Exercise 2.1.16 (Subgroup of upper triangular matrices in $\mathrm{GL}_n(F)$)

Proof. Let

be the set of upper triangular matrices in ${\mathrm{GL}}_n(F)$.

• $G\subset {\mathrm{GL}}_n(F)$ by definition. Let $I_n={(e_{\mathit{ij}})}_{(i,j)\in {[[1,n]]}^2}$ defined by $e_{\mathit{ij}}=1$ if $i=j$ and $e_{\mathit{ij}}=0$ if $i\ne j$. Then $I_n\in G$, so $G\ne \varnothing$.

• Let $A={(a_{\mathit{ij}})}_{(i,j)\in {[[1,n]]}^2},B={(b_{\mathit{ij}})}_{(i,j)\in {[[1,n]]}^2}$ be elements of $G$.

  Then $a_{\mathit{ij}}=b_{\mathit{ij}}=0$ for all $i>j$.

  Put $C=\mathit{AB}\in {\mathrm{GL}}_n(F)$, where $C={(c_{\mathit{ij}})}_{(i,j)\in {[[1,n]]}^2}$. By definition of the product, for all $(i,k)\in {[[1,n]]}^2$,

  $$\begin{array}{llll}\hfill c_{\mathit{ik}}& =\underset{j=1}{\overset{n}{\sum }}{a}_{\mathit{ij}}{b}_{\mathit{jk}}& \hfill & \\ \hfill & & \hfill \end{array}$$

  Suppose that $i>k$. Then $i>j$ or $j>k$, otherwise $i\le j$ and $j\le k$, so $i\le k$.

  • If $i>j$, then $a_{\mathit{ij}}=0$, thus $a_{\mathit{ij}}{b}_{\mathit{jk}}=0$,
  • If $j>k$ then $b_{\mathit{jk}}=0$ thus $a_{\mathit{ij}}{b}_{\mathit{jk}}=0$.

  So $a_{\mathit{ij}}{b}_{\mathit{jk}}=0$ is true for every $j\in [[1,n]]$. Hence

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->(i,k)\in {[[1,n]]}^2,i>k\Rightarrow c_{\mathit{ik}}=0.$$

  This shows that $C=\mathit{AB}\in G$.

• (For the stability by inverses, we can use minors and cofactors, but it is more easy to reason with the associate linear maps.)

  Let $A={(a_{\mathit{ij}})}_{(i,j)\in {[[1,n]]}^2}\in G\subset {\mathrm{GL}}_n(F)$.

  Let $f:F^n\to F^n$ be the linear mapping such that ${ℳ}_{ℬ}(f)=A$, where $ℬ=(e_1,e_2,\dots ,e_n)$ is the canonical basis of $F^n$.

  Since $A\in G$, $f(e_j)={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{a}_{\mathit{ij}}{e}_i={\sum }_{i=1}^j{a}_{\mathit{ij}}{e}_i$, because $i>j\Rightarrow a_{\mathit{ij}}=0$. Therefore

  $$f(e_j)\in ⟨e_1,e_2,\dots ,e_j⟩(1\le j\le n).$$

  Hence

  $$⟨f(e_1),f(e_2),\dots ,f(e_j)⟩\subseteq ⟨e_1,e_2,\dots e_j⟩(1\le j\le n).$$

  Since $f$ is a linear automorphism $f(e_1),f(e_2),\dots ,f(e_j)$ are linearly independent, therefore

  $${\dim <!--\; FUNCTION\; APPLICATION\; -->}_F⟨f(e_1),f(e_2),\dots <!--\; FUNCTION\; APPLICATION\; -->,f(e_j)⟩=j={\dim <!--\; FUNCTION\; APPLICATION\; -->}_F⟨e_1,e_2,\dots <!--\; FUNCTION\; APPLICATION\; -->e_j⟩.$$

  This shows that

  $$⟨f(e_1),f(e_2),\dots ,f(e_j)⟩=⟨e_1,e_2,\dots e_j⟩(1\le j\le n).$$

  Hence $e_j\in ⟨f(e_1),f(e_2),\dots ,f(e_j)⟩$, so $e_j={\lambda }_{1}f(e_1)+{\lambda }_{2}f(e_2)+\cdots +{\lambda }_{j}f(e_j)$ for some scalars ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_j\in F$. Since $f^{-1}$ is linear, $f^{-1}(e_j)={\lambda }_1{e}_1+{\lambda }_2{e}_2+\cdots +{\lambda }_j{e}_j$, so

  $$f^{-1}(e_j)\in ⟨e_1,e_2,\dots ,e_j⟩(1\le j\le n).$$

  This proves that $A^{-1}={ℳ}_{ℬ}(f^{-1})$ is upper triangular, i.e. $A^{-1}\in G$.

In conclusion, the set $G$ of upper triangular matrix in ${\mathrm{GL}}_n(F)$ is a subgroup of ${\mathrm{GL}}_n(F)$. □