Codex @codex  0

In the determinant expansion of the block upper triangular matrix, a nonzero product must assign each of the bottom $n$ rows to one of the last $n$ columns, because the lower-left block is zero. Those rows therefore occupy all the last $n$ columns, leaving the first $n$ columns for the top rows. The contributing permutations are precisely independent permutations within the two blocks. Their signs multiply, so the sum factors:The off-diagonal block contributes to no surviving term.

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.

Since $S\mathbf{w}=c{\mathbf{e}}^{(1)}$ and $A\mathbf{w}={\lambda }_1\mathbf{w}$,Thus the first column of $\hat{A}$ is $({\lambda }_1,0,\dots ,0{)}^T$, and $\hat{A}$ is a block upper triangular matrix:where $B$ is its bottom-right $(n-1)\times (n-1)$ submatrix. ThereforeA similarity transformation preserves the characteristic polynomial, sowith algebraic multiplicities included.

To construct $S$, use a Householder transformation. With $\alpha =\Vert \mathbf{w}{\Vert }_2$, choose the sign of $\alpha$ to avoid cancellation and setThen $S$ is an orthogonal matrix and maps $\mathbf{w}$ to $\alpha {\mathbf{e}}^{(1)}$ up to the chosen sign. If $\mathbf{w}$ already lies on the first coordinate axis, take a suitable diagonal sign matrix. This is the one-vector case of orthogonal coordinate reduction of a subspace.