The motivation for this question arises from the following: Is it possible, given two Quantum Mechanical observables $A$ and $B$ with associated operators $\hat{\mathbf{A}}$ and $\hat{\mathbf{B}}$ such that $[\hat{\mathbf{A}},\hat{\mathbf{B}}] \neq \mathbf{0}$, for $$[\hat{\mathbf{A}},\hat{\mathbf{B}}]\left|\phi\right\rangle=0 \qquad \exists \left|\phi\right\rangle \neq 0$$
We can reduce this to the following: $\exists \left|\phi\right\rangle \neq 0$ such that this is true, if and only iff $\left| \phi \right\rangle \in \operatorname{Ker}([\hat{\mathbf{A}},\hat{\mathbf{B}}])$. So my question is this:
Given two matrices $\mathbf{A}$ and $\mathbf{B}$ such that $\operatorname{Ker}(\mathbf{A}) = \{0\}$ and $\operatorname{Ker}(\mathbf{B}) = \{0\}$, is it possible for: $$\operatorname{Ker}([\mathbf{A},\mathbf{B}])\neq \{0\}$$
The kernel of a linear operator can never be $\varnothing$. The smallest possible kernel is $\{0\}$.
But assuming that is what you meant: Yes, that is easily possible. For example, $$ \mathbf A=\begin{pmatrix}1&0&0\\0&0&1\\0&1&0\end{pmatrix} \qquad \mathbf B=\begin{pmatrix}1&0&0\\0&0&-1\\0&1&0\end{pmatrix} $$ gives $$ [\mathbf A,\mathbf B]=\begin{pmatrix}0&0&0\\0&2&0\\0&0&-2\end{pmatrix} $$ which is nonzero but singular.