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Thread: Can combination coefficients converge when the vector converges?

  1. #1
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    Can combination coefficients converge when the vector converges?

    In the vector space spanned by $\displaystyle {\bf{e}}_1,{\bf{e}}_2,...,{\bf{e}}_n$ which are linearly independent (not necessarily the standard basis), $\displaystyle {\bf{y}}_1=\sum_{i=1}^na_i{\bf{e}}_i$ and $\displaystyle {\bf{y}}_2=\sum_{i=1}^nb_i{\bf{e}}_i$. My question is: if the distance of $\displaystyle {\bf{y}}_1$ and $\displaystyle {\bf{y}}_2$ can be arbitrarily small, can the distance of their combination coefficients $\displaystyle a_i$ and $\displaystyle b_i$ be arbitrarily small too? Assume that we are discussing in the Euclidean space R$\displaystyle ^n$. Thanks.
    Last edited by zzzhhh; Aug 2nd 2009 at 06:13 AM.
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  2. #2
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    Yes, it is. First we prove that if $\displaystyle \|\sum_{i = 1}^na_i{\bf{e}}_i\|\rightarrow\bf{0}$, where $\displaystyle {\bf{e}}_i$'s are linearly independent, then each $\displaystyle |a_i|\rightarrow0$ too. Let $\displaystyle {\bf{e}}_1=(e_{11},e_{12},...e_{1n}), {\bf{e}}_2=(e_{21},e_{22},...e_{2n}),...,{\bf{e}}_ n=(e_{n1},e_{n2},...e_{nn})$. Let $\displaystyle {\bf{x}}=(x_1,x_2,...,x_n)=\sum_{i = 1}^na_i{\bf{e}}_i$, then $\displaystyle x_j=\sum_{i = 1}^na_ie_{ij}$. We can write this relation in matrix: $\displaystyle \left(\begin{array}{*{20}{c}}
    e_{11}&e_{21}&...&e_{n1} \\
    e_{12}&e_{22}&...&e_{n2} \\
    ...&...&...&...\\
    e_{1n}&e_{2n}&...&e_{nn} \\
    \end{array}\right)\left(\begin{array}{cc}a_1\\a_2\ \...\\a_n\end{array}\right)$$\displaystyle ={\bf{E}}\left(\begin{array}{cc}a_1\\a_2\\...\\a_n \end{array}\right)=\left(\begin{array}{cc}x_1\\x_2 \\...\\x_n\end{array}\right)$. Since $\displaystyle {\bf{e}}_i$ is linearly independent, so $\displaystyle n \times n$ matrix $\displaystyle \bf{E}$ is non-singular, then we have $\displaystyle \left(\begin{array}{cc}a_1\\a_2\\...\\a_n\end{arra y}\right)={\bf{E}}^{-1}\left(\begin{array}{cc}x_1\\x_2\\...\\x_n\end{ar ray}\right)$, that is, each $\displaystyle a_i$ can be expressed as a linear combination of $\displaystyle x_i$ with coefficients from $\displaystyle {\bf{E}}^{-1}$. Because $\displaystyle x_i\rightarrow0$, so is $\displaystyle a_i$ and $\displaystyle |a_i|$.
    Now if the distance of $\displaystyle {\bf{y}}_1$ and $\displaystyle {\bf{y}}_2\rightarrow0$, i.e. $\displaystyle \|{\bf{y}}_1-{\bf{y}}_2\|=\|\sum_{i = 1}^n(a_i-b_i){\bf{e}}_i\|\rightarrow0$, then we have $\displaystyle |a_i-b_i|\rightarrow0$ for each $\displaystyle i$ according to the preceding paragraph, this just means that the distance of $\displaystyle a_i$ and $\displaystyle b_i$ is arbitrarily small.
    If we are discussing in $\displaystyle {\bf{R}}^N$ where $\displaystyle n<N$, then the linear independence of $\displaystyle {\bf{e}}_i$ implies that $\displaystyle \bf{E}$ has determinantal rank $\displaystyle n$, so we can shrink $\displaystyle \bf{E}$ to be the $\displaystyle n$-rowed minor whose det$\displaystyle \ne$0 and the rest discussion remains unchanged.
    Generally, I don't like to handle the low-level elements, so if there is some other better proof, please follow, thanks.
    Last edited by zzzhhh; Aug 2nd 2009 at 08:12 AM.
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