3
$\begingroup$

Answers given in this question descirbe how if we square an order-$m$ polynomial with coefficients $a_k$: $$ p(x) = \left( \sum_{k=0}^{m} a_k x^k \right)^2 = \sum_{k=0}^{2m} c_k x^k $$ The coefficients $c_k$ of the resulting order-$2m$ polynomial can be calculated as $$ c_k = \sum_{j=0}^{k} a_j a_{k-j} $$ provided we take $a_j = 0$ in cases where $j \notin \{ 0, \dots,m\}$.

Question:

If we let $\mathbf{c}$ be a length $2m+1$ vector containing all the $c_k$, and $\mathbf{a}$ be a length $m+1$ vector containing all the $a_k$, is it possible to express $\mathbf{c}$ as a matrix product of $\mathbf{a}$ with other matrices?

I've been thinking about constructing some matrices to pad the $\mathbf{a}$ with zeros up to the length of $\mathbf{c}$, then making a reversed copy, but haven't been able to get any further.

Thanks!

  • 0
    I think you would need something like a circulant matrix built from $\bf a$ (but one that didn't wrap around at the ends), which isn't exactly expressible as a matrix product. – 2017-01-01
  • 1
    You asked if $c$ depends linearly on $a$, which is in general not the case. – 2017-01-01
  • 0
    But you might interested in the connection of Convolution and (Fast) Fourier Transform (which is well known) – 2017-01-01
  • 0
    I have read this three times and I still forget the main question by the time I get my paper out. Bravo. – 2017-01-01
  • 0
    Having reread it, @user251257 is correct that the relation is not linear, so a matrix in general will not work, although you can surely find situations where one does the job. – 2017-01-01

0 Answers 0