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Let $B$ be the set of all upper triangular matrices in $GL_n$. What are the conditions for a matrix in $GL_n$ lies in $Bw_0B$ (What do the matrices in $Bw_0B$ look like)? Thank you very much.

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The first useful observation is that $w_0 B w_0 = B^-$, where $B^-$ denotes the group of all lower triangular matrices in $GL_n$. Thus $B w_0 B = w_0 B^- B$. I'll describe $B^- B$ for you and let you make the appropriate modifications to handle multiplying the entire set on the left by $w_0$.

As an illustrative first example, consider the case $n = 2$. Then an element of $B^- B$ has the form $$ M = \begin{pmatrix} a_1 & 0 \\ b_1 & c_1 \end{pmatrix} \begin{pmatrix} a_2 & b_2 \\ 0 & c_2 \end{pmatrix} = \begin{pmatrix} a_1 a_2 & a_1 b_2 \\ b_1 a_2 & b_1 b_2 + c_1 c_2 \end{pmatrix}. $$

Note that the upper-left entry of $M$ is non-zero because $a_1$, $a_2$ are, as is $\det(M) = a_1 a_2 c_1 c_2$.

In general, for arbitrary $n$, you should check that

  1. The upper left $k \times k$ submatrix of $M = b^- b$ with $b^- \in B^-$ and $b \in B$ is equal to the upper left $k \times k$ submatrix of $b^-$ times the upper left $k \times k$ submatrix of $b$.

  2. Consequently, the upper left $k \times k$ submatrix of $M$ has a non-zero determinant.

  3. If $M$ is any $n \times n$ matrix all of whose upper left $k \times k$ submatrices haave non-zero determinant (these determinants are called the leading principal minors of $M$), then $M$ can be written in the form $b^- b$. (Hint: Use Gaussian elimination. The hypothesis on the leading principal minors assures us that we will not have to do any row swaps.)

Conclusion: $B^- B$ consists of all $n \times n$ invertible matrices whose leading principal minors are all non-zero.

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  • $\begingroup$ thank you very much. Should "Bw_0B=w_0B^{-1}B" be "Bw_0B=w_0^{-1}B^{-1}B"? $\endgroup$ – LJR Dec 3 '20 at 15:52
  • $\begingroup$ $w_0$ is an involution, so $w_0^{-1} = w_0$ $\endgroup$ – Michael Joyce Dec 3 '20 at 23:06

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