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Let's say that I have a submanifold cut out of $\mathbb{R}^{n+k}$ as $f^{-1}(0)$ where $f:\mathbb{R}^{n+k} \rightarrow \mathbb{R}^k$ is smooth and $0$ is a regular value. The Lagrange multiplier criterion tells me that if $x\in f^{-1}(0)$ is a critical point of $g|_{f^{-1}(0)}$ where $g:\mathbb{R}^{n+k} \rightarrow \mathbb{R}$, then there is a linear function $\lambda:\mathbb{R}^k \rightarrow \mathbb{R}$ such that $DG(x) + \lambda DF(x) = 0$. One can use this to probe for critical points of $g$ by first considering solutions of the system of equations $DG(x) + \lambda DF(x) = 0$ and $f(x) = 0$ for varying $\lambda$.

Can you adapt this in a useful way to the scenario where $g$ is a vector valued function? i.e. $g:\mathbb{R}^{n+k} \rightarrow \mathbb{R}^{l}$ and $\lambda:\mathbb{R}^k \rightarrow \mathbb{R}^{l}$

If not, then what is a good approach to computing the sets of critical points of such vector valued functions.

AnonymousCoward
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1 Answers1

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This can be adapted to vector-valued functions in a certain sense. First, recall the definitions of a critical point:

  • A critical point for a scalar-valued function $\phi\colon\mathbb{R^n} \to \mathbb{R}$ is a point for which $D\phi = 0$.

  • A critical point for a vector-valued function $\Phi\colon\mathbb{R^n} \to \mathbb{R}^l$ is a point for which $\mathrm{rank}(D\Phi) < \min(n,l)$.

With this in mind, we can state the Lagrange multipliers method in both cases:

  • A critical point for a scalar-valued function $g\colon \mathbb{R}^{n+k}\to\mathbb{R}$ restricted to the submanifold $F^{-1}(0)$ is a point for which $Dg + \lambda\,DF = 0$ for some linear function $\lambda$ (i.e. $Dg$ is a linear combination of the rows of $DF$).

  • A critical point for a vector-valued function $G\colon\mathbb{R}^{n+k}\to\mathbb{R}^l$ restricted to the submanifold $F^{-1}(0)$ is a point for which $\mathrm{rank}(\frac{DG}{DF}) < \min(n+k,l+k)$, where $\frac{DG}{DF}$ denotes the $(l+k)\times (n+k)$ augmented matrix whose first $l$ rows are $DG$ and whose remaining $k$ rows are $DF$.

You can use any method you want to check the rank of $\frac{DG}{DF}$. For example, in the case where $n \geq l$ you could try to find a nontrivial solution to the equations $$ \mu\,DG + \lambda\,DF = 0\qquad\text{and}\qquad F=0 $$ where $\mu\colon \mathbb{R}^l\to\mathbb{R}$ and $\lambda\colon \mathbb{R}^k\to\mathbb{R}$ are linear, and "nontrivial" means that $\mu$ is not identically zero. (This is equivalent to checking whether the rows of $\frac{DG}{DF}$ are linearly independent.)

If $n \leq l$, you should check the columns instead. This involves finding nontrivial solutions to the equations $$ DG\,\lambda = 0,\qquad DF\,\lambda = 0\qquad\text{and}\qquad F=0 $$ where $\lambda\colon \mathbb{R}\to\mathbb{R}^{n+k}$ is linear, and "nontrivial" means that $\lambda$ is not identically zero.

Jim Belk
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