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## Download Stick Figure Animation Softwarel VERIFIED

Download. Free Animated Stickers for Microsoft Movie Maker Stick Figure. animated stickers to your screen using Microsoft Movie Maker.. Animated video is great for.Q: How to prove that matrix product is an isometry? Let $F:\ell^2\rightarrow\ell^2$ be defined by $F(x)=\left[\begin{matrix}x_1\\x_2\\\vdots\\x_n\end{matrix}\right]$ and $G(\left[\begin{matrix}x_1\\x_2\\\vdots\\x_n\end{matrix}\right])=\left[\begin{matrix}x_1\\x_2\\\vdots\\x_n\end{matrix}\right]$ Let $T:\ell^2\rightarrow\ell^2$ be defined by $T(x)=F(G(x))$.Show that $T$ is an isometry. I tried by writing $$T(x)=\left[\begin{matrix}x_1\\x_2\\\vdots\\x_n\end{matrix}\right]$$ $$=\left[\begin{matrix}x_1\\x_2\\\vdots\\x_n\end{matrix}\right]$$ $$=F\left(\left[\begin{matrix}x_1\\x_2\\\vdots\\x_n\end{matrix}\right]\right)$$ $$=\left[\begin{matrix}x_1\\x_2\\\vdots\\x_n\end{matrix}\right]$$ $$=x$$ Thus $$\lVert T(x)\rVert=\lVert x\rVert$$ This just shows that $T(x)=x$ how can I show that it is an isometry? A: Hint: Show that $T$ is $0$ on $(\mathbb{R}^n)^\perp$ and is injective. For example, suppose $Tx=0$. Do you see why there can be no $y e 0$ such that $y\perp\Bbb R^n$ and $y\perp x$? (Try 6d1f23a050