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I present here a detailed proof of the equality of the Schwarz inequality given a condition that two vectors are linearly dependent.

Let $p$ be an inner product of an inner product space $V$. Then $|p(\alpha,\beta)|=\| \alpha \| \|\beta\| $ if and only if $\{\alpha,\beta\}$ is linearly dependent.

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Proof:

$(\Longleftarrow):$ Suppose that $\{\alpha,\beta\}$ is linearly dependent. Without loss of generality, assume that $\beta \neq 0$. (Since if $\beta=0$, the equation $|p(\alpha,\beta)|=\| \alpha \| \|\beta\| $ is trivially satisfied,i.e,$|p(\alpha,0)|=0=\| \alpha \| \|0\| $ ) Then $\alpha =c\beta$ for some scalar $c$. Thus \begin{align} |p(\alpha,\beta)|&=|p(c\beta,\beta)| \\ &= |c p(\beta,\beta)| \\ &=|c||p(\beta,\beta)| \\ &=|c|\|\beta\|^2 \\ &= (|c|\|\beta\|)\|\beta\| \\ &=||c\beta\|\|\beta\| \\ &=\|\alpha\|\|\beta\|. \end{align}

$(\implies$): Let $|p(\alpha,\beta)|=\| \alpha \| \|\beta\| $. If $\beta=0$, then $\{\alpha,\beta\}$ is linearly dependent. Suppose that $\beta \neq 0$.
Let $$\gamma=\alpha - \frac{p(\alpha,\beta)}{p(\beta,\beta)}\cdot \beta$$ Then $\gamma$ is orthogonal to $\beta$ (This is easy to verify). Thus, \begin{align} |p(\alpha,\beta)|^2&=|p(\gamma+\frac{p(\alpha,\beta)}{p(\beta,\beta)}\cdot \beta,\beta)|^2 \\ &=\|\gamma+\frac{p(\alpha,\beta)}{p(\beta,\beta)}\cdot \beta\|^2\|\beta\|^2 \\ &=\|\gamma\|^2+\left|\frac{p(\alpha,\beta)}{p(\beta,\beta)}\right|^2\cdot \|\beta\|^2\|\beta\|^2 ,\mbox{by Pythagorean Theorem}\\ &=\|\gamma\|^2+\|\alpha\|^2\|\beta\|^2 \end{align} This implies that $\|\gamma\|^2=0$ so that $\gamma=0$. Hence $$\alpha = \frac{p(\alpha,\beta)}{p(\beta,\beta)}\cdot \beta$$ Therefore, $\{\alpha,\beta\}$ is linearly dependent.

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    $\begingroup$ Minor nit: In ($\Longleftarrow$) you also assume (without loss of generality) that $\beta \neq 0$ when you say that $\alpha = c \beta$ for some scalar $c$. // In the first part of your proof $c$ is a scalar and in the second part $c$ is a vector, maybe use $\gamma$ instead of $c$ in the second part? $\endgroup$ – Martin Mar 14 '13 at 17:40
  • $\begingroup$ I get your point Martin. I'll make an edit. Thanks much. $\endgroup$ – Philip Benj Mar 14 '13 at 17:46
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    $\begingroup$ Thanks for the edits, nice job, +1! If I were in an overly pedantic mood, I'd ask you to add a short explanation why you don't lose generality by making the assumption $\beta \neq 0$ in the first part of the argument; And maybe you could say that you use that $\gamma$ is perpendicular to $\beta$ in the third displayed equation of your argument (Pythagorean theorem). $\endgroup$ – Martin Mar 14 '13 at 17:52
  • $\begingroup$ Thank you very much Martin, I am in an overly pedantic mood so I reflected all your suggestions. LOL, thanks... $\endgroup$ – Philip Benj Mar 14 '13 at 18:05
  • $\begingroup$ :) You're welcome. I missed a typo: it should be $\lVert \gamma \rVert^2 \lVert \beta \rVert^2$ instead of just $\lVert \gamma \rVert^2$ twice. $\endgroup$ – Martin Mar 15 '13 at 7:54

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