Difference in surfaces described by equivalent quadratic forms It is fairly straightforward to prove that a quadratic form $Q(\mathbf{x})=\mathbf{x}^{T}A\mathbf{x}$ can equivalently be written $Q(\mathbf{x})=\mathbf{x}^{T}M\mathbf{x}$ for $M=\frac{1}{2}(A+A^{T})$. This would lead me to expect that the surface described by $Q(\mathbf{x})=c$ for some constant $c$ is the same regardless of whether $A$ or $M$ is used to write the quadratic form.
It is also fairly straightforward to show that, if we change basis to that made up by the orthonormal eigenvectors of the matrix used (assuming that $A$ is diagonalisable), then $Q$ takes the form $Q(\mathbf{x^{\prime}})=Q(\mathbf{x})=\sum_{i}{\lambda_{i}x_{i}^{\prime 2}}$, where $\mathbf{x}^{\prime}$ is $\mathbf{x}$ written in the new basis, and the $\lambda_{i}$ are the eigenvalues of the matrix used to describe the quadratic form. Just to make things simple, if all of the eigenvalues are positive, then in 3D this describes an ellipsoid with semi-axes of length given by the eigenvalues. 
Assuming that what I said in the first paragraph is correct, then the choice of $A$ or $M$ to write the quadratic form shouldn't make a difference to the surface described by $Q$. However, in general $A$ and $M$ will have different eigenvalues, which would lead me to believe that $Q$ describes a different surface depending on whether the eigenvector basis is chosen using $A$ or $M$, in contradiction to my previous assumption. What is wrong with my understanding here?
 A: Let $P$ be the transition matrix to some new coordinates, i.e. $P(v)_{new}=(v)_{old}$. If $Q$ is the matrix of the corresponding bilinear form and is symmetric, then there exists an orthonormal basis of eigenvector and the transition matrix from the standard basis to these eigenvector basis is an orthogonal matrix, i.e. $P^T=P^{-1}$. The matrix of a bilinear form in the new basis is $P^T Q P$. In this case, however, this equals $P^{-1} Q P$, which is the diagonal matrix of the eigenvalues. However, for non-symmetric bilinear forms $M$, there doesn't need to be an orthonormal basis of eigenvectors (and it doesn't even need to be diagonalizable). So, even if you get a basis of eigenvectors, the transition matrix $P$ will have these eigenvectors in columns, then there is no reason to believe that $P^T M P = P^{-1} M P$. 
So, it seems to me that the core is that you somehow mix different type of tensors: linear operators transform like $P^{-1} M P$ whereas bilinear forms as $P^T M P$. Just for orthonormal basis, it turns out to be the same. (I hope it answers, at least partially, your question.)
