# Thread: Can combination coefficients converge when the vector converges?

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

In the vector space spanned by ${\bf{e}}_1,{\bf{e}}_2,...,{\bf{e}}_n$ which are linearly independent (not necessarily the standard basis), ${\bf{y}}_1=\sum_{i=1}^na_i{\bf{e}}_i$ and ${\bf{y}}_2=\sum_{i=1}^nb_i{\bf{e}}_i$. My question is: if the distance of ${\bf{y}}_1$ and ${\bf{y}}_2$ can be arbitrarily small, can the distance of their combination coefficients $a_i$ and $b_i$ be arbitrarily small too? Assume that we are discussing in the Euclidean space R $^n$. Thanks.

2. Yes, it is. First we prove that if $\|\sum_{i = 1}^na_i{\bf{e}}_i\|\rightarrow\bf{0}$, where ${\bf{e}}_i$'s are linearly independent, then each $|a_i|\rightarrow0$ too. Let ${\bf{e}}_1=(e_{11},e_{12},...e_{1n}), {\bf{e}}_2=(e_{21},e_{22},...e_{2n}),...,{\bf{e}}_ n=(e_{n1},e_{n2},...e_{nn})$. Let ${\bf{x}}=(x_1,x_2,...,x_n)=\sum_{i = 1}^na_i{\bf{e}}_i$, then $x_j=\sum_{i = 1}^na_ie_{ij}$. We can write this relation in matrix: $\left(\begin{array}{*{20}{c}}
e_{11}&e_{21}&...&e_{n1} \\
e_{12}&e_{22}&...&e_{n2} \\
...&...&...&...\\
e_{1n}&e_{2n}&...&e_{nn} \\
\end{array}\right)\left(\begin{array}{cc}a_1\\a_2\ \...\\a_n\end{array}\right)$
$={\bf{E}}\left(\begin{array}{cc}a_1\\a_2\\...\\a_n \end{array}\right)=\left(\begin{array}{cc}x_1\\x_2 \\...\\x_n\end{array}\right)$. Since ${\bf{e}}_i$ is linearly independent, so $n \times n$ matrix $\bf{E}$ is non-singular, then we have $\left(\begin{array}{cc}a_1\\a_2\\...\\a_n\end{arra y}\right)={\bf{E}}^{-1}\left(\begin{array}{cc}x_1\\x_2\\...\\x_n\end{ar ray}\right)$, that is, each $a_i$ can be expressed as a linear combination of $x_i$ with coefficients from ${\bf{E}}^{-1}$. Because $x_i\rightarrow0$, so is $a_i$ and $|a_i|$.
Now if the distance of ${\bf{y}}_1$ and ${\bf{y}}_2\rightarrow0$, i.e. $\|{\bf{y}}_1-{\bf{y}}_2\|=\|\sum_{i = 1}^n(a_i-b_i){\bf{e}}_i\|\rightarrow0$, then we have $|a_i-b_i|\rightarrow0$ for each $i$ according to the preceding paragraph, this just means that the distance of $a_i$ and $b_i$ is arbitrarily small.
If we are discussing in ${\bf{R}}^N$ where $n, then the linear independence of ${\bf{e}}_i$ implies that $\bf{E}$ has determinantal rank $n$, so we can shrink $\bf{E}$ to be the $n$-rowed minor whose det $\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.