# When is the concatenation of a language $L$ with $\Sigma^*$ regular?

I've been looking at questions about the regular concatenation of two languages; one question said that the concatenation of $$\{0^n1^n|n\geq 0\}$$ and $$\Sigma^*$$ was regular (over the alphabet $$\Sigma = \{0,1\}$$). Another provided a counterexample by the concatenation of $$\{a^{n^2}b|n\geq 0\}$$ with $$\Sigma^*$$, which is irregular (over the alphabet $$\Sigma = \{a,b\}$$). Could someone provide clarification as to why some concatenations with $$\Sigma^*$$ are regular while others aren't? Both cases concatenate an irregular language unrecognisable by a DFA to one that is, and I don't understand how this produces a language that is recognisable.

• If $\epsilon \in L$ then $L\Sigma^*=\Sigma^*$. – Yuval Filmus Oct 18 '18 at 15:39

In the first case, $$n$$ can be 0, and the concatenation will produce $$\Sigma^*$$. So any other value of $$n$$ is irrelevant; no other strings could be added. If the restriction were $$n>1$$, then the concatenation would not be regular.