# Can I say independent components of ICA are also linear independent?

In other words, does the independent components form a vector basis?

EDIT:Let'me try to clarify the question

If I calculate $n$ independent components for a given set of $n$ dimensional observations, are those components also linearly independent (property required to form a vector basis)?

• You should be able to reconstruct the original data out of the data projected onto the components. So yes, unless you have some seriously degenerate data, in which case I don't know what the algorithm will spit out. – Stumpy Joe Pete Nov 20 '13 at 2:46
• May we presume that "ICA" refers to independent components analysis and not one of a hundred other possible meanings of this acronym? But where you mention a "vector basis," to which space are you referring? – whuber Nov 20 '13 at 2:59
• Yes, by ICA I refer to Independent Component Analysis. Let'me clarify the question. – RafaelLopes Nov 20 '13 at 4:14
• Suppose, to see what can happen, there is a single actual signal that is being picked up by $n$ observers. If these observers are perfect, you can solve the problem using just one component. If you calculate $n\gt 1$ components, what solution does your algorithm give you? For another useful thought experiment, what happens when this set of observations has less than full rank (i.e., does not span a space of $n$ dimensions)? – whuber Nov 20 '13 at 13:56