Trace of the energy-momentum tensor

I had to find the canonical energy-momentum tensor defined by this Lagrangian density
$ mathcal{L} = – {1 over 4} F_{mu nu} F^{mu nu}$
and I got the result of $ T^{mu nu} = – F^{mu lambda} partial^{nu} A_{lambda} + {1 over 4} eta^{mu nu} F^{rho sigma} F_{rho sigma}$
In order to make it a symmetric tensor, I had to add total derivative term and knowing from equation of motion that $ partial_{lambda} F^{mu lambda}=0$, I got new tensor
$$ hat{T}^{mu nu} = F^{mu lambda}F^{nu}_{lambda} + {1 over 4} eta^{mu nu} F^{rho sigma} F_{rho sigma}$$ which is symmetric.
The question is – what is the trace of $hat{T}^{mu nu}$?

I know that, as far as it comes to the stress-energy tensor $T^{mu nu}$, we can write the trace as $T^{mu}_{mu}$ and using the formula for the energy-momentum tensor, do we get sth like that $T^{mu}_{mu} = {{partial mathcal{L}} over {partial (partial_{mu} A_{lambda})}} partial_{nu} A_{lambda} – delta^{mu}_{nu} mathcal{L} $? Honestly, I can’t see what would the result for $hat{T}^{mu nu}$ be. Would it be that the canonical stress-energy tensor is traceless? But how actually prove it?