Angle of \ddots in an uneven matrix

I have the following triangle matrix with zeroes on the diagonal

\documentclass[a4paper]{memoir}
\usepackage{amsmath}
\begin{document}
$M = \begin{pmatrix} 0 & a_{1,2} & a_{1,3} & \dots & a_{1,n} \\ & 0 & a_{2,3} & \dots & a_{2,n} \\ & & 0 & \ddots & \vdots \\ & & & & 0 \end{pmatrix}$
\end{document}

The small problem I have now, is that the dots from \ddots point from a_{2,3} to the last zero. This could lead to the impression that the diagonal is not given by the zeroes. Anyone an idea?

\documentclass[a4paper]{memoir}
\usepackage{amsmath}
\begin{document}
$M = \begin{pmatrix} 0 & a_{1,2} & a_{1,3} & \dots & a_{1,n} \\ & 0 & a_{2,3} & \dots & a_{2,n} \\ & & 0 & \ddots & \vdots \\ & & & 0 & a_{n-1,n} \\ & & & & 0 \end{pmatrix}$
\end{document}

BTW: please post small compilable docs instead of sniplets. The less others have to add the more likely it is to get help.

• Thanks for the advise I added a MWE to my post. Your solution is fine, thanks (sometimes it is so easy that one didn't see the obvious ;)) – user2653422 Jul 4 '14 at 13:30

With nicematrix:

\documentclass[a4paper]{memoir}
\usepackage{amsmath}
\usepackage{nicematrix}
\begin{document}
$\renewcommand{\arraystretch}{1.3} M = \begin{pNiceArray}{ccwc{4mm}wc{7mm}wc{7mm}}[xdots/shorten=1mm] 0 & a_{1,2} & a_{1,3} & \Cdots & a_{1,n} \\ & 0 & a_{2,3} & \Cdots & a_{2,n} \\ & & \Ddots & \Ddots & \Vdots \\ & & & 0 & a_{n-1,n} \\ & & & & 0 \end{pNiceArray}$
\end{document}