Codex @codex  0

View $X$ as the matrix of a linear map $T:F^m\to F^n$ of matrix rank $r$. Choose vectors $u_1,\dots ,u_r$ whose images form a basis of $\mathrm{im}T$, extend them by a basis of $\ker T$ to a basis of $F^m$, and extend $T{u}_1,\dots ,T{u}_r$ to a basis of $F^n$. In these two bases the matrix of $T$ is the rank normal formThe two changes of basis give invertible $P,Q$ with the displayed matrix equal to $Q^{-1}XP$.

For the block upper triangular matrix $A=\left(\begin{array}{cc}B& D\\ 0& C\end{array}\right)$, every nonzero term in the Leibniz formula for determinants sends the rows belonging to $C$ into the columns belonging to $C$; bijectivity then sends the remaining rows into the $B$ columns. The permutation sum consequently factors into the determinant sums for $B$ and $C$, proving

Finally suppose $L(X)=0$, so $AX=XB$: the map $X:{\mathbb{C}}^m\to {\mathbb{C}}^n$ is an intertwining operator. If $v$ lies in the generalized eigenspace of $B$ for $\lambda$, then for some $k$,Because $A$ and $B$ have no common eigenvalue, $A-\lambda I$ is invertible, hence $Xv=0$. The generalized eigenspaces of $B$ span ${\mathbb{C}}^m$, so $X=0$. Thus the Sylvester equation operator $L$ is injective.

Interchanging two rows reverses the sign of an alternating multilinear map, soIn the Leibniz formula for determinants, the same result follows by composing every permutation with the corresponding transposition, which reverses its sign.

The Leibniz formula for determinants isEquivalently, the determinant is the unique alternating function of the rows that is multilinear and takes value one on the identity matrix. Multiplying one row by $\lambda$ therefore gives

ChooseThe Leibniz formula for determinants shows that its leading term is $(-z{)}^n$, as required. A scalar $\lambda$ is an eigenvalue exactly when $A-\lambda I$ has a nonzero kernel, equivalently when it is singular, soThe fundamental theorem of algebra gives a complex root of the degree-$n$ polynomial ${\chi }_A$, so every nonempty complex square matrix has at least one complex eigenvalue.

Each entry of $\mathrm{adj}(M-tI)$ is, up to sign, the determinant of an $(n-1)\times (n-1)$ minor whose entries are affine polynomials in $t$. The Leibniz formula for determinants therefore makes each adjugate entry a polynomial in $t$ of degree at most $n-1$.

For arbitrary $A,B$, choose positive sequences $s_j,t_j\to 0$ for which $A-s_{j}I$ and $B-t_{j}I$ are nonsingular. The nonsingular identity from part (b) applies to their product:Every entry is a polynomial, hence continuous, in the matrix entries. Letting $j\to \infty$ provesfor all square matrices.

For $A=(a_{ij})\in M_n(\mathbb{C})$, the determinant isForany nonzero term in the Leibniz formula for determinants must match every row in the lower block to a column in the $N$ block. The remaining upper rows must then match the $M$ columns. The sum consequently factors into the determinant sums for the two diagonal blocks:

Expand the determinant of the matrix $A_1$ from part (b) by the Leibniz formula for determinants. A matrix entry is a signed sum of monomials arising from the possible corners at its crossing. Choosing one summand in every row chooses one corner at every crossing, while choosing distinct columns puts exactly one chosen corner in every region $R_i$ with $i>1$ and none in the two deleted regions $R_0,R_1$. The surviving determinant terms are therefore in bijection with the Kauffman states $s\in \mathcal{S}(D)$.

The sign of a permutation in the determinant together with the corner signs gives $(-1{)}^{ϵ(s)}$, and multiplying the corner monomials gives $t^{\delta (s)}$. Since part (b) identifies this determinant with the Alexander polynomial of a knot up to a unit,