿_{0})$ is the image of $𝐿_{0}$ , and $𝑇 (𝑆)$ is the image of the whole triangle.