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If $x$, $y$ in $R^n$ are such that $\lvert x+ty\rvert \geq \lvert x\rvert$ for all $t \in R$, then how do I show that $x\cdot y=0$?

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  • $\begingroup$ In $t(\|y\|^2 t - 2 x \cdot y) \ge 0$ let $t \to \pm 0$ (depending on the sign of $x \cdot y$). $\endgroup$
    – njguliyev
    Commented Sep 28, 2013 at 20:41

2 Answers 2

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$\newcommand{\abs}[1]{\left\vert #1\right\vert}$ $\large x\ \mbox{and}\ y\ \mbox{are}\ \underline{vectors}\ \mbox{in}\ {\mathbb R}^{n}$. $$ \abs{x + ty} \geq \abs{x} \quad\Longrightarrow\quad t\left(ty^{2} + 2x\bullet y\right) \geq 0 $$

If $y \not= 0$, we'll get $$ \left(\vphantom{\Huge A}\left(\vphantom{\Large A}~t \leq 0~\right) \quad\vee\quad \left( t \geq -\,{2x\bullet y \over y^{2}}\right)\right) \qquad\vee\qquad \left(\vphantom{\Huge A}\left(~t \leq -\,{2x\bullet y \over y^{2}} \right) \quad\vee\ \left(\vphantom{\Large A}~t \geq 0~\right)\right) $$

Then, $\abs{x + ty} \geq \abs{x}$ will not be true $\forall\ t \in {\mathbb R}$. Then, $y = 0\quad\Longrightarrow\quad \color{#ff0000}{\large x\bullet y = 0}$

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  • $\begingroup$ $x$ and $y$ are vectors, not numbers ... $\endgroup$ Commented Sep 29, 2013 at 8:57
  • $\begingroup$ @MichaelHoppe Sorry. I was misleading. I'll check tomorrow. Now, it's too late. $\endgroup$ Commented Sep 29, 2013 at 9:08
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If $y=0$ the claim is correct. Let's $y\neq0$ and $t=-\frac{\langle x,y\rangle}{\|y\|^2}$. Now verify that $$\|x+ty\|^2=\|x\|^2-\frac{\langle x,y\rangle^2}{\|y\|^2}.$$

Michael

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