How can this matrix be geometrically described I am revising basic linear algebra concepts and I came across the concept of projecting vectors. Recall

The matrix that (orthogonally) projects the vector to which it is applied onto the vector $v$ is given by :
$$
\mathcal{T}:=\frac{vv^{\top}}{\|v\|_{2}^{2}}
$$
while :
$$
\mathcal{S}:= I-\mathcal{T}=I-\frac{vv^{\top}}{\|v\|_{2}^{2}}
$$
is the matrix that (orthogonally) projects the vector to which it is applied onto the space orthogonal to the vector $v$

I wanted to visualize both $\mathcal{S,T}$ for some $u,v,\in\mathbb{R}^{2}$. I have chosen $\mathbf{u}:=\begin{bmatrix}1&2 \end{bmatrix}^{\top}$ and $\mathbf{v}:=\begin{bmatrix}3&5\end{bmatrix}^{\top}$. On MATLAB I plotted $u,v,\mathcal{T}u, $ and $\mathcal{S}u$ all together and I obtained the following plot :
$\hspace{2cm}$
As seen above, $\mathcal{T}u$ (red) is a projection of $u$ to align on the same direction of $v$. Furthermore, $\mathcal{S}u$ (green) is an orthogonal projection of $u$. What I am curious about is how can we interpret the geometrical structure of $\mathcal{T}$ and $\mathcal{S}$ which I plotted both of them :

The above surface is $\mathcal{T}$ which appears to be an oriented plane for some specific degree of orientation with respect to the $xy$ plane. Furthermore :

is the surface $\mathcal{S}$ which I don't know how to describe it but I do notice symmetry in the middle. I would hope to know more about $\mathcal{S,T}$ as to how do they perform this reflection operation of $u$ and $v$ and what properties do they have they seem to be similar to the concept of Householder reflection.
Note that some textbooks define $\mathcal{T}u$ as $\widehat{u}$ and $\mathcal{S}u$ as $u^{\perp}$
Update : After doing some calculation, it appeared that $\mathcal{T}^{2}=\mathcal{T}$, which appears to be a property of $\mathcal{T}$ I believe this is meant to describe $\mathcal{T}$ geometrically, thus I hope someone can interpret this result for me.
 A: Let $V=\operatorname{span}(v)$ and $U=V^{\bot}=\{u\in \mathbb R^n\mid\langle u,v\rangle=0\}$. Then $V$ and $U$ are the eigenspaces of $\mathcal S$ as well as $\mathcal T$. We have $\mathcal S v=v-v\frac{v^\top v}{v^\top v}=0$ and $\mathcal T v=v\frac{v^\top v}{v^\top v}=v$. For $u\in U$, we get $\mathcal S u=u-v\frac{v^\top u}{v^\top v}=u$ and $\mathcal T u=v\frac{v^\top u}{v^\top v}=0$.
Hence, $\mathcal S$ is the orthogonal projection on $U$ and $\mathcal T$ is the orthogonal projection on $V$. Clearly, $\mathcal S^2=\mathcal S$ and $\mathcal T^2=\mathcal T$. Since after one orthogonal projection, we are already in the eigen space, so a second projection does not change anything. Further observe $\mathcal S\mathcal T=\mathcal T\mathcal S=0$, since $U\cap V=\{0\}$ and one operator maps on $U$ and the other maps on $V$.
You can visualize it, if you draw a plane and a line, which is orthogonal to the plane. The line has the direction $v$. Every point in the space will be mapped on the plane by $\mathcal S$. And it will be mapped on the line by $\mathcal T$.
