Can TDT^T transform a square matrix D into diag{D_0,0} where D_0 is nonsingular?

Question

If $D\in R^{n\times n} $ is a square matrix with rank $m$ ($m < n$), can we always find a nonsingular matrix $T\in R^{n\times n}$ such that $$ TDT^T = \left[\begin{matrix}D_0 & 0\\ 0& 0\end{matrix}\right], $$ where $D_0 \in R^{m\times m}$ is nonsingular? If such $T$ does not always exist, what conditions does $D$ need to satisfy in order to let such $T$ exist? (Note that here $T^T$ denotes the transpose of the matrix $T$.)

Answer 1

The answer is No in general. Consider for example the matrix $$\pmatrix{0&1\\0&0}.$$ The key is given by the notion of Jordan canonical forms.