complex numbers- proving the equality part in the Cauchy–Schwarz inequality using Lagrange identity I need to discuss the equality case of:
$$ \left | \sum_{k=1}^{n} z_{k}w_{k} \right |^{2} \leq \left (  \sum_{k=1}^{n}\left | z_{k} \right |^{2}\right )\left (  \sum_{k=1}^{n}\left | w_{k} \right |^{2}\right )
$$
when we know:
$$ \left | \sum_{k=1}^{n} z_{k}w_{k} \right |^{2} = \left (  \sum_{k=1}^{n}\left | z_{k} \right |^{2}\right )\left (  \sum_{k=1}^{n}\left | w_{k} \right |^{2}\right ) - \sum_{1\leq k<  j\leq n} \left | z_{k}w_{j} - z_{j}w_{k} \right |^2
$$
$$
\forall z_{1},...,z_{n}, w_{1},...,w_{n}
$$
 A: If you're just trying to prove the equality in the Cauchy Schwarz equality then we have:
$$ \left | \sum_{k=1}^{n} z_{k}w_{k} \right |^{2} = \left (  \sum_{k=1}^{n}\left | z_{k} \right |^{2}\right )\left (  \sum_{k=1}^{n}\left | w_{k} \right |^{2}\right ) - \sum_{1\leq k<  j\leq n} \left | z_{k}w_{j} - z_{j}w_{k} \right |
$$
so as consequence:
$$\begin{align} \left | \sum_{k=1}^{n} z_{k}w_{k} \right |^{2} = \left (  \sum_{k=1}^{n}\left | z_{k} \right |^{2}\right )\left (  \sum_{k=1}^{n}\left | w_{k} \right |^{2}\right ) & \text{ iff } \sum_{1\leq k<  j\leq n} \left | z_{k}w_{j} - z_{j}w_{k} \right |=0\\
& \text{ iff }\forall (i,j)\in [1,n]^2 z_{k}w_{j} - z_{j}w_{k}=0\\
& \text{ iff } (z_1,\cdots,z_n)\text{ and } (w_1,\cdots,w_n) \text{ are linearly dependent } \end{align}$$
A: In order that equality holds we need that for every $k,j$
$$ z_k w_j = z_j w_k\tag{1} $$
holds, or:
$$ \frac{z_k}{w_k}=\frac{z_j}{w_j}.\tag{2} $$
That leads to:
$$\forall k\in[1,n],\qquad \lambda = \frac{z_k}{w_k}\tag{3}$$
hence:
$$ (z_1,\ldots,z_n) = \lambda (w_1,\ldots,w_n).\tag{4}$$
