这是一份northeastern东北大学(美国) MATH 1213作业代写的成功案
If we transform a line:
$$
\begin{aligned}
L(t) &=(1-t) P_{0}+t P_{1} \
\mathcal{T}(L(t)) &=\mathfrak{J}\left((1-t) P_{0}+t P_{1}\right) \
&=(1-t) \mathcal{J}\left(P_{0}\right)+t \mathfrak{J}\left(P_{1}\right)
\end{aligned}
$$
The result is clearly still a line (assuming $\mathcal{T}\left(P_{0}\right)$ and $\mathcal{T}\left(P_{1}\right)$ aren’t coincident). Similarly, if we transform a plane:
$$
\begin{aligned}
P(t) &=(1-s-t) P_{0}+s P_{1}+t P_{2} \
\mathcal{T}(P(t)) &=\mathfrak{J}\left((1-s-t) P_{0}+s P_{1}+t P_{2}\right) \
&=(1-s-t) \mathcal{T}\left(P_{0}\right)+s \mathcal{T}\left(P_{1}\right)+t \mathcal{T}\left(P_{2}\right)
\end{aligned}
$$
MATH 1213COURSE NOTES ：
We know that the transformed vector will be $\mathbf{v}{\perp}-\mathbf{v}{|}$. Substituting equations and into this gives us
$$
\begin{aligned}
\mathcal{T}(\mathbf{v}) &=\mathbf{v}{\perp}-\mathbf{v}{|} \
&=\mathbf{v}-2 \mathbf{v}_{|} \
&=\mathbf{v}-2(\mathbf{v} \cdot \hat{\mathbf{n}}) \hat{\mathbf{n}}
\end{aligned}
$$
From Chapter 2, we know that we can perform the projection of $\mathbf{v}$ on $\hat{\mathbf{n}}$ by multiplying by the tensor product matrix $\hat{\mathbf{n}} \otimes \hat{\mathbf{n}}$, so this becomes
$$
\begin{aligned}
\mathcal{T}(\mathbf{v}) &=\mathbf{v}-2(\hat{\mathbf{n}} \otimes \hat{\mathbf{n}}) \mathbf{v} \
&=[\mathbf{I}-2(\hat{\mathbf{n}} \otimes \hat{\mathbf{n}})] \mathbf{v}
\end{aligned}
$$