If you find any mistakes, please make a comment! Thank you.

## Set of all strictly upper triangular matrices is a subgroup of general linear group

Let $n \in \mathbb{Z}^+$ and let $F$ be a field.Prove that the set $$UT_n^1(F) = \{ [a_{i,j}]_{i,j = 1}^n \in GL_n(F) \ |\ a_{i,j} = 0\ \mathrm{if}\ i > j, a_{i,i} = 1\ \mathrm{for\ all}\ i \}$$ is a subgroup of $GL_n(F)$.

Solution: First we prove some technical lemmas.

Lemma 1. If $A, B \in UT_n^1(F)$, then $AB \in UT_n^1(F)$.

Proof: We saw in a lemma to Exercise 2.1.16 that $AB \in UT_n(F)$. Now for all $i$ the $(i,i)$-entry of $AB$ is $\sum_{k=1}^n a_{i,k} b_{k,i}$. Note that for $k \neq i$, either $a_{i,k}$ or $b_{k,i}$ is 0, so all terms except $a_{i,i}b_{i,i} = 1$ drop out. So the $(i,i)$-entry of $AB$ is 1.

Lemma 2. If $A \in UT_n^1(F)$ and $AB = I$, then $B \in UT_n^1(F)$.

Proof: Proof: We saw in a lemma Exercise 2.1.16 that $B \in UT_n(F)$. Now we have $\sum_{k=1}^n a_{i,k}b_{k,i} = 1$ for all $i$. Note that for $k \neq i$, either $a_{i,k} or b_{k,i}$ is $0$, so all terms in the sum except $a_{i,i}b_{i,i} = 1 \cdot b_{i,i}$ drop out. Then $b_{i,i} = 1$.

Now to the main result. Note that $UT_n^1(F)$ is not empty since $I \in UT_n^1(F)$.

Now let $A,B \in UT_n^1(F)$. By Lemma 2, $B^{-1} \in UT_n^1(F)$, and by Lemma 1, $AB^{-1} \in UT_n^1(F)$. By the subgroup criterion, $UT_n^1(F) \leq GL_n(F)$.