Suppose n is a positive integer and Ai, j ∈𝐅 for i, j=1, …, n . Prove that the following are equ

Suppose n is a positive integer and Ai, j ∈𝐅 for i, j=1, …,...

Suppose $n$ is a positive integer and $A_{i, j} \in \mathbf{F}$ for $i, j=1, \ldots, n .$ Prove that the following are equivalent (note that in both parts below, the number of equations equals the number of variables): (a) The trivial solution $x_{1}=\dots=x_{n}=0$ is the only solution to the homogeneous system of equations \[ \begin{array}{c} \sum_{k=1}^{n} A_{1, k} x_{k}=0 \\ \vdots \\ \sum_{k=1}^{n} A_{n, k} x_{k}=0. \end{array} \] (b) For every $c_{1}, \ldots, c_{n} \in \mathbf{F},$ there exists a solution to the system of equations \[ \begin{array}{l} \sum_{k=1}^{n} A_{1, k} x_{k}=c_{1} \\ \sum_{k=1}^{n} A_{n, k} x_{k}=c_{n}. \end{array} \]

Suppose $n$ is a positive integer and $A_{i, j} \in \mathbf{F}$ for $i, j=1, \ldots, n .$ Prove that the following are equivalent (note that in both parts below, the number of equations equals the number of variables):
(a) The trivial solution $x_{1}=\dots=x_{n}=0$ is the only solution to the homogeneous system of equations
\[
\begin{array}{c}
\sum_{k=1}^{n} A_{1, k} x_{k}=0 \\
\vdots \\
\sum_{k=1}^{n} A_{n, k} x_{k}=0.
\end{array}
\]
(b) For every $c_{1}, \ldots, c_{n} \in \mathbf{F},$ there exists a solution to the system of equations
\[
\begin{array}{l}
\sum_{k=1}^{n} A_{1, k} x_{k}=c_{1} \\
\sum_{k=1}^{n} A_{n, k} x_{k}=c_{n}.
\end{array}
\]

Key Concepts

Equivalence of Uniqueness and Existence in Linear Systems

The problem addresses the fundamental equivalence in linear algebra: a homogeneous system having only the trivial solution is equivalent to the coefficient matrix being invertible, which in turn ensures that every non-homogeneous system has exactly one solution. This highlights the deep connection between the properties of linear transformations (injectivity and surjectivity) and the solution sets of linear systems.

Non-Homogeneous Linear System

A non-homogeneous linear system has at least one non-zero constant term. The existence of an invertible coefficient matrix in such systems implies that there is a unique solution to the system for every choice of constant terms. This guarantees the consistency of the system since the mapping defined by the matrix is surjective.

Invertible Matrix

An invertible matrix is a square matrix that has an inverse such that when it is multiplied by its inverse, the result is the identity matrix. This property is crucial in linear algebra because it guarantees that the associated linear transformation is both one-to-one and onto, thereby ensuring the existence of a unique solution for every system of linear equations, whether homogeneous or non-homogeneous.

Homogeneous Linear System

A homogeneous linear system is one in which all the constant terms (right-hand side values) are zero. The trivial solution, where all unknowns are zero, always satisfies such a system. When this trivial solution is the only solution, it indicates that the system has no free parameters, which implies that the coefficient matrix is of full rank and thus invertible.

Transcript

00:01 Hello there.

00:03 So for this exercise we have the following linear system where a is a square matrix with integer entries and b is also a vector with integer entries.

00:38 So we need to show or actually prove that if the determinant of the matrix a is equals to one then that implies that the solution is composed by integers or that the vector x has integer entries so let's this let's analyze how it's how to prove this so we know that the solution for this system will be due to a is a square matrix and the determinant is different from zero because we have that the determinant of a is equals to 1 which is different from 0 that means that a is an is invertible so the solution will be a inverse times b so we know that v is composed by integers so we need to show to do to get that x is uh is an integral vector we need to show that the inverse of a will be composed by an integral entries.

01:50 Because then that multiplication will lead that x is formed by integral entries.

01:56 So we just need to focus on the inverse of a.

02:01 So using the adjoin method, the adjun definition, we know that the inverse of a matrix of a square matrix is equals to the adjoin of the matrix over the determinant.

02:14 Of a but here the determinant is equal to one that means that the inverse will be just the adjoin matrix now what is the definition of adjoin so the adjoin matrix of a is equals to the co -factor matrix transpose so the transposition of the matrix will not change the type of will not change the entries we'll just change the position so we need to focus on the coefficients of the cofacture matrix so let's rewrite the cofacture matrix the entries ij as c ij so how is given these coefficients of the of the of the of the cofactor matrix well each c ij will be given by minus 1 to the power of a plus j times the minor ij.

03:22 So just to remind you that the minor is obtained by the determinant of this matrix m obtained by eliminating the i -th row so if you have here your matrix a and then here is your row i and here is your column j so you eliminate those and then you left with some some entries here.

03:49 That form a sub matrix and you take the determinant of that new sub matrix here that you obtain by eliminating the i -th row and the jade column so the determinant of of integral matrices so you have here a sub -matrix let's say s ij so s ij will be composed by integers because a is a with integral coefficients...

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