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Mathematics LibreTexts

5: Proof of Theorem 1

  • Page ID
    17320
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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

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