2. Block type Bruhat decomposition

For the convenience of the reader, we include some elementary, well-known facts from linear algebra which we shall need.

 

(2.1) Fix an integer $k \in \{ 1, \ldots , r\! -\! 1 \}$ and let

$${\,I\!\!\!\!G}={\,I\!\!\!\!G}_{(k)}= \left\{ \left( \be... ... \bigg\vert \ S \in Gl_k (R), \ W \in Gl_{r-k}(R) \right\}.$$

After a permutation of rows (or columns), any regular matrix will belong to ${\,I\!\!\!\!G}$.
Consider the following subgroups of Gl_r contained in ${\,I\!\!\!\!G}$:

$$\begin{eqnarray*}I\!\!D= I\!\!D^{(k)} & = & \left\{ \left( \begin{array}{rl}... ... \right) \bigg\vert V \in \mbox{Mat}\,(r\!-\!k,k) \right\}, \end{eqnarray*}$$

where I always denotes the unit matrix of the corresponding size.

 

(2.2) The following lema will be verified by direct computation:

Lemma 1  
  Using the notation above,

(i1) $I\!\!D\subseteq {\em N}({I\!\!B}_+) \cap {\em N}({I\!\!B}_-)$,
(i2) $\,I\!\!\!\!G= {\,I\!\!\!\!G}^{-1}$,
(i3) $\,I\!\!\!\!G\subseteq I\!\!D\cdot I\!\!B_+ \cdot I\!\!B_- \cap I\!\!D\cdot I\!\!B_+ \cdot I\!\!B_-$.

Here $N( \ )$ denotes the normalizer in Gl_r(R).
(i1) Just check the following identities:

$$\left( \begin{array}{rl} S & 0 \\ 0 & W \end{array} \r... ...( \begin{array}{rl} I & 0 \\ V & I \end{array} \right)$$

$$= \left( \begin{array}{cr} I & 0 \\ WVS^{-1} & I \end{ar... ...left( \begin{array}{rl} S & 0 \\ 0 & W \end{array} \right)$$

(i2) From the identity

$$\begin{eqnarray*}\left( \begin{array}{rl} S & T \\ V & W \end{array} \rig... ...t( \begin{array}{rl} I & 0 \\ 0 & I \end{array} \right), \end{eqnarray*}$$

assuming that the first matrix is in ${\,I\!\!\!\!G}$, we obtain T'=-S^-1 T W' and V'=-W^-1VS'; hence, SS'-TW^-1 VS'=I. Thus, S' is invertible and $\tilde S :=S-TW^{-1}V$ is its inverse. Similarly, $\tilde W :=W-VS^{-1}T$ is the inverse of W' .
(i3) Using the notation of (i2) we obtain

$$\begin{eqnarray*}\left( \begin{array}{rr} S & T \\ V & W \end{array} \right... ...begin{array}{rc} I & S^{-1}T \\ 0 & I \end{array} \right). \end{eqnarray*}$$

 

(2.3) Fix a k-subset J of row indices $\{ 1,...,r \}$. Let Q_J be the submatrix of Q formed by those rows of Q whose index belongs to J. Then ${\cal M}_J (Q):=\langle Q_J \rangle \subseteq R^k$ is invariant under equivalence of matrices. After a certain permutation of rows of Q, it is enough to consider for simplicity only the special cases $J_0=\{1,...,k\}$, $J'_0=\{k\!+\!1,...,r\}$