$T_1$ and $T_1$ are any real, square, lower-triangular, toeplitz matrices of dimension $p>1$.
Let $\left[ \begin{array} [c]{cc}% T_1 & T_2 \end{array} \right] _{\mathcal{B}}$ denote the $p\times p$ sub-matrix of $\left[ \begin{array} [c]{cc}% T_1 & T_2 \end{array} \right] $ made up of the columns of $\left[ \begin{array} [c]{cc}% T_1 & T_2 \end{array} \right] $ determined by $\mathcal{B} :=\left[ i_{1}% ,i_{2},\ldots,i_{p}\right] $. The vector complementary to $\mathcal{B},$ denoted $\mathcal{B}^{\ast},$ is the vector whose components are the $p$ integers from $1,2,\ldots,2p$ not in $\mathcal{B}.$
Similarly, let $\left[ \begin{array} [c]{c}% T_2\\ -T_1 \end{array} \right] _{\mathcal{B}}$ denote the $p\times p$ sub-matrix made up of the rows of $\left[ \begin{array} [c]{c}% T_2\\ -T_1 \end{array} \right] $ determined by $\mathcal{B}.$
Define
$M_1=\left[ \begin{array} [c]{cc}% T_1 & T_2 \end{array} \right] _{\mathcal{B}}$, $M_2=\left[ \begin{array} [c]{cc}% T_1 & T_2 \end{array} \right] _{\mathcal{B}^{\ast}}$, $M_3= \left[ \begin{array} [c]{c}% T_2\\ -T_1 \end{array} \right] _{\mathcal{B}}$ and $M_4= \left[ \begin{array} [c]{c}% T_2\\ -T_1 \end{array} \right] _{\mathcal{B}^{\ast}}$.
I conjecture that the following matrix identity holds:
$M_1M_3+M_2M_4 \equiv 0_{p \times p}$.
Is this result known ?
I have just realised that it is very easy to show this. It follows from two elementary observations:
For any two matrices $A$ and $B$ of compatible size, the product $AB$ remains the same if the columns of A and the rows of B are permuted in the same way, and
Lower-triangular toeplitz matrices commute.
In the question statement $\left[ \begin{array} [c]{cc}% M_1 & M_2 \end{array} \right]$ is obtained from $\left[ \begin{array} [c]{cc}% T_1 & T_2 \end{array} \right]$ by a permutation of columns. The same permutation applied to the rows of $\left[ \begin{array} [c]{c}% T_2\\ -T_1 \end{array} \right]$ gives $\left[ \begin{array} [c]{c}% M_3\\ M_4 \end{array} \right]$.
Hence $M_1M_3+M_2M_4 = T_1T_2-T_2T_1$ by 1). But $T_1T_2-T_2T_1=0$ by 2). So $M_1M_3+M_2M_4=0$ which was the conjecture statement.