Let $u$ and $v$ be non-zero vectors in a real inner product space $X.$ Show that $\| u+v \|= \| u\|+\|v\|$ if and only if $v=au,$ for some $a>0.$
I only know the proof for $\|u+v\| ≤ \|u\|+\|v\|.$ How can the equality be established?
Let $u$ and $v$ be non-zero vectors in a real inner product space $X.$ Show that $\| u+v \|= \| u\|+\|v\|$ if and only if $v=au,$ for some $a>0.$
I only know the proof for $\|u+v\| ≤ \|u\|+\|v\|.$ How can the equality be established?
Let $r=\|u\|$ and $s=\|v\|$. The hypothesis that $\|u+v\|=\|u\|+\|v\|$ means that $(r+s)^2=\|u+v\|^2=r^2+2\langle u,v\rangle+s^2$, that is, that $\langle u,v\rangle=rs$. Hence, $ \|su-rv\|^2=s^2\|v\|^2-2rs\langle u,v\rangle+r^2\|u\|^2=s^2r^2-2rs(rs)+r^2s^2=0. $ Thus $su=rv$, hence $u$ and $v$ are colinear and pointing to the same direction.
The other implication is obvious.
If $v=au$, then $\left\| u+v \right\|=\left\| u+au \right\|=\left\| (a+1)u \right\|=|a+1|\left\| u \right\|=(a+1)\left\| u \right\|$ and, on the other hand, $\left\| u \right\|+\left\| v \right\|=\left\| u \right\|+\left\| au \right\|=\left\| u \right\|+|a|\left\| u \right\|=(a+1)\left\| u \right\|$.