The following is taken from exercise 12.4 in D'Inverno. We wish to compute the transformation properties of the electric field and magnetic induction under a Lorentz boost.
Given the following boost transformation (working in units where $c=1$):
$t' = \beta(t-vx), \quad x' = \beta(x-vt), \quad y'=y, \quad z'=z$,
wish to find the components of $F'^{ab}$, where
$F^{ab} = \begin{pmatrix}0&E_x&E_y&E_z\\-E_x&0&B_z&-B_y\\-E_y&-B_z&0&B_x\\-E_z&B_y&-B_x&0 \end{pmatrix}$.
Resolve to use the transformation law: $F'^{ab} = \frac{\partial x'^a}{\partial x^c} \frac{\partial x'^b}{\partial x^d} F^{cd}$.
Worked out the Jacobian matrix as $\frac{\partial x'^a}{\partial x^c} = \begin{pmatrix}\beta&-\beta v&0&0\\-\beta v&\beta&0&0\\0&0&1&0\\0&0&0&1 \end{pmatrix}$.
Then proceeded to square the matrix and apply it to the matrix $F^{ab}$ above. Rather than going through all the calculations I will instead give the component $F'^{01} = \beta^2(1+v^2)E_x$. Is it wrong to equate this with $E'_x?$ If not, where have I gone wrong? The book seems to indicate that $E'_x = E_x$.
It appears I have gone wrong in squaring the matrix. It is post- and premultiplied instead which results in the expected answer. You pre and post multiply (post by transposes) because you are summing over columns vs rows.