I'm sorry for this silly question. Several times I faced with optimization problems which can be expressed as

$$\begin{array}{ll} \text{minimize} & \mathrm x^T \mathrm A \mathrm x\\ \text{subject to} & \|\mathrm x\| = 1\end{array}$$

where $\mathrm A$ is square matrix and $\mathrm x$ is vector.

(This kind of problem arises in PCA or rigid alignment of 3D point sets)

I know that solution of this problem is eigenvalue decomposition of $\mathrm A$. Smallest eigenvalue should be minimum of optimization problem and corresponding eigenvector should be solution.

I want to know is there exist special name for such optimization problem?

PS: Sorry for my poor English.

  • $\begingroup$ Thank you for reply! I thought that well-known problems have own names like Least Squares, Linear/Quadratic/Convex programming, etc. So this fact is surprise for me. Can you rewrite your comment as answer? $\endgroup$
    – Daiver
    Oct 13, 2016 at 16:06
  • $\begingroup$ Done! Glad to help. $\endgroup$ Oct 13, 2016 at 16:19

2 Answers 2


Since the optimization problem has a well-known closed-form solution, it is rarely used in itself and hence usually not given a name. The objective function $x^TAx$ (using $\|x\|=1$), however, is known under the name Rayleigh quotient; the set of its values is called field of values, or numerical range (of $A$).

More generally, this is a textbook example of a quadratic programming problem with a single differentiable equality constraints (if written as $\|x\|^2-1=0$).

  • $\begingroup$ Due to the equality constraint and that $1-\|x\|$ is not a convex function, this isn't a convex constraint. $\endgroup$ Nov 1, 2016 at 15:41
  • $\begingroup$ @Deathbreath Actually, it is convex if written as $g(x):=\|x\|^2-1=0$. I've made this explicit now. $\endgroup$ Nov 1, 2016 at 16:52
  • $\begingroup$ that still isn't convex. To make it precise, a convex constraint is $g(x) \le 0$, where $g$ is a convex function. The above equality constraint is only a convex constraint for linear $g$. Since equality implies $g(x)\le 0$ and $-g(x) \le 0$ and $-g(x)=1-\|x\|^2$ is concave, this is not a convex problem. Or more to the point, the surface of a sphere is not a convex set, hence the feasible set here isn't convex. $\endgroup$ Nov 1, 2016 at 18:20
  • $\begingroup$ But it's not an inequality constraint, but an equality constraint defined by a convex function. I never claimed that the problem itself is convex (because I know that the nomenclature here is not fully consistent, for one). Be that as it may, I have removed the offending claim. $\endgroup$ Nov 1, 2016 at 18:27
  • 1
    $\begingroup$ I'm fully aware of the difference; I claimed the latter but not the former. But I agree that the formulation was misleading, which is why I edited the answer. $\endgroup$ Nov 1, 2016 at 18:33

I know this should be a comment but I do not have enough reputation to comment. I just want to point out that for an arbitrary $A$ with real entries the solution (minimum) to this problem is the smallest eigenvalue of the symmetric part of $A$, namely $S(A)=\frac{(A+A^T)}{2}$, and in the special case when $A$ is symmetric this indeed (as in the original question) gives the smallest eigenvalue of $A$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.