# 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.

• You can put the $x_i$ in a long vector after each other and then $\beta$ in another matrix $I_p \otimes \beta$ where $\otimes$ is Kronecker product. I can try write answer later today if you want. – mathreadler Oct 15 '18 at 5:50
• wouldn't $(x_{i}^{t}\beta)$ be the inner product and be a scalar? – Shogun Oct 15 '18 at 5:52
• hey @mathreadler thanks for introducing me to a new operation! Unfortunately I'm not sure that would work. The $T_i (x):\mathbb{R}^p \to \mathbb{R}$ are statistics and so must output a scalar, while in your form (if I understand correctly), $T_1 (x)$ would be a single vector of each $x$ end to end. This problem is related to sufficiency and the neyman factorisation theorem. I've editedmy post to clarify this – Xiaomi Oct 15 '18 at 7:03
• @Xiaomi yes it is only part of solving the problem. like a hint that maybe can help you finish. i dont have time to write it all now as I am on work. I can write more later. – mathreadler Oct 15 '18 at 7:06
• Oh, I see what you mean now. Thanks, I'll give that a think! – Xiaomi Oct 15 '18 at 7:07

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 $$\mathcal{FINISHED}$$ :)

• that's pretty cool – Shogun Oct 15 '18 at 22:22
• Very interesting! Thank you! – Xiaomi Oct 16 '18 at 7:25
• Pff you didn't really think I finished, did you? It would be 16 times overkill at this rate. – mathreadler Oct 16 '18 at 8:48