# 5: Proof of Theorem 1

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For notational convenience, let $$r_{\ell}=\ell$$, $$1\le \ell\le m$$, so Equation \ref{eq:6} becomes

$\label{eq:30} \left|\begin{array}{ccccccc} \displaystyle{\frac{\partial{g_{1}({\bf X}_{0})}}{\partial{x_{1}}}}& \displaystyle{\frac{\partial{g_{1}({\bf X}_{0})}}{\partial{x_{2}}}}& \cdots& \displaystyle{\frac{\partial{g_{1}({\bf X}_{0})}}{\partial{x_{m}}}}\\ \\ \displaystyle{\frac{\partial{g_{2}({\bf X}_{0})}}{\partial{x_{1}}}}& \displaystyle{\frac{\partial{g_{2}({\bf X}_{0})}}{\partial{x_{2}}}}& \cdots& \displaystyle{\frac{\partial{g_{2}({\bf X}_{0})}}{\partial{x_{m}}}}\\ \vdots&\vdots&\ddots&\vdots\\ \displaystyle{\frac{\partial{g_{m}({\bf X}_{0})}}{\partial{x_{1}}}}& \displaystyle{\frac{\partial{g_{m}({\bf X}_{0})}}{\partial{x_{2}}}}& \cdots& \displaystyle{\frac{\partial{g_{m}({\bf X}_{0})}}{\partial{x_{m}}}}\\ \\ \end{array}\right|\ne0$

Denote

${\bf U}=(x_{m+1},x_{m+2},\dots x_{n})\text{\; and\;\;} {\bf U}_{0}=(x_{m+1,0},x_{m+2,0},\dots x_{n0}).$

From Equation \ref{eq:30}, the Implicit Function Theorem implies that there are unique continuously differentiable functions $$h_{\ell}=h_{\ell}({\bf U})$$, $$1\le \ell\le m$$, defined on a neighborhood $$N$$ of $${\bf U}_{0}$$, such that

$(h_{1}({\bf U}),h_{2}({\bf U}),\dots, h_{m}({\bf U}), {\bf U})\in D, \text{\; for all\;\;} {\bf U}\in N, \nonumber$

$\label{eq:31} (h_{1}({\bf U_{0}}),h_{2}({\bf U_{0}}),\dots, h_{m}({\bf U_{0}}),{\bf U}_{0})={\bf X}_{0},$

and

$\label{eq:32} g_{\ell}(h_{1}({\bf U}),h_{2}({\bf U}),\dots, h_{m}({\bf U}),{\bf U})=0,\quad {\bf U}\in N, \quad 1\le \ell\le m.$

Again from Equation \ref{eq:30}, the system

$\label{eq:33} \left[\begin{array}{ccccccc} \displaystyle{\frac{\partial{g_{1}({\bf X}_{0})}}{\partial{x_{1}}}}& \displaystyle{\frac{\partial{g_{1}({\bf X}_{0})}}{\partial{x_{2}}}}& \cdots& \displaystyle{\frac{\partial{g_{1}({\bf X}_{0})}}{\partial{x_{m}}}}\\ \\ \displaystyle{\frac{\partial{g_{2}({\bf X}_{0})}}{\partial{x_{1}}}}& \displaystyle{\frac{\partial{g_{2}({\bf X}_{0})}}{\partial{x_{2}}}}& \cdots& \displaystyle{\frac{\partial{g_{2}({\bf X}_{0})}}{\partial{x_{m}}}}\\ \vdots&\vdots&\ddots&\vdots\\ \displaystyle{\frac{\partial{g_{m}({\bf X}_{0})}}{\partial{x_{1}}}}& \displaystyle{\frac{\partial{g_{m}({\bf X}_{0})}}{\partial{x_{2}}}}& \cdots& \displaystyle{\frac{\partial{g_{m}({\bf X}_{0})}}{\partial{x_{m}}}}\\ \\ \end{array}\right] \left[\begin{array}{ccccccc} \lambda_{1}\\\lambda_{2}\\ \vdots\\\lambda_{m} \end{array}\right]= \left[\begin{array}{ccccccc} f_{x_{1}}({\bf X}_{0})\\f_{x_{2}}({\bf X}_{0})\\\vdots\\ f_{x_{m}}({\bf X}_{0}) \end{array}\right]$

has a unique solution. This implies that

$\label{eq:34} \frac{\partial{f({\bf X}_{0})}}{\partial x_{i}} -\lambda_{1}\frac{\partial{g_{1}({\bf X}_{0})}}{\partial x_{i}} -\lambda_{2}\frac{\partial{g_{2}({\bf X}_{0})}}{\partial x_{i}}-\cdots -\lambda_{m}\frac{\partial{g_{m}({\bf X}_{0})}}{\partial x_{i}}=0$

for $$1\le i\le m$$.

