Is it possible to diagonalize a singular matrix?

A matrix is singular if and only if $0$ is one of its eigenvalues. A singular matrix can be either diagonalizable or not diagonalizable. For example, $$ \pmatrix{ 1&0\\0&0 } $$ Is diagonalizable (since it is diagonal), whereas $$ \pmatrix{ 0&1\\0&0 } $$ is not diagonalizable.

Yes, diagonalize the zero matrix.

THM Let $A$ be a matrix associated to a linear transformation $T:V\to V$, $V$ a $k$-vector space. Let $\lambda_1,\ldots,\lambda_l$ be it's distinct eigenvalues. Then $A$ is diagonalizable iff $$V=E(\lambda_1)\oplus\cdots \oplus E(\lambda_l)$$ where $E(\lambda_i)=\ker(A-\lambda_i1)$.

The sum is always direct, but it might not comprise all of $V$. This is equivalent to the easier claim that $V$ admits a basis of eigenvectors of $A$, also equivalent to the fact that the minimal polinomial splits over $k$ and has all simple roots. In turn, this is equivalent to the fact that $\chi_A$ splits and the multiplicity of $\lambda_i$ as a root equals the dimension of $E(\lambda_i)$. We always have $\dim E(\lambda_i)\leqslant {\rm mult}\;(\chi_A,\lambda_i)$, and this is in fact a good thing to use when checking for diagonalizability, in some cases. For example, $$\begin{pmatrix}1&1\\0&1\end{pmatrix}$$ is not diagonalizable, since $\chi=(X-1)^2$ but $\dim E(1)=1<2$.