Assume the conjunction and obtain a contradiction in each of the two cases supplied by the premise:

The two case assumptions disappear at line 9, while the conjunction assumption remains available until line 10. The proof uses only the primitive rules printed in the paper.

Derive each negated component in its own subproof, then combine them. Assumption lines below begin local scopes; the listed discharge ranges close those scopes.

Only the original premise remains open. No classical principle is needed for this direction of De Morgan's laws.

The boxes delimit a subproof and identify the lifetime of its temporary assumptions. A result depending on a boxed assumption is available inside that box and its nested boxes; it cannot simply be exported outside. The enclosing rule specifies what may be exported and which assumptions are discharged. Other premises on which the derivation depends remain undischarged. This assumption discharge discipline distinguishes a conditional argument from an unconditional proof.

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.

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.

For the first matrix choose its single independent column:Their product has every column equal to $B_1$, and both factors have rank one.

For the second matrix, the first and third columns are independent and the second equals the first, givingMultiplication reconstructs the stated matrix. The first factor has independent columns and the second has independent rows, so both have matrix rank two, as required for a rank factorization.

True for these square matrices. The determinant multiplication identity gives $\det (XY)=\det X\det Y$. Since $XY$ is invertible, its determinant is nonzero; both factors are therefore nonzero, so both matrices have matrix inverses.

Equivalently, $Y$ must be injective and $X$ must be surjective, and either property implies invertibility for an endomorphism of a finite-dimensional vector space. The equal square dimensions matter: a rectangular matrix and a right inverse can have identity product without being mutually invertible square matrices.