In Einstein's "The Meaning of Relativity" https://en.wikisource.org/wiki/The_Meaning_of_Relativity/Lecture_1 by putting $λ = 1$, (2b) and (3a) should furnish the conditions:
$ \sum \limits _{\nu }b_{\nu \alpha }b_{\nu \beta }=\delta _{\alpha \beta }$
but If I do myself the passages by replacing (3a) in (2b) I get:
$\sum_v (\sum_\alpha b_{\nu \alpha } \Delta x_{\alpha})^2 = \sum_\nu \Delta x_{\nu }^{2}$
so that the equality condition should be:
$b_{\nu\alpha} = \delta_{\nu\alpha}$
What am I doing wrong?
Plugging $\Delta x_\nu' = \sum_\alpha b_{\nu\alpha}\Delta x_\alpha$ into $\sum_\nu \Delta x_\nu'^2 = \sum_\nu \Delta x_\nu^2$ gives $$\sum_\nu \left(\sum_\alpha b_{\nu\alpha}\Delta x_\alpha\right)^2 = \sum_\nu \Delta x_\nu^2$$ Note that the sum on $\alpha$ belongs inside the square. Now expand the square on the left side.
Added:
In general to expand something like this, you do:
$$(\sum_\alpha c_{\alpha})^2 = \sum_{\alpha,\beta} c_{\alpha}c_{\beta}$$
See if you can get it from there.