Number of elements in the set of invertible lower triangular matrices over a finite field


Problem:

Let $ F_q$ be a finite field with $ q$ elements.

$ T_n(F_q) := \{ A = (a_{ij}) \in F^{n \times n}$ | $ a_{ij} = 0$ for $ i < j,$ and $ a_{ij} \neq 0$ $ \forall i \}$ .

Determine the number of elements in $ T_n(F_q)$ .

My solution is as follows:

Starting with the last row going upwards, there are:

$ q-1$ possibilities for the last row;

$ (q-1)q$ possibilities for the row before the last;

.

.

.

$ (q-1)q^{n-1}$ possibilities for the first row.

Therefore, in total there are $ (q-1)^nq^{\sum_{i=1}^{n-1} i} = (q-1)^nq^{\frac{n(n-1)}{2}}$ elements.

Could you, please, check my solution?

The set of all invertible operators need not be dense

I want to show that the set of all invertible operators $ \mathcal{G}(\ell^2)$ is not dense in $ \mathcal{B}(\ell^2)$ .

Consider the right shift operator $ T\in \mathcal{B}(\ell^2)$ . We also know that $ T\notin \mathcal{G}(\ell^2)$ . I want to show now that $ $ B=\{R\in \mathcal{B}(\ell^2):\|R-T\|<1\}$ $ is disjoint with $ \mathcal{G}(\ell^2)$ .

Suppose not. Then there exists $ R\in \mathcal{G}(\ell^2)$ such that $ \|R-T\|<1$ . How to arrive at a contradiction from here? Any help is appreciated.

if A is an invertible matrix and B is an invertible matrix, is A+B also an invertible matrix?

the answer is obviously false. for example: $ $ A=\begin{pmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 1/2 \ \end{pmatrix} A^{-1}=\begin{pmatrix} 1 & 0 & 0 \ 0 & 1/2 & 0 \ 0 & 0 & 2 \ \end{pmatrix} \ B=\begin{pmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & -1/2 \ \end{pmatrix} B^{-1}=\begin{pmatrix} 1 & 0 & 0 \ 0 & 1/2 & 0 \ 0 & 0 & -2 \ \end{pmatrix} \ A+B=\begin{pmatrix} 2 & 0 & 0 \ 0 & 4 & 0 \ 0 & 0 & 0 \ \end{pmatrix}$ $ A+B dose not have an inverse matrix(rank is not n). is there a better way to prove it other then an example?

How can we prove that if we can change a matrix into an identity matrix by doing some elementary row exchanges, then the matrix is invertible?

I started linear algebra, and I encountered the part that using Gauss-Jordan method to compute invertible matrix. Then, how can we prove that a matrix is invertible if we can change that matrix into an identity matrix by doing some elementary row exchanges?

Show that the functor that takes $R$ to the set of invertible elements of $R[X]/(X^2-a)$ is representable.

The following question is from the Fall 2016 UCLA algebra qualifying exam:

Let $ F$ be a field and $ a\in F$ . Show that the functor that takes $ R$ , commutative $ F$ -algebras to the invertible elements of $ R[X]/(X^2-a)$ is representable.

What I have so far: If $ a\in F$ , then then we have that $ R[X]/(X^2-a)\cong R\times R$ . Hence we get that $ $ Hom_{F-\text{alg}}(F[t,t^{-1}]\otimes_FF[t,t^{-1}],R)\cong Hom_{F-\text{alg}}(F,R\times R)\cong U(R)$ $ where $ U$ is the functor that take $ R$ to units of $ R[X]/(X^2-a)$ . Hence in this case, the functor is representable.

I’m unsure how to extend this to the general case.

Null spaces and invertible matrix

The question is: “If A and B are n x n matrices, show that they have the same null space if and only if A = UB for some invertible matrix U.

I started the question by saying Ax = 0 for some vector x in null(A). Now I’m lost. Could someone please help me out with this question? Thank you very much.

Product of two left invertible elements is also left invertible in Semigroup

Consider a Semigroup $ (M, \ast)$ with a neutral element $ e$ . Now I have to prove that all left invertible elements of $ (M, \ast)$ form a sub-semigroup.

As a left invertible element is an element who’s left inverse exist. i.e. $ a$ is left invertible if there exist an element $ a’$ such that $ $ a’ \ast a=e$ $ Let us consider a set $ A\subset M$ containing all left invertible elements. Let $ a,b \in A$ then there exist $ a’,b’ \in M$ such that $ $ a’ \ast a=e \ \ \text{and } \ b’ \ast b=e$ $ Now I have to prove that $ a \ast b$ is also left invertible i.e. there exist some $ c \in M$ such that $ $ c \ast (a \ast b) = e$ $ Now I don’t know how to prove it. May be my statement is wrong i.e. set of invertible is sub-semigroup or may there be another way to prove it.