Q: A basic question on adjoint operator Suppose we have a non-zero vector $v$ for which $Tv=0$. Then can we say that there exist a non-zero vector $w$ such that $T^*w=0$ where $T^*$ is adjoint operator.

A: If your inner product space is finite dimensional then yes. The idea is that if $Tv=0$ then
$\langle Tv, w\rangle=\langle v, T^*w\rangle=0$ for any $w$. So $T$ send the whole space into $v^\perp$, the orthogonal complement of $v$, which is a proper subspace. Thus $T^*$ does not have full rank so by rank-nullity it has non-trivial kernel.
However, if the space if infinite dimensional this is not longer true. Consider the left shift operator on $l^2(\mathbb{N})$. Given by
$T(a_0, a_1, a_2, ...)=(a_1, a_2, ...)$. Then $T(1, 0, 0, ...)=0$ but $T^*(a_0, a_1, ...)=
(0, a_0, a_1, ...)$ which is injective.  
