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).
Best Answer
\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 $\vec u\cdot\vec v$.