Prob 15, Chap. 1 in Baby Rudin: If this condition also sufficient for equality? Here's Prob. 15, Chap. 1 in the book Principles of Mathematical Analysis by Wlater Rudin, 3rd edition:

Under what conditions does equality hold in the Schwarz inequality?

Now the Schwarz inequality, which is Theorem 1.35 in Rudin, is as follows:

If $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ are complex numbers, then $$ \left\vert \sum_{j=1}^n a_j \overline{b_j} \right\vert^2 \leq \sum_{j=1}^n \left\vert a_j \right\vert^2 \sum_{j=1}^n \left\vert b_j \right\vert^2.$$
If there is a complex number $z$ such that $a_j = z b_j$ for each $j=1, \ldots, n$, then we have
$$
\begin{align}
\left\vert \sum_{j=1}^n a_j \overline{b_j} \right\vert^2 &= \left\vert \sum_{j=1}^n z b_j \overline{b_j} \right\vert^2 \\ 
&= \left\vert z \ \sum_{j=1}^n \left\vert b_j \right\vert^2 \right\vert^2 \\
&= \vert z \vert^2 \cdot \left\vert  \sum_{j=1}^n \left\vert b_j \right\vert^2 \right\vert^2 \\
&= \vert z \vert^2 \cdot \left( \sum_{j=1}^n \left\vert b_j \right\vert^2 \right)^2,  
\end{align}
$$
whereas
$$
\begin{align}
\sum_{j=1}^n \left\vert a_j \right\vert^2 \sum_{j=1}^n \left\vert b_j \right\vert^2 &= \sum_{j=1}^n \left\vert z b_j \right\vert^2 \sum_{j=1}^n \left\vert b_j \right\vert^2 \\ 
&=  \sum_{j=1}^n  \vert z \vert^2 \left\vert  b_j \right\vert^2 \sum_{j=1}^n \left\vert b_j \right\vert^2 \\
&= \vert z \vert^2 \cdot \left( \sum_{j=1}^n \left\vert b_j \right\vert^2 \right)^2 \\ 
&= \left\vert \sum_{j=1}^n a_j \overline{b_j} \right\vert^2.
\end{align}
$$

Now is this condition also a necessary condition for the equality to hold in the Schwarz "inequality"?
 A: The Schwarz inequality follows from expanding
$$ \begin{aligned}
0 &\le \sum_{i=1}^n \sum_{j=1}^n \left\vert a_i b_j - a_j b_i \right\vert^2 \\
 &= \sum_{i=1}^n \sum_{j=1}^n (a_i b_j - a_j b_i)\overline{(a_i b_j - a_j b_i)} \\
 &= \quad ... \\
 &= 2 \sum_{j=1}^n \left\vert a_j \right\vert^2 \sum_{j=1}^n \left\vert b_j \right\vert^2
 - 2 \left\vert \sum_{j=1}^n a_j \overline{b_j} \right\vert^2 
\end{aligned}
$$
and therefore equality holds if and only if
$$ \tag{*}
 a_i b_j - a_j b_i = 0 \quad
 \text{ for all } i, j = 1, \dots n \, .
$$
If at least one $b_i$ is not zero then you can define $z = a_i/b_i$
and conclude from $(*)$ that
$a_j = z b_j$ for each $j=1, \ldots, n$.
Another way to express $(*)$ is that one of the vectors
$(a_1, \ldots, a_n)$ and $(b_1, \ldots, b_n)$ is a constant multiple
of the other, or that they are linearly dependent over $\Bbb C$.
A: Yes: define $\lVert x\rVert:=\sum_{j=1}^n|x_j|^2$; call $a=(a_1,\dots,a_n)$ and $b=(b_1,\dots,b_n)$ (that we assume both non-zero). Then $\left\lVert \lVert a\rVert b-\lVert b\rVert a\right\rVert^2$ equals $0$ if and only if we have equality in Schwarz inequality.
