On the equality of the Schwarz inequality 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.

 A: 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.
