# 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.

• Could you explain those second and third pictures? How have you taken these two matrices and turned them into these surfaces? Aug 4 at 8:04
• I used MATLAB's command $\mathsf{surf()}$ to plot them @TheoBendit Aug 4 at 9:07
• 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. Aug 4 at 18:06
• $S-T$ 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 $T^2=T$. Aug 6 at 7:32
• 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. 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 $$\mathcal T$$.