When does Schwarz inequality become an equality?

Solution 1:

$|\langle \mathbf{x},\mathbf{y}\rangle|=\|\mathbf{x}\|\,\|\mathbf{y}\| \iff \mathbf{x}$ and $\mathbf{y}$ are linearly dependent, i.e. if $\mathbf{x}=\lambda\mathbf{y}$ (or, trivially, if either $\mathbf{x}$ or $\mathbf{y}$ is zero; handle this first to get it out of the way).

($\implies$): Follow the normal Schwarz inequality argument, but equality will force $\mathbf{x}={\langle \mathbf{x},\mathbf{y}\rangle\over \langle\mathbf{y},\mathbf{y}\rangle}\mathbf{y}$

($\Longleftarrow$): Substitute $\mathbf{x}=\lambda\mathbf{y}$.

Solution 2:

Since $$ \left(x_1^2+x_2^2\right)\left(y_1^2+y_2^2\right)-(x_1y_1+x_2y_2)^2=(x_1y_2-x_2y_1)^2 $$ we get equality if and only if $x_1y_2=x_2y_1$.