# Are there any numerical advantages in solving symmetric matrix compared to matrices without symmetry?

I'm applying finite-difference method to a system of 3 coupled equations. Two of the equations are not coupled, however the third equation couples to both the other two. I noticed that by changing the order of equations, say from $(x, y, z)$ to $(x, z, y)$ that the coefficient matrix becomes symmetric.

Is there any advantage to doing this? For example, in terms of stability or efficiency/speed of solution. The matrices are highly sparse, if that is important, the non-zero terms are along central diagonals.

• Yes, it takes much less effort to solve a symmetric system than an unsymmetric one. If, in addition, you can show that your coefficient matrix is positive-definite, then you are in a good place. May 9 '13 at 11:04

A typical thing that many people will think of are Krylov subspace methods for the solution of equation systems $Ax=b$: If your problem is symmetric, you know that you don't need methods for nonsymmetric problem like GMRES, and can reside to something less memory-intensive like MINRES, or -- if your matrix is also positive-definite -- CG. The convergence behavior of Krylov methods is not influenced by permutations though, so you could even use symmetric methods for your unpermuted system.
Another example is the factorization of your matrix $A=LU$ in to an lower-triangular part $L$ and an upper triangular part $U$. If $A$ is symmetric, then $A=LL^T$, and you only have to store one factor (Cholesky decomposition).