Results 1 to 2 of 2

Math Help - Can combination coefficients converge when the vector converges?

  1. #1
    Member
    Joined
    Jul 2009
    Posts
    120

    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.
    Last edited by zzzhhh; August 2nd 2009 at 06:13 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jul 2009
    Posts
    120
    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}}<br />
e_{11}&e_{21}&...&e_{n1} \\<br />
e_{12}&e_{22}&...&e_{n2} \\<br />
...&...&...&...\\<br />
e_{1n}&e_{2n}&...&e_{nn} \\<br />
\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<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 \ne0 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; August 2nd 2009 at 08:12 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] {a_n} and {a_n+b_n} converge. Prove {b_n} converges.
    Posted in the Differential Geometry Forum
    Replies: 8
    Last Post: June 24th 2011, 02:02 PM
  2. [SOLVED] Linear combination of a vector S
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 27th 2010, 09:47 AM
  3. Why coefficients in affine combination should add up to 1
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: March 5th 2010, 01:52 AM
  4. Writing a vector as a linear combination.
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 23rd 2009, 11:22 AM
  5. Replies: 3
    Last Post: May 4th 2009, 05:53 AM

Search Tags


/mathhelpforum @mathhelpforum