If $$m+1\le i\le n$$, differentiating Equation \ref{eq:32} with respect to $$x_{i}$$ and recalling Equation \ref{eq:31} yields

$\frac{\partial g_{\ell}({\bf X}_{0})} {\partial x_{i}} +\sum_{j=1}^{m} \frac{\partial g_{\ell}({\bf X}_{0})}{\partial x_{j}} \frac{\partial h_{j}({\bf X}_{0})}{\partial x_{i}}=0, \quad 1\le \ell\le m.$

If $${\bf X}_{0}$$ is local extreme point $$f$$ subject to $$g_{1}({\bf X})=g_{2}({\bf X})= \cdots =g_{m}({\bf X})=0$$, then $${\bf U}_{0}$$ is an unconstrained local extreme point of $$f(h_{1}({\bf U}),h_{2}({\bf U}), \dots h_{m}({\bf U}),{\bf U})$$; therefore,

$\frac{\partial f({\bf X}_{0})} {\partial x_{i}} +\sum_{j=1}^{m} \frac{\partial f({\bf X}_{0})}{\partial x_{j}} \frac{\partial h_{j}({\bf X}_{0})}{\partial x_{i}}=0. \nonumber$

The last two equations imply that

$\left|\begin{array}{ccccccc} \displaystyle{\frac{\partial{f({\bf X}_{0})}}{\partial{x_{i}}}}& \displaystyle{\frac{\partial{f({\bf X}_{0})}}{\partial{x_{1}}}}& \displaystyle{\frac{\partial{f({\bf X}_{0})}}{\partial{x_{2}}}}& \cdots& \displaystyle{\frac{\partial{f({\bf X}_{0})}}{\partial{x_{m}}}}\\ \\ \displaystyle{\frac{\partial{g_{1}({\bf X}_{0})}}{\partial{x_{i}}}}& \displaystyle{\frac{\partial{g_{1}({\bf X}_{0})}}{\partial{x_{1}}}}& \displaystyle{\frac{\partial{g_{1}({\bf X}_{0})}}{\partial{x_{2}}}}& \cdots& \displaystyle{\frac{\partial{g_{1}({\bf X}_{0})}}{\partial{x_{m}}}}\\ \\ \displaystyle{\frac{\partial{g_{2}({\bf X}_{0})}}{\partial{x_{i}}}}& \displaystyle{\frac{\partial{g_{2}({\bf X}_{0})}}{\partial{x_{1}}}}& \displaystyle{\frac{\partial{g_{2}({\bf X}_{0})}}{\partial{x_{2}}}}& \cdots& \displaystyle{\frac{\partial{g_{2}({\bf X}_{0})}}{\partial{x_{m}}}}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ \displaystyle{\frac{\partial{g_{m}({\bf X}_{0})}}{\partial{x_{i}}}}& \displaystyle{\frac{\partial{g_{m}({\bf X}_{0})}}{\partial{x_{1}}}}& \displaystyle{\frac{\partial{g_{m}({\bf X}_{0})}}{\partial{x_{2}}}}& \cdots& \displaystyle{\frac{\partial{g_{m}({\bf X}_{0})}}{\partial{x_{m}}}}\\ \\ \end{array}\right|=0, \nonumber$

so

$\left|\begin{array}{ccccccc} \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_{i}}}& \displaystyle{\frac{\partial g_{1}({\bf X}_{0})}{\partial x_{i}}}& \displaystyle{\frac{\partial g_{2}({\bf X}_{0})}{\partial x_{i}}}&\dots& \displaystyle{\frac{\partial g_{m}({\bf X}_{0})}{\partial x_{i}}}\\\\ \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_{1}}}& \displaystyle{\frac{\partial g_{1}({\bf X}_{0})}{\partial x_{1}}}& \displaystyle{\frac{\partial g_{2}({\bf X}_{0})}{\partial x_{1}}}&\dots& \displaystyle{\frac{\partial g_{m}({\bf X}_{0})}{\partial x_{1}}}\\\\ \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_{2}}}& \displaystyle{\frac{\partial g_{1}({\bf X}_{0})}{\partial x_{2}}}& \displaystyle{\frac{\partial g_{2}({\bf X}_{0})}{\partial x_{2}}}&\dots& \displaystyle{\frac{\partial g_{m}({\bf X}_{0})}{\partial x_{2}}}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_{m}}}& \displaystyle{\frac{\partial g_{1}({\bf X}_{0})}{\partial x_{m}}}& \displaystyle{\frac{\partial g_{2}({\bf X}_{0})}{\partial x_{m}}}&\dots& \displaystyle{\frac{\partial g_{m}({\bf X}_{0})}{\partial x_{m}}} \end{array}\right|=0. \nonumber$

