I am trying to understand the following argument concerning this problem : we have $x_1=(x_{11}, ..., x_{1d})$ and $x_{2}=(x_{21}, ..., x_{2d})$ two vectors in $\mathbb{R}^d$. We want to find the rank of the matrix $B=\frac{1}{4}(x_1 - x_2)(x_1 - x_2)^{T}$ where $T$ denotes the transpose.
To show that the rank is $1$, the following argument has been proposed : the rank is $1$ because it is proportional to the projection matrix on the vector $(x_1 -x_2)$.
To be honest, I don't know what is a projection matrix on a vector ? I thought of looking at the image of this matrix but considering the face it has it would probably be a bad idea.
For a vector $v$ in $\mathbb{R}^{d}$, I know that we can project (orthogonal) him on a vector $w\in\mathbb{R}^d$ and this projection is equal to $\frac{\langle w,v\rangle}{\langle w, w\rangle}v$ using the orthogonality condition. Then, if we go further and consider a subspace $F=span(u_1, ..., u_k)$ we can generalize the previous formula to say that the orthogonal projection of $v$ on $F$ is $\sum_{i=1}^{k}\frac{\langle u_i, v\rangle}{\langle u_i, u_i\rangle}u_i$ or $\sum_{i=1}^{k}\langle u_i, v\rangle u_i$ if we assume orthonormality for the $u_i$ . However, I cannot see how to translate that in a matrix form since I have never been exposed to this way of projecting.
Thank you a lot !
