Find eigenvalues with given multiplicity in presence of errors

Question

Let $A^0$ a $3 \times 3$ real symmetric matrix with eigenvalues $\lambda_1^0 = \lambda_2^0 \neq \lambda_3^0$. Hence, $\lambda_1^0$ has multiplicity $2$.

In real-world applications, I will have a perturbed matrix $A=A^0 + \Delta A $.

Hence, I will find 3 different eigenvalues, but, if the errors are "small", two of them will be similar: $\lambda_1 \simeq \lambda_2 \neq \lambda_3$.

Is it possible to find the "best" approximation of the "original" eigenvalue $\lambda_1^0$?

The first idea I came up was to use the average of $\lambda_1$ and $\lambda_2$, but I would like to have a sort of optimality criterion.

Answer 1

If $A^0$ is symmetric, then taking the arithmetic mean of the two closest eigenvalues gives the closest matrix with repeated eigenvalues in the Euclidean norm. Indeed, by the Bauer-Fike theorem, every perturbation with $\|\Delta A\|_2 < \frac{|\lambda_1-\lambda_2|}{2} =: \delta$ has distinct eigenvalues (because there must be an eigenvalue such that $|\lambda - \lambda_1| \leq \delta$, one such that $|\lambda - \lambda_2| \leq \delta$, and one such that $|\lambda - \lambda_3| \leq \delta$, and these three circle can't overlap).

In the Frobenius norm (as suggested in michael's answer), it can be proved using the Weyl inequalities that the same matrix achieves the minimum over all symmetric perturbations $\Delta A$. I guess that it is also the minimum over general $\Delta A$, but I don't have an immediate proof.

For a non-normal matrix, that matrix nearness problem isn't as easy to solve, unfortunately.

Answer 2

Let $M$ be the set of symmetric $3\times3$ matrices that admit a multiple eigenvalue. It is a closed subset, a codimension-$2$ submanifold (with a singular line along the homotheties). A reasonnable approach is to project $A$ over $M$, in the euclidian structure defined by the Frobenius norm.

Let $S\in M$ be generic (not a homothety). It has a double eigenvalue, corresponding to an eigenplane $P$. The tangent space to $M$ at $S$ is made of those symmetric $T$ such that the restriction of $T$ to $P$ (in the sense of quadratic forms) is proportional to the identity. That is $v^TTv\equiv\alpha|v|^2$ over $P$, for some $\alpha$.

Projecting $A$ on $M$ consists therefore in finding a plane $P$ on which the quadratic form $q_A(v)=v^TAv$ coincides with $\lambda|v|^2$ ; and this $\lambda$ is the eigenvalue you are searching for. In practice, you consider a level set of $q_A$ and look for a plane that cuts it along a circle.

Edit. There exists orthonormal coordinates in which $q_A$ writes $ax^2+by^2+cz^2$, where $a,b,c$ are the eigenvalues of $A$. A rather simple calculation shows that if a plane cuts a level set of $q_A$ along a circle, then its equation has to be of the form $z=\alpha x$, up to a permutation of the coordinates (i.e. the plane contains a coordinate axis). Then $q_A(v)\equiv b|v|^2$ on this plane and $\alpha$ is given by $(b-c)\alpha^2=(a-b)$. This is possible only if $b$ lies between $a$ and $c$.

In conclusion, we are led to choose one eigenvalue of $A$, precisely the intermediate one $\lambda_2$ (with the order $\lambda_1\le\lambda_2\le\lambda_3$). This is the best approximation of $\lambda_1^0=\lambda_2^0$ in the sense of the Frobenius-projection onto the set of symmetric matrices having a double eigenvalue.

Answer 3

Suppose the entries of $\Delta A$ are $N(0, \sigma^2)$. Then you have a fairly standard estimation problem where you are estimating the mean of a $N(\mu, \Sigma)$ observation, and $\mu$ lies in the space of matrices with exactly 2 distinct eigenvalues. The mle in this problem is to pick the closest such matrix in the regular Euclidean metric, and the mle for the multiple eigenvalue is its multiple eigenvalue. OTOH, the mle does optimize something, but maybe not what you want, any other error structure is different and probably harder and it is an awkward space to work in anyway.