Path from an invertible matrix $A$ to a diagonal matrix $E=(e_{ij})$ with $e_{11}=\mbox{sgn}(\det (A))$

Question

Can you help me to show, using Gaussian elimination, that there exist a path in $GL_n{\mathbb{R}}$ from an invertible matrix $A$ to a diagonal elemental matrix $E=(e_{ij})$ with $e_{11}=\mbox{sgn}(\det (A))$?.

Thanks in avance.

Answer 1

Let $r_1,\dots,r_n$ denote the rows of $A$. There are two operations which successively transform $A$ into a diagonal matrix $E$ such that $e_{11 }= \text{sgn}(\det(A))$ and $e_{ii} = 1$ for $i > 1$:

(1) For $i \ne j$ and $m \in\mathbb{R}$ replace $r_i$ by $r_i + m\cdot r_j$, written as $A \mapsto A(i;j,m)$.

(2) For $m > 0$ replace $r_i$ by $m\cdot r_i$, written as $A \mapsto A(i;m)$.

Both operations can be realized by paths in $GL_n{\mathbb{R}}$:

(1) $u_1(t) = (1-t)A + tA(i;j,m) = A(i;j,(1-t) + tm)$. Then $u_1(0) = A, u_1(1) = A(i;j,m)$ and $\det(u_1(t)) = \det(A) \ne 0$.

(2) $u_2(t) = (1-t)A + tA(i;m) = A(i;(1-t) + tm)$. Then $u_2(0) = A, u_2(1) = A(i;m)$ and $\det(u_2(t)) = ((1-t) + tm)\det(A) \ne 0$ (recall $m > 0$).

Note that that both paths preserve the sign of the determinant.

Two more transformations are generated by successive application of transformations of type (1):

(3) For $i \ne j$ simultaneously replace $r_i$ by $r_j$ and $r_j$ by $-r_i$: $$\begin{matrix} r_i \\ r_j \end{matrix} \mapsto \begin{matrix} r_i + r_j \\ r_j \end{matrix} \mapsto \begin{matrix} r_i + r_j \\ -r_i \end{matrix} \mapsto \begin{matrix} r_j \\ -r_i \end{matrix} $$

(4) For $i \ne j$ simultaneously replace $r_i$ by $-r_i$ and $r_j$ by $-r_j$: $$\begin{matrix} r_i \\ r_j \end{matrix} \mapsto \begin{matrix} r_j \\ -r_i \end{matrix} \mapsto \begin{matrix} -r_i \\ -r_j \end{matrix} $$

By transformations of type (1) and (3) $A$ is transformed into a diagonal matrix $D$. By transformations of type (4) $D$ is transformed into a diagonal matrix $D'$ such that $d'_{ii} > 0$ for $i > 1$. Finally by transformations of type (2) $D'$ is transformed into $E$.