size of the embedding to $$m \geq 2 \sqrt{6n^2 + \kappa_2(T) \frac{3 \cdot 2^{11/2}
    \, n^{5/2}}{5^{5/2}}}.
  \label{4.8}$$ Note that among all positive semidefinite matrices $C$ of order greater or equal $m$, we can choose one of prime order. This embedding will be positive definite according to Theorem \[thmGKPS\]. It is obvious that embeddings (\[4.7\])–(\[4.8\]) are bounded by a function of the condition number of $T$, i.e., the quantity which we are trying to bound.

Ferreira in [@FerreiraP-94] embeds a Töplitz matrix $T$ of order $n$, defined by the symbol $t = (t_0, \ldots, t_r, 0, \ldots, 0) \in \mathbb{R}^n$, into the circulant $C$ of order $m = 2n$, $$C = \begin{bmatrix}
      T & S \\
      S & T
      \end{bmatrix},
  \label{4.9}$$ where the symbol of the Töplitz matrix $S$ is $s = (0, \ldots, 0, t_r, \ldots, t_1) \in \mathbb{R}^n$.

If we take $T = T_n^d$ from (\[3.1\]), the only difference between embeddings (\[4.1\]) and (\[4.9\]) is in exactly $n - d - r$ zero diagonals, added as the first diagonals of $S$. A sufficient condition for positive semidefiniteness of $C$ is given by the next result.

Let $C$ be defined as in (\[4.9\]), and let $b^T = [t_0, \ldots, t_{n 