I am reading Herbert Federer's book called "Geometric Measure Theory", in chapter one of Grassmann algebra, on pages 36-37, he says that for $f$ being an endomorphism of a finite dimensional inner product space we have the relations:
$$ 2trace[\bigwedge_2 f ] = (trace(f))^2-trace(f\circ f)$$
$$trace(\bigwedge_2 (f+f^*)) = 2(trace(f))^2-trace(f\circ f ) - trace(f^*\circ f)$$
Now the first identity I proved as he points in the book on page 37, but I am stuck with the second identity, he says "we obtain the second formula by applying the first to $f+f^*$", (BTW, $f^*$ is the adjoint of $f$).
I don't see how to apply the first formula to the second, I mean we don't have $trace(\bigwedge_2(f+f^*)) = trace((\bigwedge_2 f + \bigwedge_2 f^*))$, right?
You have left out the most important part: That your space is real, i.e. defined over $\mathbb{R}$ and not over $\mathbb{C}$. This has the following consequence:
$$(1) \text{Tr}(f^k) = \text{Tr}((f^{*})^k)$$ where $f^k$ denotes $k$-fold composition. (Without knowing the space is real, the RHS must be complex-conjugated.)
Anyway, let's follow the hint provided in the book: Plug $f+f^{*}$ in the first formula. We get:
$$(2) 2\text{Tr}[\bigwedge_2 f+f^{*} ] = (\text{Tr}(f +f^{*}))^2-\text{Tr}((f + f^{*}) \circ (f+ f^{*}))$$
Making use of $(1)$ and of the linearity of the trace, we find:
$$(3) (\text{Tr}(f +f^{*}))^2 = (\text{Tr}(f) + \text{Tr}(f))^2 = 4\text{Tr}^2(f)$$
Additionally, linearity of the trace shows:
$$\text{Tr}((f + f^{*}) \circ (f+ f^{*})) = \text{Tr}(f^2) + \text{Tr}((f^{*})^2) + \text{Tr}(f \circ f^{*}) + \text{Tr}(f^{*} \circ f)$$
Now we apply $(1)$ and also the fact $\text{Tr}(AB)=\text{Tr}(BA)$ to deduce:
$$(4) \text{Tr}((f + f^{*}) \circ (f+ f^{*})) = 2(\text{Tr}(f^2) + \text{Tr}(f^{*} \circ f))$$
Now combine $(2),(3),(4)$.