- #1

DeadOriginal

- 274

- 1

## Homework Statement

Let ##v_{1},...,v_{m}## be an orthonormal set of vectors in ##V##. Let ##v\not\in S(v_{1},...,v_{m})##. Show that the vector ##v'=v-\sum\limits_{i=1}^{n}(v,v_{i})v_{i}## given by the Gram-Schmidt process has the shortest length among all vectors of the form ##v-x## for ##x\in Sv_{1},...,v_{n})##.

## The Attempt at a Solution

Since the problem wants me to deal with the length of vectors I thought that I would probably have to use something along the lines of

$$

||v'||\leq ||v-x||\Leftrightarrow \left|\left|v-\sum\limits_{i=1}^{n}(v,v_{i})v_{i}\right|\right|\leq \left|\left|v-\sum\limits_{i=1}^{n}\lambda_{i}v_{i}\right|\right|

$$

Since ##x\in S(v_{1},...,v_{n})## ##x## must be a linear combination of ##v_{1},...,v_{n}## so ##x=\sum\limits_{i=1}^{n}\lambda_{i}v_{i}##.

I then tried to show that

$$

0\leq \left|\left|v-\sum\limits_{i=1}^{n}\lambda_{i}v_{i}\right|\right|-\left|\left|v-\sum\limits_{i=1}^{n}(v,v_{i})v_{i}\right|\right|.

$$

To get rid of the square roots I squared both sides to get

$$

0\leq \left|\left|v-\sum\limits_{i=1}^{n}\lambda_{i}v_{i}\right|\right|^{2}-2\left(\left|\left|v-\sum\limits_{i=1}^{n}\lambda_{i}v_{i}\right|\right|\right) \left(\left|\left|v-\sum\limits_{i=1}^{n}(v,v_{i})v_{i}\right|\right|\right)+\left|\left|v-\sum\limits_{i=1}^{n}(v,v_{i})v_{i}\right|\right|^{2}

$$

and this is equivalent to

$$

0\leq (v-\sum\limits_{i=1}^{n}\lambda_{i}v_{i},v-\sum\limits_{i=1}^{n}\lambda_{i}v_{i})-2 \left(\left|\left|v-\sum\limits_{i=1}^{n}\lambda_{i}v_{i}\right|\right|\right)

\left(\left|\left|v-\sum\limits_{i=1}^{n}(v,v_{i})v_{i}\right|\right|\right)+(v-\sum\limits_{i=1}^{n}(v,v_{i})v_{i},v-\sum\limits_{i=1}^{n}(v,v_{i})v_{i}).

$$

I am not sure how to proceed from here.