Therefore, there are constant $$c_{0}$$, $$c_{1}$$, …$$c_{m}$$, not all zero, such that

$\label{eq:35} \left[\begin{array}{ccccccc} \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_{i}}}& \displaystyle{\frac{\partial g_{1}({\bf X}_{0})}{\partial x_{i}}}& \displaystyle{\frac{\partial g_{2}({\bf X}_{0})}{\partial x_{i}}}&\dots& \displaystyle{\frac{\partial g_{m}({\bf X}_{0})}{\partial x_{i}}}\\\\ \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_{1}}}& \displaystyle{\frac{\partial g_{1}({\bf X}_{0})}{\partial x_{1}}}& \displaystyle{\frac{\partial g_{2}({\bf X}_{0})}{\partial x_{1}}}&\dots& \displaystyle{\frac{\partial g_{m}({\bf X}_{0})}{\partial x_{1}}}\\\\ \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_{2}}}& \displaystyle{\frac{\partial g_{1}({\bf X}_{0})}{\partial x_{2}}}& \displaystyle{\frac{\partial g_{2}({\bf X}_{0})}{\partial x_{2}}}&\dots& \displaystyle{\frac{\partial g_{m}({\bf X}_{0})}{\partial x_{2}}}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_{m}}}& \displaystyle{\frac{\partial g_{1}({\bf X}_{0})}{\partial x_{m}}}& \displaystyle{\frac{\partial g_{2}({\bf X}_{0})}{\partial x_{m}}}&\dots& \displaystyle{\frac{\partial g_{m}({\bf X}_{0})}{\partial x_{m}}}\\ \end{array}\right] \left[\begin{array}{ccccccc} c_{0}\\c_{1}\\c_{3}\\\vdots\\c_{m} \end{array}\right]= \left[\begin{array}{ccccccc} 0\\0\\0\\\vdots\\0 \end{array}\right].$

If $$c_{0}=0$$, then

$\left[\begin{array}{ccccccc} \displaystyle{\frac{\partial{g_{1}({\bf X}_{0})}}{\partial{x_{1}}}}& \displaystyle{\frac{\partial{g_{1}({\bf X}_{0})}}{\partial{x_{2}}}}& \cdots& \displaystyle{\frac{\partial{g_{1}({\bf X}_{0})}}{\partial{x_{m}}}}\\ \\ \displaystyle{\frac{\partial{g_{2}({\bf X}_{0})}}{\partial{x_{1}}}}& \displaystyle{\frac{\partial{g_{2}({\bf X}_{0})}}{\partial{x_{2}}}}& \cdots& \displaystyle{\frac{\partial{g_{2}({\bf X}_{0})}}{\partial{x_{m}}}}\\ \vdots&\vdots&\ddots&\vdots\\ \displaystyle{\frac{\partial{g_{m}({\bf X}_{0})}}{\partial{x_{1}}}}& \displaystyle{\frac{\partial{g_{m}({\bf X}_{0})}}{\partial{x_{2}}}}& \cdots& \displaystyle{\frac{\partial{g_{m}({\bf X}_{0})}}{\partial{x_{m}}}}\\ \\ \end{array}\right] \left[\begin{array}{ccccccc} c_{1}\\c_{2}\\\vdots\\c_{m} \end{array}\right]= \left[\begin{array}{ccccccc} 0\\0\\\vdots\\0 \end{array}\right]$

and Equation \ref{eq:30} implies that $$c_{1}=c_{2}=\cdots = c_{m}=0$$; hence, we may assume that $$c_{0}=1$$ in a nontrivial solution of Equation \ref{eq:35}. Therefore,

