Tensors: Simplifying system of linear equations by choosing new coordinate system My textbook starts with the following linear system, where $\mathbf{J}$ and $\mathbf{E}$ are tensors:

$$J_x = \sigma_{xx} E_x + \sigma_{xy}E_y + \sigma_{xz}E_z$$
$$J_y = \sigma_{yx}E_x + \sigma_{yy}E_y + \sigma_{yz}E_z$$
$$J_z = \sigma_{zx}E_x + \sigma_{zy}E_y + \sigma_{zz}E_z$$

It then goes on to say the following:

Based on the basic symmetry of the equations of motion, Onsager demonstrated that the tensor is symmetric, i.e., $\sigma_{ik} = \sigma_{ki}$, so that the nine coefficients are always found to reduce to six. Taking advantage of this symmetry argument and multiplying the expressions by $E_x$, $E_y$, and $E_z$, respectively, one obtain on adding:
$$J_x E_x + J_y E_y + J_z E_z = \sigma_{xx} E^2_x + \sigma_{yy}E^2_y + \sigma_{zz} E^2_z + 2 \sigma_{xy} E_x E_y + 2\sigma_{yz} E_y E_z + 2 \sigma_{zx} E_z E_x$$
To make the mixed terms on the right side of the equation disappear, one chooses a new coordinate system with the coordinates along the principal axes of the quadratic surface represented by this right-hand side (rhs) term, and in this new coordinate system one obtains:
$$J_x = \sigma_1 E_x; J_y = \sigma_2 E_y; J_z = \sigma_3 E_z,$$
where $\sigma_1$, $\sigma_2$, and $\sigma_3$ are the principal conductivities.

I don't understand how the author went from
$$J_x E_x + J_y E_y + J_z E_z = \sigma_{xx} E^2_x + \sigma_{yy}E^2_y + \sigma_{zz} E^2_z + 2 \sigma_{xy} E_x E_y + 2\sigma_{yz} E_y E_z + 2 \sigma_{zx} E_z E_x$$
to
$$J_x = \sigma_1 E_x; J_y = \sigma_2 E_y; J_z = \sigma_3 E_z$$?
I would greatly appreciate it if people could please take the time to clarify this.
 A: Another way to phrase this is that $J = \Sigma\cdot E$ where $J$ and $E$ are 3-vectors and $\Sigma$ is a $3\times3$ symmetric matrix: the $\cdot$ represents the matrix-vector product.
What the argument reduces to in terms of linear algebra (assuming this is familiar to you) is that, since $\Sigma$ is symmetric, it can be diagonalised: ie, written on the form $\Sigma = UDU^{T}$ where $U$ is a rotation matrix and $D$ is diagonal. Another way to formulate this is that there are eigenvectors $u_1, u_2, u_3$ (which can be picked to be orthogonal and of unit length) with eigenvalues $\sigma_1, \sigma_2, \sigma_3$, ie $\Sigma\cdot u_i = \sigma_i u_i$.
If you express the vectors in the basis $u_1, u_2, u_3$, ie $E=E'_1u_1+E'_2u_2+E'_3u_3$ where $E'_i$ are the coefficient of $E$ in this basis, you get $\Sigma\cdot E = \sigma_1E'_1u_1+\sigma_2E'_2u_2+\sigma_3E'_3u_3$.
Also writing $J=J'_1u_1+J'_2u_2+J'_3u_3$ in this basis, you see that $J'_i=\sigma_i E'_i$.
What may make the presentation in the book a bit confusing (which I have changed) is that they use $E_x, E_y, E_z$ as the coordinates first in the regular coordinate system, and then again in the eigenvector-based coordinate system: ie, the $E_x$ in the original equation is not the same as the $E_x$ in the final equation.
The use of the square expression for which they find the principal axes is just a way to derive at the existence of eigenvectors and eigenvalues: the principal axes correspond to the eigenvectors.
