# Factorize laplacian in terms of first derivative matrix

I am trying to factorize the following Laplacian matrix in terms of $$D^TD$$, D is the first derivative matrix. The tridiagonal form of the secon derivative matrix using Neumann boundary condition is given by,

$$\begin{bmatrix} -1 & 1 & 0 & 0 & 0 \\ 1 & -2 & 1 & 0 &0\\ 0 & 1 & -2 &1&0\\0&0&1&-2&1\\0&0&0&1&-1\end{bmatrix}$$.

In order to write the above in terms of the gradient( first derivative) operator, I tried factorizing the Laplacian using Cholesky factorization. But, I couldn't succeed in factorizing.

Any suggestions on how to find a decomposition of the form $$D^TD$$?

• naïve Cholesky won't work, because your Laplacian matrix is not positive definite. there are many square roots (=choices of $D$) to select from. to recover the "upwind" discretization of the first order derivative operator, you'll want to use the LDL decomposition. see here for a more thorough discussion. – GoHokies Nov 14 '18 at 20:34
• Isn't it just D=[-1 +1 0 0 0; 0 -1 +1 0 0; 0 0 -1 +1 0; 0 0 0 -1 +1]? I think you can also extend to higher dimensions, using the (signed) edge-to-vertex adjacency matrix. (Essentially a discrete version of the gradient operator). – rchilton1980 Nov 14 '18 at 20:49
• @rchilton1980 yes, and that's exactly what the LDL decomposition produces. – GoHokies Nov 14 '18 at 21:17
• But the Laplace matrix is not just $D^T D$. That's because the Laplacian operator is the divergence of the gradient, so the two operators are transposes of each other. But the Laplace matrix is not of this form. – Wolfgang Bangerth Nov 15 '18 at 4:26
• Even if the inverse does not exist, take a pseudo-inverse. It will also not be sparse. – Wolfgang Bangerth Nov 15 '18 at 14:17