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? $\endgroup$ – Dark_Daiver Oct 13 '16 at 16:06
  • $\begingroup$ Done! Glad to help. $\endgroup$ – Christian Clason Oct 13 '16 at 16:19

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 $\|x\|^2-1=0$).

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  • $\begingroup$ Due to the equality constraint and that $1-\|x\|$ is not a convex function, this isn't a convex constraint. $\endgroup$ – Deathbreath Nov 1 '16 at 15:41
  • $\begingroup$ @Deathbreath Actually, it is convex if written as $g(x):=\|x\|^2-1=0$. I've made this explicit now. $\endgroup$ – Christian Clason Nov 1 '16 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. $\endgroup$ – Deathbreath Nov 1 '16 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. $\endgroup$ – Christian Clason Nov 1 '16 at 18:27
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    $\begingroup$ 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. $\endgroup$ – Christian Clason Nov 1 '16 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 $A$.

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