# Solving discrete convolution linear functions

Consider we have samples $$\mathbf{X} \in \mathcal{R}^{n\times p}$$ and we aim to find a "regression" coefficients $$\beta \in \mathcal{R}^{q \times 1}$$ ($$q>p$$), but the regression is defined as a convolution function as follows:

For every $$\mathbf{x}$$, we have:

$$\mathbf{x}\odot\beta = \begin{cases} \mathbf{x}_1\beta_1+\mathbf{x}_2\beta_2 + \dots + \mathbf{x}_p\beta_p \\ \mathbf{x}_1\beta_2+\mathbf{x}_2\beta_3 + \dots + \mathbf{x}_p\beta_{p+1} \\ \vdots\\ \mathbf{x}_1\beta_{q-p+1}+\mathbf{x}_2\beta_{q-p+2} + \dots + \mathbf{x}_p\beta_q \end{cases}$$ therefore, $$\mathbf{x}\odot\beta$$ leads to a vector.

Given a data set $$\mathbf{X}$$, can we solve a non-trivial $$\beta$$ so that for every $$\mathbf{x} \in \mathbf{X}$$, $$\mathbf{x}\odot\beta = \mathbf{0}$$?

• There's a trivial solution: set all $\beta$ to zero. – user20160 Jul 24 '19 at 10:49
• Thanks, I edited the question a little bit to specify that we need a non-trivial solution @user20160 – Haohan Wang Jul 24 '19 at 17:26