## 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?

## Why are the Cartier divisors on a variety $X$ isomorphic to invertible subsheaves of $\mathcal{K}_{X}?$

$$\mathcal{K}_{X}$$ denotes the sheaf of total rings of fractions of $$\mathcal{O}_{x}.$$

## f(x) is invertible polynomial function of degree ‘n’ {n≥3} then f”(x) = 0 has exactly ‘n – 2’ distinct real roots if

A)f′(x)=0 has n−1/2 distinct real roots B)f′(x)=0 has n−1 distinct real roots C)all the roots of f′(x)=0 are distinct D)none of these

## 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.

## A function strictely positif almost everywhere on [a,b], is it invertible?

A real function $$f$$ strictely positif almost everywhere on the interval $$[a,b]$$, is it invertible on $$[a,b]$$?

## 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.