How to prove FD and FE will give the same estimates when T = 2 We simply use the time-demeaning of Ti observations in time for each cross-section i and FE is equivalent to an FE on balanced panel.How to prove it?
 A: The fixed effects model is defined as:
$$ y_{it} = \alpha_i + X_{it}^\prime \beta + \epsilon_{it} $$
where the $ \alpha_i $ defines the unobserved indvidual-specific effects.
The Fixed Effects (FE) estimator is obtained by eliminating $\alpha_i$ and time-invariant regressors via subtraction of the time average:
$$ y_{it} - \bar{y}_i = (X_{it} - \bar{X}_i)^\prime \beta + (\epsilon_{it} - \bar{\epsilon}_i) $$
Alternatively, one can subtract the one-period lag of the model to obtain the First-Differences (FD) estimator:
$$ y_{it} - y_{it-1} = (X_{it} - X_{it-1})^\prime \beta + (\epsilon_{it} - \epsilon_{it-1}) $$ 
Now we have to show that in the two-period case (T=2) both estimators are equivalent. In this case we can write the equation for the FD estimator as follows (omitting subscript i for simplicity):
$$ y_2 - y_1 = (X_2 - X_1)^\prime \beta + \epsilon_2 - \epsilon_1 $$ 
The equation for FE estimator is defined as:
$$ y_{1} - \bar{y} = (X_{1} - \bar{X})^\prime \beta + (\epsilon_{1} - \bar{\epsilon}) $$
where we can substitute the time averages:
$$ y_{1} - (y_{1}+y_{2})\frac{1}{2} = (X_{1} - (X_{1}+X_{2})\frac{1}{2})^\prime \beta + (\epsilon_{1} - (\epsilon_{1}+\epsilon_{2})\frac{1}{2}) $$ 
which yields the following relationship:
$$ y_{1} - y_{2} = (X_{1} - X_{2})^\prime \beta + (\epsilon_{1} - \epsilon_{2}) $$
Finally, multiplying both sides by -1:
$$ y_{2} - y_{1} = (X_{2} - X_{1})^\prime \beta + (\epsilon_{2} - \epsilon_{1}) $$
Note that the last equation is equivalent to the model for the FD estimator.
