Show that one gets equality in the Schwarz inequality if and only if $\overrightarrow v$,$\overrightarrow w$ are linearly dependent.

(I am supposing they want me to prove it in an inner product space we call V)

The $\Leftarrow$ part of the proof was pretty straightforward but I don't know how to go about the $\Rightarrow$ part because linear dependence requires us to show $\overrightarrow v$=$\lambda \overrightarrow w$ for some $\lambda \in \mathbb{R}$ but the Cauchy-Schwarz inequality only involves norms\lengths of vectors and not relations between the vectors themselves. I would appreciate tips on starting points.

  • $\begingroup$ @MiloBrandt, yes my bad, edited. $\endgroup$
    – Omrane
    Commented Apr 9, 2016 at 20:09
  • $\begingroup$ How did you proof of the Cauchy-Schwarz inequality go? $\endgroup$ Commented Apr 9, 2016 at 20:16

2 Answers 2


Assume $\overrightarrow v\neq0$(if $\overrightarrow v=0$ the proof is trivial), and take any norm of $V$. Fix an arbitrary $\lambda$ in the scalar field ($\mathbb{R}$in this case) and note that $$0\leq \|\overrightarrow u-\lambda \overrightarrow v \|^2=\langle \overrightarrow u-\lambda \overrightarrow v ,\overrightarrow u-\lambda \overrightarrow v \rangle$$ Is the first step in deriving the C.S inequality. Thus equality holds iff $$\|\overrightarrow u-\lambda \overrightarrow v \|=0$$ From our properties of norms we know then that $$\overrightarrow u-\lambda \overrightarrow v=0 $$ and we have shown linear dependence.

  • $\begingroup$ It'd be more symmetrical (and closer to the definition of linear dependence) to take $0\leq \|\alpha u - \beta v\|^2$ and proceed. Then you don't need to consider the case where one is zero - it'd be built in to the proof naturally. $\endgroup$ Commented Apr 9, 2016 at 20:18
  • $\begingroup$ This is the proof I'm familiar with, but I'm sure your proof would have been along similar lines and you can adjust your inequality to equality accordingly. $\endgroup$
    – K.Power
    Commented Apr 9, 2016 at 20:19
  • $\begingroup$ @MiloBrandt I'd have to look at that. In my proof I end up dividing by $\|v\|$ so that's why I have two cases. $\endgroup$
    – K.Power
    Commented Apr 9, 2016 at 20:21
  • $\begingroup$ @K.Power, Thanks for your answer. I don't understand how you use C.S going from first line of operations to second. $|<\overrightarrow u,\overrightarrow v>| \le ||\overrightarrow u|| . ||\overrightarrow v||$ is the equation I have, we need to turn that into an equality as a supposition right? $\endgroup$
    – Omrane
    Commented Apr 9, 2016 at 20:33
  • $\begingroup$ @Sadem my first line of operations is how I derive C.S in the first place. It's the starting point of my proof for C.S. What proof did you use to derive C.S? $\endgroup$
    – K.Power
    Commented Apr 9, 2016 at 20:37

From the proof of C-S. With $0\ne x=(x_1,..,x_n)$ and $0\ne y=(y_1,..,y_n).$ $$\|x\|^2\cdot \|y\|^2-(x\cdot y)^2=\sum_{1\leq i<j\leq n}(x_i y_j-x_j y_i)^2.$$ Now $x_i y_j=x_j y_i$ whenever $1\leq i<j\leq n$ iff there exists $k\ne 0$ such that $x_i=k y_i$ for $i=1,..,n.$ Use induction on $n$ to prove this.

The case $x=0\lor y=0$ is trivial.


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