independant proof

**transgalactic** · July 26th 2011, 02:50 AM

there is $\{y_{1}..y_{n}\}$ a group of vectors in $R^{n}$ ,there is $\{u_{1}..u_{n}\}$ orthonormal basis of $R^{n}$ .
the first part i have proved that for $\lambda_{1}..\lambda_{n}\$ in R
$\parallel\sum_{i=1}^{n}\lambda_{i}y_{i}\parallel^{ 2}\leq(\sum_{i=1}^{n}|\lambda_{i}|^{2})(\sum_{i=1} ^{n}\parallel y_{i}\parallel^{2})$
the second part is my question
prove that if $\{y_{1}..y_{n}\}$ is a set for which $\sum_{i=1}^{n}\parallel y_{i}\parallel^{2}<1$ then $\{y_{1}+u_{1,}..,y_{n}+u_{n}\}$ is independant?
how i tried to solve it:
a set is independant if for $k_{1}(y_{1}+u_{1})+..+k_{n}(y_{n}+u_{n})=0$ the only way it to be zero is if k1=..=kn=0
i opened it and got $k_{1}y_{1}+..+k_{n}y_{n}=k_{1}u_{1}+..+k_{n}u_{n}$
so on both side we have equal vectors so is their norm
$(\sum_{i=1}^{n}|k_{i}|^{2})(\sum_{i=1}^{n}\paralle l y_{i}\parallel^{2})=(\sum_{i=1}^{n}|k_{i}|^{2})(\s um_{i=1}^{n}\parallel u_{i}\parallel^{2})<br />$ what next how to prove that $k_{1}=..=k_{n}=0$
??

**FernandoRevilla** · July 26th 2011, 04:23 AM

We have $\sum_{i=1}^{n}k_iy_i=-\sum_{i=1}^{n}k_iu_i$ . Taking norms we obtain $||{\sum_{i=1}^n{k_iy_i}}||^2=\sum_{i=1}^n|k_i|^2$ . But $\sum_{i=1}^n|k_i|^2\leq (\sum_{i=1}^n|k_i|^2)(\sum_{i=1}^n||y_i||^2)$ . As $\sum_{i=1}^n||y_i||^2\right$ < $1$ , necessarily $\sum_{i=1}^n|k_i|^2=0$ which implies $k_i=0$ for all $i=1,\ldots,n$ .

**transgalactic** · July 26th 2011, 06:09 AM

Originally Posted by FernandoRevilla

We have $\sum_{i=1}^{n}k_iy_i=-\sum_{i=1}^{n}k_iu_i$ . Taking norms we obtain $||{\sum_{i=1}^n{k_iy_i}}||^2=\sum_{i=1}^n|k_i|^2$ . But $\sum_{i=1}^n|k_i|^2\leq (\sum_{i=1}^n|k_i|^2)(\sum_{i=1}^n||y_i||^2)$ . As $\sum_{i=1}^n||y_i||^2\right$ < $1$ , necessarily $\sum_{i=1}^n|k_i|^2=0$ which implies $k_i=0$ for all $i=1,\ldots,n$ .

why if orthonormal then $||{\sum_{i=1}^n{k_iy_i}}||^2=\sum_{i=1}^n|k_i|^2$
?

norm of one vector is 1
but here we have a sum

**FernandoRevilla** · July 26th 2011, 08:05 AM

Originally Posted by transgalactic

why if orthonormal then $||{\sum_{i=1}^n{k_iy_i}}||^2=\sum_{i=1}^n|k_i|^2$ norm of one vector is 1 but here we have a sum

If $\{u_i:i=1,\ldots,n\}$ is an orthonormal system then, $||{\sum_{i=1}^n{k_iu_i}}||^2 =<\sum_{i=1}^n{k_iu_i},\sum_{i=1}^n{k_iu_i}>=...$

Conclude.

**transgalactic** · August 9th 2011, 02:36 PM

Originally Posted by FernandoRevilla

If $\{u_i:i=1,\ldots,n\}$ is an orthonormal system then, $||{\sum_{i=1}^n{k_iu_i}}||^2 =<\sum_{i=1}^n{k_iu_i},\sum_{i=1}^n{k_iu_i}>=...$

Conclude.

$<\sum_{i=1}^n{u_i},\sum_{i=1}^n{u_i}>$
again we have here a sum of vectors not a single vector,
the norm of the single vector is 1

why the resolt is 1 too?

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