Help factorising this matrix series Let $x_i$ be a series of vectors of equal length, and let $\beta$ be a constant vector of equal length to $x_i$'s
I have the following sum
$$\sum_{i=1}^p (x_i^T \beta)^2 = \sum_{i=1}^p x_i^T \beta \beta^T x_i = x_1^T \beta \beta^T x_1 + x_2^T \beta \beta^T x_2 + \dots + x_p^T \beta \beta^T x_p$$
In order to apply a statistical theorem, I need to factorise this into a form
$$\sum_{i=1}^d T_i (x) g_i (\beta)$$
where each $T_i (x): \mathbb{R}^n \to \mathbb{R}$ out puts a single scalar $x$. and where the lower the $d$ the better. ie, I want to find the simplist factorisation of the above sum such that the $x$ terms and $\beta$ terms are separated.
My attempt:
I tried writing out the matrix multiplication $x_i^T \beta$ as the sum $\sum_j x_{ij}\beta_j$ but this didn't get me anywhere since it leads me to
$$\sum_j \sum_j \beta_j \beta_k \sum_i x_{ij}x_{ik}$$
which gives a total of $d=p^2$ summands... which is terrible considering $\beta$ is only of length $p$.
Any help here finding a simpler factorisation is much appreciated, thank you.

Let us give a small example for
$$I_p \otimes \beta$$
will be if p=3 and $\beta = [1,2,3]$:
$$\left[\begin{array}{ccccccccc}1&2&3&0&0&0&0&0&0\\0&0&0&1&2&3&0&0&0\\0&0&0&0&0&0&1&2&3\end{array}\right]$$
We see that if we stuff $[x_1,x_2,x_3]^T$ into column vector we can do
$$(I_p \otimes \beta)[x_1,x_2,x_3]^T$$
and then we will get the 3 scalar products you have sought in resulting product vector. The only thing that remains is to take the squared two-norm of this vector.
$$\|(I_p \otimes \beta)[x_1,x_2,x_3]^T\|_2^2$$
It is known that $\|a\|_2^2=a^Ta$ so we can calculate this with for example:
$$([x_1,x_2,x_3](I_p \otimes \beta)^T)((I_p \otimes \beta)[x_1,x_2,x_3]^T)$$
And we have $\mathcal{FINISHED}$ :)
