# Weyl's inequality

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In mathematics, there are at least two results known as Weyl's inequality.

## Weyl's inequality in number theory

In number theory, Weyl's inequality, named for Hermann Weyl, states that if M, N, a and q are integers, with a and q coprime, q > 0, and f is a real polynomial of degree k whose leading coefficient c satisfies

$\displaystyle{ |c-a/q|\le tq^{-2}, }$

for some t greater than or equal to 1, then for any positive real number $\displaystyle{ \scriptstyle\varepsilon }$ one has

$\displaystyle{ \sum_{x=M}^{M+N}\exp(2\pi if(x))=O\left(N^{1+\varepsilon}\left({t\over q}+{1\over N}+{t\over N^{k-1}}+{q\over N^k}\right)^{2^{1-k}}\right)\text{ as }N\to\infty. }$

This inequality will only be useful when

$\displaystyle{ q \lt N^k, }$

for otherwise estimating the modulus of the exponential sum by means of the triangle inequality as $\displaystyle{ \scriptstyle\le\, N }$ provides a better bound.

## Weyl's inequality in matrix theory

#### Weyl's inequality about perturbation

In linear algebra, Weyl's inequality is a theorem about the changes to eigenvalues of a Hermitian matrix that is perturbed. It is useful if we wish to know the eigenvalues of a Hermitian matrix but there is an uncertainty about its entries. We let H be the exact matrix and P be a perturbation matrix that represents the uncertainty. The matrix we 'measure' is $\displaystyle{ M \,=\, H \,+\, P }$.

The theorem says that if any two of M, H and P are n by n Hermitian matrices, where M has eigenvalues

$\displaystyle{ \mu_1 \ge \cdots \ge \mu_n }$

and H has eigenvalues

$\displaystyle{ \nu_1 \ge \cdots \ge \nu_n }$

and P has eigenvalues

$\displaystyle{ \rho_1 \ge \cdots \ge \rho_n }$

then the following inequalities hold for $\displaystyle{ i \,=\, 1,\dots ,n }$:

$\displaystyle{ \nu_i + \rho_n \le \mu_i \le \nu_i + \rho_1 }$

More generally, if $\displaystyle{ j+k-n \,\ge\, i \,\ge\, r+s-1 }$, we have

$\displaystyle{ \nu_j + \rho_k \le \mu_i \le \nu_r + \rho_s }$

If P is positive definite (that is, $\displaystyle{ \scriptstyle\rho_n \,\gt \, 0 }$) then this implies

$\displaystyle{ \mu_i \gt \nu_i \quad \forall i = 1,\dots,n. }$

Note that we can order the eigenvalues because the matrices are Hermitian and therefore the eigenvalues are real.

#### Weyl's inequality between eigenvalues and singular values

Let $\displaystyle{ A \in \mathbb{C}^{n \times n} }$ have singular values $\displaystyle{ \sigma_1(A) \geq \cdots \geq \sigma_n(A) \geq 0 }$ and eigenvalues ordered so that $\displaystyle{ |\lambda_1(A)| \geq \cdots \geq |\lambda_n(A)| }$. Then

$\displaystyle{ |\lambda_1(A) \cdots \lambda_k(A)| \leq \sigma_1(A) \cdots \sigma_k(A) }$

For $\displaystyle{ k = 1, \ldots, n }$, with equality for $\displaystyle{ k=n }$. 

## Applications

### Estimating perturbations of the spectrum

Assume that we have a bound on P in the sense that we know that its spectral norm (or, indeed, any consistent matrix norm) satisfies $\displaystyle{ \|P\|_2 \le \epsilon }$. Then it follows that all its eigenvalues are bounded in absolute value by $\displaystyle{ \epsilon }$. Applying Weyl's inequality, it follows that the spectra of M and H are close in the sense that

$\displaystyle{ |\mu_i - \nu_i| \le \epsilon \qquad \forall i=1,\ldots,n. }$

### Weyl's inequality for singular values

The singular values {σk} of a square matrix M are the square roots of eigenvalues of M*M (equivalently MM*). Since Hermitian matrices follow Weyl's inequality, if we take any matrix A then its singular values will be the square root of the eigenvalues of B=A*A which is a Hermitian matrix. Now since Weyl's inequality hold for B, therefore for the singular values of A.

This result gives the bound for the perturbation in the singular values of a matrix A due to perturbation in A.