# Prove Cauchy-Schwarz equality.

My professor asked me to prove the equality in Cauchy-Schwarz inequality. The equality holds iff the vectors $v$ and $u$ are linearly dependent.

I am able to show the equality using the fact $v$ and $u$ are linearly dependent.

But I don't know how to show the converse (i.e Showing the linear dependence using the equality).

• A good start would be to look at your proof of this inequality. Given that, we can show you where to tinker a little to obtain the linear dependence from the equality. – TZakrevskiy Mar 11 '15 at 17:52

\begin{align} \|\vec u+t\vec v\|^2 & = t^2 \|\vec v\|^2 + 2t\vec u\cdot\vec v + \|\vec u\|^2 \\[10pt] & = at^2+bt+c. \end{align} This quadratic polynomial with $a>0$ is positive for every $t$ if the discriminant $b^2-4ac$ is negative, and is positive for all except one value of $t$ if the discriminant is $0$. Express the discriminant in terms of $\|\vec u\|$, $\|\vec v\|$, and $\vec u\cdot\vec v$.

• That's a nice solution. I wonder however why the discriminant has to be equal to zero? Why does the quadratic polynomial must have only one zero? – Hendrra Jun 3 '18 at 20:06
• @Hendrra : The point is that it has no zeros except in the case where there is some value of $t$ for which $\vec u + t\vec v$ is $\vec0,$ i.e. except where $\vec u$ is a scalar multiple of $\vec v. \qquad$ – Michael Hardy Jun 3 '18 at 23:08

Hint: Let $z = pr_u (v) = \frac{\langle u,v\rangle}{\|u\|^2} u$ then we may write $v = z + w$, with $w \perp z$. By Pythagoras Theorem $$\|v\|^2 = \|z\|^2 + \|w\|^2$$ Then $$\|v\| \geq \|z\|$$

And $\|v\| = \|z\| \implies\|w\| = 0$.

• Notice my edit and the different appearance of the notation afterwards. That is standard usage. – Michael Hardy Mar 11 '15 at 18:28
• @MichaelHardy Thanks! I didn't know that! – Aaron Maroja Mar 11 '15 at 18:29

The CS inequality is: $|u\cdot v|^2 \leq|u|^2 |v|^2$ .

Observe that : $\left(|u| - t|v|\right)^2 \geq 0, \forall t \in \mathbb{R}$ expanding this and treat it as a quadratic function in variable $t$. This is true for all $t$ implies $\triangle \leq 0 \to$ CS inequality follows, with equality when $|u| = t|v| \to u = \pm tv \to$ linear dependent.

Suppose that $$|u\cdot v|=\|u\|\,\|v\|\tag{1}$$ This means that $u\cdot v=\pm\|u\|\,\|v\|$. Then \begin{align} \|u-tv\|^2 &=(u-tv)\cdot(u-tv)\\ &=\|u\|^2\pm2t\|u\|\,\|v\|+t^2\|v\|\\ &=(\|u\|\pm t\|v\|)^2\tag{2} \end{align} If $\|v\|=0$, then $u$ and $v$ are dependent (vacuously).

If $\|v\|\ne0$, then, by division, there is a $t$ so that $\|u\|\pm t\|v\|=0$. Using that $t$, $(2)$ says that $\|u-tv\|=0$, that is $$u=tv\tag{3}$$ in which case $u$ and $v$ are dependent.