Find critical points of $\langle Av,v\rangle$

Question

Let $A$ be an $n \times n$ real symmmetric matrix. Find the critical points of the function $\langle Av,v\rangle$ restricted to the unit sphere in $\mathbb{R}^n$.

I would think you just use Lagrange multipliers, and $\nabla\langle Av,v\rangle=2Ax$, since $A$ is symmetric. But this only gives the possible locations for the local max/min. So the gradient is zero at these points, but doesn't necessarily include all points were the gradient is zero. Is there something I'm missing here?

Answer 1

Since we are dealing here with a symmetric $n\times n$ real matrix $A$, we can use the spectral theorem. So, we know that there are $n$ eigenvectors $v_1,\ldots,v_n$ of $A$ such that $(v_1,\ldots,v_n)$ is an orthogonal basis of $\mathbb{R}^n$. Therefore, after a change of variables, the fuction that we are dealing with here is just$$(x_1,x_2,\ldots,x_n)\mapsto\lambda_1{x_1}^2+\lambda_2{x_2}^2+\cdots+\lambda_n{x_n}^2$$ restricted to the unit sphere, where, for each $j\in\{1,2,\ldots,n\}$, $A.v_j=\lambda_jv_j$. The gradient of this map is$$(x_1,x_2,\ldots,x_n)\mapsto2(\lambda_1x_1,\lambda_2x_2,\ldots,\lambda_nx_n)$$and, within the unit sphere, this is $0$ if and only if $(x_1,x_2,\ldots,x_n)$ is an eigenvector (of $A$) with eigenvalue $0$.

So, going back to the original problem, if $E_0=\{v\in\mathbb{R}^n\,|\,A.v=0\}$, then the set of critical points of your function is $E_0\cap S^{n-1}$.

Added note: Please take into account the remark provided by skyking below.

Answer 2

Actually the restriction of a function to the unit sphere does not have directional derivates perpendicular to the sphere surface. This means that the relevant gradient is the gradient of the extended function projected onto the tangent plane of the surface that is $\nabla A - \langle \nabla A, \hat n\rangle$:

$$2Ax - \langle 2Ax, x\rangle x = 0$$

or

$$(A-\langle Ax, x\rangle I)x = 0$$

which only have non-trivial solutions when $\langle Ax, x\rangle$ is an eigenvale of $A$ with the solutions being eigenvectors with that eigenvalue.

Answer 3

You forgot the constraint that $v$ should be on the unit sphere. The Lagrangian is $L(v,\lambda)=\langle Av,v\rangle -\lambda(\langle v,v\rangle -1)$. The gradient wrt $v$ is $2Av-2\lambda v$ and it is 0 when $v$ is an eigenvector to A with eigenvalue $\lambda$. The constraint then gives that eigenvector should be normalized to unit length. The function value is then the eigenvalue. Thus the max is the largest eigenvalue and is obtained in the eigenspace restricted to the unit sphere, in case eigenspace is onedimensional it will just be two antipodal points. Similarly for min and smallest eigenvalue. Any intermediate eigenvalue will be saddle point because moving along a great circle in the plane spanned by the eigenvector with the intermediate eigenvalue and an eigenvector with a larger/smaller eigenvalue will increase/decrease the function value towards the larger/smaller eigenvalue.