Why does this automation accept the empty string?

Problem Detail: 

I am new to Automata Theory. I am studying the book thoroughly and understanding it well. I have got a confusion regarding NFA. Why does this automaton accept the empty string λ?

What I think is that: implicitly δ(q,λ)=q for all states q. But here also explicitly defined δ(q0,λ)=q2. As there is a walk to the final state using δ(q0,λ)=q0, so δ(q0,λ)=q2 will not be followed. Am I correct?

Asked By : tanmoy

Answered By : jmite

An automaton accepts $\lambda$, the empty string, if and only if the starting state is a final state, or there is some final state in the empty-string-closure of the starting state (in the non-deterministic case).

You don't need to walk from the start state, since it is already final. The fact that you have the option to doesn't mean you are obligated to.

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Question Source : http://cs.stackexchange.com/questions/20143