# Relating a lower triangular Toeplitz matrix to a diagonal matrix

Suppose I had a lower triangular Toeplitz matrix:

T=[x100…0x2x10…0⋮⋮⋮⋱⋮xnxn−1xn−2…x1]
T=
\begin{bmatrix}
x_{1} & 0 & 0 & \dots & 0 \\
x_{2} & x_{1} & 0 & \dots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
x_{n} & x_{n-1} & x_{n-2} & \dots & x_{1}
\end{bmatrix}

and a diagonal matrix:
D=[x100…00×20…0⋮⋮⋮⋱⋮000…xn]
D=
\begin{bmatrix}
x_{1} & 0 & 0 & \dots & 0 \\
0 & x_{2} & 0 & \dots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \dots & x_{n}
\end{bmatrix}

where the diagonal elements are the elements of TT.

Is it possible to relate TT and DD in either of the following forms:

D=ATBD=ATB

where AA and BB are square matrices that do not have xx’s as entries? If not, is there another way to relate them using matrix equations?

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I suspect that it is not possible to write D=ATB or T=ADB since the determinant of D is x1*x2*…xn, and the determinant of T is x1*x1*…x1 which implies the determinant of AB is either (x1*x2*…xn)/(x1^n) or (x1^n)/(x1*x2*…xn). This suggests that A and/or B must have x’s as entries.
– InquisitivePerson
Oct 20 at 23:28

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