Eigenvalue bounds
I am looking for any information on the bounds of the eigenvalues of the cubic spline interpolation matrix when all the values are positive. The bounds in question are:
$$frac{1}{6} Min(x_n) le |lambda| le Max(x_n) $$
This should hold for any eigenvalue of the matrix and the matrix looks like
$$
begin{bmatrix} frac{x_0}{3} & frac{x_0}{6} & 0 & ... \ frac{x_0}{6} & frac{x_0 + x_1}{6} & frac{x_1}{6} & 0 & ... \ 0 & frac{x_1}{6} & frac{x_1 + x_2}{3} & frac{x_2}{6} & 0 & ... \ 0 & 0 & ... \ 0 & 0 & 0 & frac{x_{n-2}}{6} & frac{x_{n-2} + x_{n-1}}{3} & frac{x_{n-1}}{6} \ 0 & 0 & 0 & 0 & frac{x_{n-1}}{6} & frac{x_{n-1}}{3}
end{bmatrix}$$
Is there a name for this bounds and where it comes from would be helpful.
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 1 at 4:51
user621595
11
add a comment |
I am looking for any information on the bounds of the eigenvalues of the cubic spline interpolation matrix when all the values are positive. The bounds in question are:
$$frac{1}{6} Min(x_n) le |lambda| le Max(x_n) $$
This should hold for any eigenvalue of the matrix and the matrix looks like
$$
begin{bmatrix} frac{x_0}{3} & frac{x_0}{6} & 0 & ... \ frac{x_0}{6} & frac{x_0 + x_1}{6} & frac{x_1}{6} & 0 & ... \ 0 & frac{x_1}{6} & frac{x_1 + x_2}{3} & frac{x_2}{6} & 0 & ... \ 0 & 0 & ... \ 0 & 0 & 0 & frac{x_{n-2}}{6} & frac{x_{n-2} + x_{n-1}}{3} & frac{x_{n-1}}{6} \ 0 & 0 & 0 & 0 & frac{x_{n-1}}{6} & frac{x_{n-1}}{3}
end{bmatrix}$$
Is there a name for this bounds and where it comes from would be helpful.
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 1 at 4:51
user621595
11
You seem to have at least one typo.
– Ian
Dec 1 at 4:56
In any case the upper bound is quite suboptimal, you can easily use Gerschgorin's theorem to make the coefficient be $2/3$ instead. The lower bound is more problematic.
– Ian
Dec 1 at 4:58
add a comment |
I am looking for any information on the bounds of the eigenvalues of the cubic spline interpolation matrix when all the values are positive. The bounds in question are:
$$frac{1}{6} Min(x_n) le |lambda| le Max(x_n) $$
This should hold for any eigenvalue of the matrix and the matrix looks like
$$
begin{bmatrix} frac{x_0}{3} & frac{x_0}{6} & 0 & ... \ frac{x_0}{6} & frac{x_0 + x_1}{6} & frac{x_1}{6} & 0 & ... \ 0 & frac{x_1}{6} & frac{x_1 + x_2}{3} & frac{x_2}{6} & 0 & ... \ 0 & 0 & ... \ 0 & 0 & 0 & frac{x_{n-2}}{6} & frac{x_{n-2} + x_{n-1}}{3} & frac{x_{n-1}}{6} \ 0 & 0 & 0 & 0 & frac{x_{n-1}}{6} & frac{x_{n-1}}{3}
end{bmatrix}$$
Is there a name for this bounds and where it comes from would be helpful.
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 1 at 4:51
user621595
11
I am looking for any information on the bounds of the eigenvalues of the cubic spline interpolation matrix when all the values are positive. The bounds in question are:
$$frac{1}{6} Min(x_n) le |lambda| le Max(x_n) $$
This should hold for any eigenvalue of the matrix and the matrix looks like
$$
begin{bmatrix} frac{x_0}{3} & frac{x_0}{6} & 0 & ... \ frac{x_0}{6} & frac{x_0 + x_1}{6} & frac{x_1}{6} & 0 & ... \ 0 & frac{x_1}{6} & frac{x_1 + x_2}{3} & frac{x_2}{6} & 0 & ... \ 0 & 0 & ... \ 0 & 0 & 0 & frac{x_{n-2}}{6} & frac{x_{n-2} + x_{n-1}}{3} & frac{x_{n-1}}{6} \ 0 & 0 & 0 & 0 & frac{x_{n-1}}{6} & frac{x_{n-1}}{3}
end{bmatrix}$$
Is there a name for this bounds and where it comes from would be helpful.
linear-algebra matrices eigenvalues-eigenvectors
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 1 at 4:51
user621595
11
asked Dec 1 at 4:51
user621595
11
edited Dec 1 at 4:58
asked Dec 1 at 4:51
user621595
11
asked Dec 1 at 4:51
user621595
11
asked Dec 1 at 4:51
user621595
11
11
You seem to have at least one typo.
– Ian
Dec 1 at 4:56
In any case the upper bound is quite suboptimal, you can easily use Gerschgorin's theorem to make the coefficient be $2/3$ instead. The lower bound is more problematic.
– Ian
Dec 1 at 4:58
add a comment |
You seem to have at least one typo.
– Ian
Dec 1 at 4:56
In any case the upper bound is quite suboptimal, you can easily use Gerschgorin's theorem to make the coefficient be $2/3$ instead. The lower bound is more problematic.
– Ian
Dec 1 at 4:58
You seem to have at least one typo.
– Ian
Dec 1 at 4:56
You seem to have at least one typo.
– Ian
Dec 1 at 4:56
In any case the upper bound is quite suboptimal, you can easily use Gerschgorin's theorem to make the coefficient be $2/3$ instead. The lower bound is more problematic.
– Ian
Dec 1 at 4:58
In any case the upper bound is quite suboptimal, you can easily use Gerschgorin's theorem to make the coefficient be $2/3$ instead. The lower bound is more problematic.
– Ian
Dec 1 at 4:58
add a comment |
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You seem to have at least one typo.
– Ian
Dec 1 at 4:56
In any case the upper bound is quite suboptimal, you can easily use Gerschgorin's theorem to make the coefficient be $2/3$ instead. The lower bound is more problematic.
– Ian
Dec 1 at 4:58