$\label{eq:36} \left[\begin{array}{ccccccc} \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_{i}}}& \displaystyle{\frac{\partial g_{1}({\bf X}_{0})}{\partial x_{i}}}& \displaystyle{\frac{\partial g_{2}({\bf X}_{0})}{\partial x_{i}}}&\dots& \displaystyle{\frac{\partial g_{m}({\bf X}_{0})}{\partial x_{i}}}\\\\ \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_{1}}}& \displaystyle{\frac{\partial g_{1}({\bf X}_{0})}{\partial x_{1}}}& \displaystyle{\frac{\partial g_{2}({\bf X}_{0})}{\partial x_{1}}}&\dots& \displaystyle{\frac{\partial g_{m}({\bf X}_{0})}{\partial x_{1}}}\\\\ \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_{2}}}& \displaystyle{\frac{\partial g_{1}({\bf X}_{0})}{\partial x_{2}}}& \displaystyle{\frac{\partial g_{2}({\bf X}_{0})}{\partial x_{2}}}&\dots& \displaystyle{\frac{\partial g_{m}({\bf X}_{0})}{\partial x_{2}}}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_{m}}}& \displaystyle{\frac{\partial g_{1}({\bf X}_{0})}{\partial x_{m}}}& \displaystyle{\frac{\partial g_{2}({\bf X}_{0})}{\partial x_{m}}}&\dots& \displaystyle{\frac{\partial g_{m}({\bf X}_{0})}{\partial x_{m}}}\\ \end{array}\right] \left[\begin{array}{ccccccc} 1\\c_{1}\\c_{2}\\\vdots\\c_{m} \end{array}\right]= \left[\begin{array}{ccccccc} 0\\0\\0\\\vdots\\0 \end{array}\right],$

which implies that

$\left[\begin{array}{ccccccc} \displaystyle{\frac{\partial{g_{1}({\bf X}_{0})}}{\partial{x_{1}}}}& \displaystyle{\frac{\partial{g_{1}({\bf X}_{0})}}{\partial{x_{2}}}}& \cdots& \displaystyle{\frac{\partial{g_{1}({\bf X}_{0})}}{\partial{x_{m}}}}\\ \\ \displaystyle{\frac{\partial{g_{2}({\bf X}_{0})}}{\partial{x_{1}}}}& \displaystyle{\frac{\partial{g_{2}({\bf X}_{0})}}{\partial{x_{2}}}}& \cdots& \displaystyle{\frac{\partial{g_{2}({\bf X}_{0})}}{\partial{x_{m}}}}\\ \vdots&\vdots&\ddots&\vdots\\ \displaystyle{\frac{\partial{g_{m}({\bf X}_{0})}}{\partial{x_{1}}}}& \displaystyle{\frac{\partial{g_{m}({\bf X}_{0})}}{\partial{x_{2}}}}& \cdots& \displaystyle{\frac{\partial{g_{m}({\bf X}_{0})}}{\partial{x_{m}}}}\\ \\ \end{array}\right] \left[\begin{array}{ccccccc} -c_{1}\\-c_{2}\\ \vdots\\-c_{m} \end{array}\right]= \left[\begin{array}{ccccccc} f_{x_{1}}({\bf X}_{0})\\f_{x_{2}}({\bf X}_{0})\\\vdots\\ f_{x_{m}}({\bf X}_{0}) \end{array}\right] \nonumber$

Since Equation \ref{eq:33} has only one solution, this implies that $$c_{j}=-\lambda_{j}$$, $$1\le j\le n$$, so Equation \ref{eq:36} becomes

$\left[\begin{array}{ccccccc} \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_{i}}}& \displaystyle{\frac{\partial g_{1}({\bf X}_{0})}{\partial x_{i}}}& \displaystyle{\frac{\partial g_{2}({\bf X}_{0})}{\partial x_{i}}}&\dots& \displaystyle{\frac{\partial g_{m}({\bf X}_{0})}{\partial x_{i}}}\\\\ \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_{1}}}& \displaystyle{\frac{\partial g_{1}({\bf X}_{0})}{\partial x_{1}}}& \displaystyle{\frac{\partial g_{2}({\bf X}_{0})}{\partial x_{1}}}&\dots& \displaystyle{\frac{\partial g_{m}({\bf X}_{0})}{\partial x_{1}}}\\\\ \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_{2}}}& \displaystyle{\frac{\partial g_{1}({\bf X}_{0})}{\partial x_{2}}}& \displaystyle{\frac{\partial g_{2}({\bf X}_{0})}{\partial x_{2}}}&\dots& \displaystyle{\frac{\partial g_{m}({\bf X}_{0})}{\partial x_{2}}}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_{m}}}& \displaystyle{\frac{\partial g_{1}({\bf X}_{0})}{\partial x_{m}}}& \displaystyle{\frac{\partial g_{2}({\bf X}_{0})}{\partial x_{m}}}&\dots& \displaystyle{\frac{\partial g_{m}({\bf X}_{0})}{\partial x_{m}}}\\ \end{array}\right] \left[\begin{array}{ccccccc} 1\\-\lambda_{1}\\-\lambda_{2}\\\vdots\\-\lambda_{m} \end{array}\right]= \left[\begin{array}{ccccccc} 0\\0\\0\\\vdots\\0 \end{array}\right]. \nonumber$

Computing the topmost entry of the vector on the left yields yields Equation \ref{eq:34}, which completes the proof.

