TITLE: Find eigenvalues with given multiplicity in presence of errors
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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.