# Why does sign ambiguity occur in ICA?

I do not really understand the source of sign ambiguity in ICA. First, my understanding that If I apply ICA on a signal $$X$$ and I got 3 ICs which are represented by a set $$IC^1$$. Then, applying ICA on the same signal, I will get 3 ICs but with different orders and signs, let us call them $$IC^2$$. So the correlation of the two sets might look like this: $$\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & -1 \\ 0 & -1 & 0 \\ \end{bmatrix}$$

Where the permutation is due to the different orders and the negative sign is because of the sign ambiguity. My question is why does that (the sign ambiguity) happen?

We can see the ambiguity of sign as a special case of the ambiguity of scale, as explained in the Independent Component Analysis book of Hyvärinen et al. (Section 7.2.3). If we use their notation, $$\mathbf x=(x_1, x_2,\dots, x_K)$$ is our $$K$$-dimensional observed signal and $$\mathbf s=(s_1, s_2, \dots, s_K)$$ is the unknown source signal$${}^\dagger$$, and they are mixed by a (non-random) and unknown matrix $$\mathbf A$$ $$\mathbf{x=As}.$$ Another way to express the same relationship is to write $$\mathbf x$$ as the sum of the columns of the matrix $$\mathbf A$$, namely the vectors $$\mathbf a_i$$: $$\mathbf x = \sum_i \mathbf a_i s_i.$$ But the latter can be re-written as $$\mathbf{x}=\sum_i\left(\frac{1}{\alpha_i}\mathbf a_i\right)(s_i\alpha_i),$$ where the coefficients $$\alpha_i$$ are any (non-zero) scalars. Here we can see why sign ambiguity occurs: we could put $$\alpha_i=-1$$, and the resulting $$\mathbf x$$ wouldn't change, thus the model is not capable of distinguishing whether the source signals are $$s_i$$ or $$-s_i$$.
$${}^\dagger$$For simplicity, we are assuming that the matrix $$\mathbf A$$ is square (as is in the question), hence the sizes of $$\mathbf x$$ and $$\mathbf s$$ are the same.