Example $$\PageIndex{1}$$

[example:9] Minimize $$\displaystyle{\sum_{i=1}^{n}x_{i}^{2}}$$ subject to

$\label{eq:37} \sum_{i=1}^{n}a_{r i}x_{i}=c_{r}, \quad 1\le r\le m,$

where

$\label{eq:38} \sum_{i=1}^{n}a_{ri}a_{si}= \begin{cases} 1 &\text{if } r=s,\\ 0 &\text{if }r\ne s. \end{cases}$

Solution

Let

$L =\frac{1}{2} \sum_{i=1}^{n}x_{i}^{2}-\sum_{s=1}^{m}\lambda_{s} \sum_{i=1}^{n}a_{s i}x_{i}. \nonumber$

Then

$L_{x_{i}}=x_{i}-\sum_{s=1}^{m}\lambda_{s}a_{si},\quad 1\le i\le n, \nonumber$

so

$\label{eq:39} x_{i0}=\sum_{s=1}^{m}\lambda_{s}a_{s i}\quad 1\le i\le n, \nonumber$

and

$a_{ri}x_{i0}=\sum_{s=1}^{m}\lambda_{s}a_{ri}a_{s i}. \nonumber$

Now Equation \ref{eq:38} implies that

$\sum_{i=1}^{n}a_{ri}x_{i0}=\sum_{s=1}^{m}\lambda_{s} \sum_{i=1}^{n}a_{ri}a_{s i}=\lambda_{r}. \nonumber$

From this and Equation \ref{eq:37}, $$\lambda_{r}=c_{r}$$, $$1\le r\le m$$, and Equation \ref{eq:39} implies that

$x_{i0}=\sum_{s=1}^{m}c_{s}a_{s i},\quad 1\le i\le n. \nonumber$

Therefore,

$x_{i0}^{2}=\sum_{r,s=1}^{m}c_{r}c_{s}a_{r i}a_{si},\quad 1\le i\le n, \nonumber$

and Equation \ref{eq:38} implies that

$\sum_{i=1}^{n}x_{i0}^{2}=\sum_{r,s=1}^{m}c_{r}c_{s} \sum_{i=1}^{n}a_{r i}a_{si}=\sum_{r=1}^{m}c_{r}^{2}. \nonumber$

The next theorem provides further information on the relationship between the eigenvalues of a symmetric matrix and constrained extrema of its quadratic form. It can be proved by successive applications of Theorem [theorem:1]; however, we omit the proof.

Theorem $$\PageIndex{1}$$

Suppose that $${\bf A}=[a_{rs}]_{r,s=1}^{n}\in {\mathbb R}^{n\times n}$$ is symmetric and let

$Q({\bf x})=\sum_{r,s=1}^{n}a_{rs}x_{r}x_{s}.$

Suppose also that

${\bf x}_{1}= \left[\begin{array}{ccccccc} x_{11}\\x_{21}\\\vdots\\x_{n1} \end{array}\right] \nonumber$

minimizes $$Q$$ subject to $$\sum_{i=1}^{n}x_{i}^{2}$$. For $$2\le r\le n$$, suppose that

${\bf x}_{r}= \left[\begin{array}{ccccccc} x_{1r}\\x_{2r}\\\vdots\\x_{nr} \end{array}\right], \nonumber$

minimizes $$Q$$ subject to

$\sum_{i=1}^{n}x_{i}^{2} =1 \text{\; and\;\;} \sum_{i=1}^{n}x_{is}x_{i}=0,\quad 1\le s\le r-1. \nonumber$

Denote

$\lambda_{r}=\sum_{i,j=1}^{n}a_{ij}x_{ir}x_{jr}, \quad 1\le r\le n. \nonumber$

Then

$\lambda_{1}\le \lambda_{2}\le \cdots\le \lambda_{n} \text{\; and\;\;} Ax_{r}=\lambda_{r}x_{r},\quad 1\le r\le n.$

This page titled 5: Proof of Theorem 1 is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.