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 : $\hspace{2cm}$enter image description here

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 : enter image description here

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 : enter image description here

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.

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    $\begingroup$ Could you explain those second and third pictures? How have you taken these two matrices and turned them into these surfaces? $\endgroup$ Aug 4 at 8:04
  • $\begingroup$ I used MATLAB's command $\mathsf{surf()}$ to plot them @TheoBendit $\endgroup$
    – SPARSE
    Aug 4 at 9:07
  • $\begingroup$ Matlab is not one of my strong suits, but it sounds to me like you're plotting these matrices as functions of the form $x \mapsto Ax$. If this is the case, then something went wrong with your final picture. The function is linear, so the surface output should be a subspace. The fold that you're seeing is evidence that something has gone wrong. Or, it's evidence that I have totally misunderstood how surf would represent a matrix. $\endgroup$ Aug 4 at 18:06
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    $\begingroup$ $S-T$ is a reflection, namely the Householder reflection about the hyperplane orthogonal to $v$ ($v \neq 0$). So "geometrically", $S$ is a reflection plus a projection. A bunch of good things happens for $S$, and geometrically $T$ is a projection as $T^2=T$. $\endgroup$ Aug 6 at 7:32
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    $\begingroup$ I guessed and managed to recreate your plots. You are just doing surf(T) so matlab will interpret T as a heightmap over the indexed grid x=[1,2]; y=[1,2];. It's not a meaningful way to look at this data at all. $\endgroup$ Aug 7 at 0:40

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 $\mathcal T$.


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