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The two square products can have different sizes, but their matrix traces are both the same finite double sum:Interchanging two finite sums and commuting their scalar factors is valid over the real numbers. No commutation of the matrix products themselves is asserted. This proves the cyclic property of the trace in the rectangular-factor case.

Let $r=\mathrm{rank}(A)$. If $r=0$, the matrix is zero and the identity is immediate. Otherwise choose the rank factorization $A=BC$ from part b. The column space of $B$ is the image of $A$.

An idempotent matrix acts as the identity on its image: if $u=Av$, then $Au=A^{2}v=Av=u$. Applying this to each column of $B$ gives $AB=B$. Substituting $A=BC$ yields $B(CB-I_r)=0$. Since $B$ has full column rank, the represented map is injective, so each column of $CB-I_r$ is zero. Thus $CB=I_r$.

Now the rectangular cyclic property of the trace givesThis proves that the trace of an idempotent equals its rank using the supplied factorization method. Orthogonality of the projection is unnecessary: the argument applies to every real idempotent matrix.

For an invertible matrix $S$, the cyclic property of the trace gives $\mathrm{tr}(SA{S}^{-1})=\mathrm{tr}(A{S}^{-1}S)=\mathrm{tr}A$. Thus the diagonal sum defining the trace does not depend on the choice of basis for an endomorphism. Cyclic permutation of factors is valid inside a trace; arbitrary permutation of factors need not be.

For linear maps $A,B$ on a finite-dimensional vector space, the cyclic property of the trace gives $\mathrm{tr}(AB-BA)=\mathrm{tr}(AB)-\mathrm{tr}(BA)=0$. In matrix entries both traces are ${\sum }_{i,j}{A}_{ij}{B}_{ji}$ after exchanging indices. Thus the trace of a finite-dimensional Lie algebra representation vanishes on the derived algebra. In particular, for a perfect Lie algebra, every representing matrix has zero trace.