First we focus our attention on the spectral properties of
\(T_n(f_\alpha ).\) Since its generating function
\(f_\alpha\) is an even, real-valued and nonnegative function, the matrix
\(T_n(f_\alpha )\) is real, symmetric and positive semidefinite. Consequently,
$$\begin{aligned} T_n(f_\alpha ) = Q_{n,\alpha } \, \varLambda _{n,\alpha } \, Q_{n,\alpha }^T, \end{aligned}$$
(24)
where
\(\varLambda _{n,\alpha } ={\rm{ diag} } (\lambda _{1,\alpha },\lambda _{2,\alpha },\dots , \lambda _{n,\alpha })\) is the diagonal matrix containing the eigenvalues of
\(T_n(f_\alpha )\) sorted in nondecreasing order and
\(Q_{n,\alpha }\) is the orthogonal matrix whose columns are the eigenvectors associated to the eigenvalues
\(\lambda _{j,\alpha },\, j=1,2,\dots ,n.\) In addition, considering that
\({\mathrm{ess}}\inf f_\alpha =0, {\mathrm{ess}}\sup f_\alpha >0\) and
\(f_\alpha \sim |\theta |^\alpha ,\) from Theorem 3 we deduce that
-
\(T_n(f_\alpha )\) is positive definite for each \(n \in \mathbb {N};\)
-
the minimum eigenvalue of \(T_n(f_\alpha )\) behaves as follows: \(\lambda _{1,\alpha } \sim 1/n^\alpha .\)
Now, using (
20) and (
24) we obtain that
$$\begin{aligned} T_{n,n}(\psi _\alpha )& = (Q_{n,\alpha } \, \varLambda _{n,\alpha } \, Q_{n,\alpha }^T) \otimes I_{n} + I_{n} \otimes (Q_{n,\alpha } \, \varLambda _{n,\alpha } \, Q_{n,\alpha }^T) \\& = (Q_{n,\alpha } \otimes Q_{n,\alpha }) (\varLambda _{n,\alpha } \otimes I_n + I_n \otimes \varLambda _{n,\alpha }) (Q_{n,\alpha } \otimes Q_{n,\alpha })^T. \end{aligned}$$
The matrix
\((\varLambda _{n,\alpha } \otimes I_n + I_n \otimes \varLambda _{n,\alpha })\) is block diagonal with diagonal blocks, say
\(D_j,\) having the following form:
$$\begin{aligned} D_j = {\rm{ diag } }(\lambda _{1,\alpha } + \lambda _{j,\alpha }, \lambda _{2,\alpha } + \lambda _{j,\alpha }, \dots ,\lambda _{n,\alpha } + \lambda _{j,\alpha }),\quad j = 1,2,\dots ,n. \end{aligned}$$
In particular, this implies that the minimum and the maximum eigenvalues of
\(T_{n,n}(\psi _\alpha )\) are, respectively,
$$\begin{aligned} \lambda _{\min }(T_{n,n}(\psi _\alpha )) = 2\, \lambda _{1,\alpha }, \qquad \lambda _{\max }(T_{n,n}(\psi _\alpha )) = 2 \,\lambda _{n,\alpha }. \end{aligned}$$
It is worth to observe that since
\(\lambda _{1,\alpha }\) and
\(\lambda _{n,\alpha }\) are the minimum and the maximum eigenvalues of
\(T_{n}(f_\alpha ),\) the condition number in Euclidean norm of the matrix
\(T_{n,n}(\psi _\alpha )\) coincides with the one of the matrix
\(T_{n}(f_\alpha ),\) i.e.,
$$\begin{aligned} \mu _2(T_{n,n}(\psi _\alpha )) = \mu _2(T_{n}(f_\alpha )). \end{aligned}$$
In addition, the behavior of
\(\lambda _{1,\alpha }\) leads to deduce that
$$\begin{aligned} \lambda _{\min }(T_{n,n}(\psi _\alpha )) \sim 1/n^\alpha . \end{aligned}$$
By virtue of the fact that
\(f_2 \sim |\theta |^2\) (see (
16)), the two generating functions defined in (
18) and (
19) satisfy, respectively,
$$\begin{aligned} \varphi _\alpha (\theta _1,\theta _2) \sim \left( |\theta _1|^2 + |\theta _2|^2 \right) ^{\alpha /2}, \qquad \psi _\alpha (\theta _1,\theta _2) \sim |\theta _1|^\alpha + |\theta _2|^\alpha . \end{aligned}$$
(25)
Using the polar coordinates
\((\rho , \eta )\) we can write
$$\begin{aligned} |\theta _1| = \rho |{\cos \eta }|, \qquad |\theta _2| = \rho |{\sin \eta }|. \end{aligned}$$
Therefore, from the previous relations we have
$$\begin{aligned} \left( |\theta _1|^2 + |\theta _2|^2 \right) ^{\alpha /2} = \rho ^\alpha , \qquad |\theta _1|^\alpha + |\theta _2|^\alpha = \rho ^\alpha ( |{\cos \eta }|^\alpha + |{\sin \eta }|^\alpha ). \end{aligned}$$
Since
\((|{\cos \eta }|^\alpha + |{\sin \eta }|^\alpha )\) is never identically zero, we conclude that
$$\begin{aligned} \left( |\theta _1|^2 + |\theta _2|^2 \right) ^{\alpha /2} \sim |\theta _1|^\alpha + |\theta _2|^\alpha . \end{aligned}$$
(26)
By applying the block version of Theorem 4 (see Theorem 3.9 in
[
40] for a more general result), we have
$$\begin{aligned} \lambda _{\min }(T_{n,n}(\varphi _\alpha ))\sim \lambda _{\min }(T_{n,n}(\psi _\alpha )). \end{aligned}$$
Consequently,
$$\begin{aligned} \lambda _{\min }(T_{n,n}(\varphi _\alpha ))\sim 1/n^\alpha . \end{aligned}$$
Notice that because of (
25) and (
26), both
\(\varphi _\alpha\) and
\(\psi _\alpha\) have a zero at the origin of order
\(\alpha\) in the sense of Definition 2.