1. ## LU-decomposition

The LU-decomposition of a matrix A can be used to solve the system of equations Ax=v for the vector x. But...why would you do this? Is it more computationally effective? Or is the reason more subtle than that?...

2. Suppose for example $Ax=v$ with $A$ invertible and you know $L$ and $U$. The system $Ly=v$ is triangular and let $y_0$ be its solution. The system $Ux=y_0$ is also triangular and let $x_0$ be its solution. Then, $Ax_0=L(Ux_0)=Ly_0=v$ . That is, we have solved $Ax=v$ by means of two triangular systems.

Fernando Revilla

3. Originally Posted by FernandoRevilla
Suppose for example $Ax=v$ with $A$ invertible and you know $L$ and $U$. The system $Ly=v$ is triangular and let $y_0$ be its solution. The system $Ux=y_0$ is also triangular and let $x_0$ be its solution. Then, $Ax_0=L(Ux_0)=Ly_0=v$ . That is, we have solved $Ax=v$ by means of two triangular systems.

Fernando Revilla
Yes, I know this. My question, however, is why you would bother to do it this way? Why not just triangularize your matrix and go from there?

4. The benefit is more when you have multiple systems of the form $Ax = b_{k},$ for multiple $b_{k}$'s. Then, once you've done the LU factorization once, solving each system doesn't require any more row reductions, just back substitutions, which are fast. If you were going to solve only one system, then for exact methods, you'd do Gaussian elimination with back substitution, and for iterative methods (such as with very large, sparse matrices), you'd do something fancy like Gauss-Seidel (but more advanced).