Why not always make a linear system's matrix symmetric? This may be quite a naive question, but as I'm reading about different methods to solve linear systems of the type Ax=b with A a n x n matrix, I wonder why we should not always solve an equivalent system with a symmetric matrix obtained by multiplying the equation with the transpose of A, i.e. $$A^T A x = A^T b$$
(especially in the context of large, sparse matrices and matrix free methods where symmetric matrices have quite desirable properties)?
Edit: Lets consider that the problem is well posed so that the square matrix A is invertible and the solution x is unique.
 A: For a dense, square $A$, cost of forming $A^TA$ is of order $2n^3$, and cost of performing LU factorization is of order $2/3n^3$. Thus, solving $A^TAx = A^Tb$ instead of $Ax=b$ is more expensive.
For a sparse $A$, $A^TA$ is denser than $A$ and can even be a dense matrix. Thus, cost of performing the LU factorization of $A$ can be much lower than cost of performing the Cholesky factorization of $A^TA$.
If equation $A^TAx = A^Tb$ is solved using the Cholesky factorization, then this solution is computed in a backward stable way as long as $\|A\|_2\|b\|_2\approx \|A^Tb\|_2$ and the a priori normwise backward error is proportional to $\kappa_2(A) = \|A\|_2\|A^{-1}\|_2$. The normwise forward error is proportional to $\kappa_2(A)^2$. When $Ax=b$ is solved using the LU factorization, then the normwise backward error is proportional to $\eta = \frac{\||L||U|\|_2}{\|A\|_2}$ and the normwise forward error is proportional to $\kappa_2(A)\times\eta$. If $\eta$ is close to $1$ (which is usually true), then from the error analysis perspective one should solve $Ax = b$.
If equations $Ax = b$ and $A^TAx = A^Tb$ are solved using iterative methods with the same tolerance in terms of residual norm $\|A\hat{x}-b\|_2$, then from the error analysis perspective both methods are equivalent. If solving $A^TAx = A^Tb$ is faster, then this method should be chosen. It is hard to predict apriori, which method will be faster, since it strongly depends on solver and preconditioner used, required tolerance, and the matrix $A$.
