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Products below combine matrix multiplication with the wedge product of differential forms of differential forms. Differentiate the projection identity $p^2=p$ to getMultiplying by $p$ on both sides also gives $pdpp=0$; applying the same argument to $1-p$ gives $(1-p)dp(1-p)=0$. Thus at each point $dp$ is off-diagonal relative to $\mathrm{im}p\oplus \ker p$, whereas $(dp{)}^2$ is diagonal and commutes with $p$. For $\omega =p(dp{)}^2$,because $d(dp)=0$.

The graded Leibniz rule and cyclicity of the matrix trace give, since $\omega$ has even degree,Here $(d\omega ){\omega }^{k-1}$ is off-diagonal: $(dp{)}^3$ is off-diagonal and every power of $\omega$ is diagonal. Its matrix trace is zero. Hence $\boxed{d\mathrm{tr}({\omega }^k)=0}$ for every $k\ge 1$; for $k=0$ the trace is the constant matrix size.

For the determinant, work over the commutative algebra of even differential forms. If $f(0)=1$, write $f(\omega )=I+N$, where $N$ has strictly positive degree and is nilpotent for dimensional reasons. The finite power series identityholds over this algebra, as follows either from the formal determinant identity over characteristic-zero commutative rings or by differentiating $\det (I+sN)$ in the scalar variable $s$. Each $N^j$ is a linear combination of positive powers of $\omega$. The matrix trace of the logarithm is therefore a closed differential form, and its finite exponential is closed by the graded Leibniz rule. Thus $\boxed{d\det f(\omega )=0}$. The projection connection interprets $\omega$ as the curvature of $pd$ on $\mathrm{im}p$, but the proof above directly establishes both closedness claims.