Why the equation Ax = 0 has fewer than n pivot positions when it has a nontrivial solution: Understanding the Relationship between Augmented Matrix and Pivot Positions

If the equation Ax = 0 has a nontrivial solution, then A has fewer than n pivot positions.

To understand why the equation Ax = 0 has fewer than n pivot positions when it has a nontrivial solution, let’s first define what pivot positions are in the context of an augmented matrix

To understand why the equation Ax = 0 has fewer than n pivot positions when it has a nontrivial solution, let’s first define what pivot positions are in the context of an augmented matrix.

In linear algebra, an augmented matrix represents a system of linear equations. The pivot positions in an augmented matrix are the positions in each row where the leading non-zero entry (pivot) is located. These pivot positions define the pivot columns in the corresponding coefficient matrix.

Now, for the equation Ax = 0, let’s assume that A is an m × n matrix, meaning it has m rows and n columns. The matrix A can be written as [A₁ | A₂ | … | Aₙ], where A₁, A₂, …, Aₙ are the column vectors of A.

If the equation Ax = 0 has a nontrivial solution, it means there exists a non-zero vector x (not the zero vector) such that when multiplied by A, the result is the zero vector. In other words, A₁x₁ + A₂x₂ + … + Aₙxₙ = 0, where x₁, x₂, …, xₙ are the entries of vector x.

Now, consider the augmented matrix [A | 0]. In order for the equation Ax = 0 to have a nontrivial solution, this matrix must have infinitely many solutions or at least one free variable. If the matrix has infinitely many solutions, it means that at least one of the columns of A is a linear combination of the other columns. This implies that the matrix A does not have full column rank, and consequently, it cannot have n pivot positions.

If the matrix has at least one free variable, it means that there are fewer linearly independent columns than the total number of columns. Again, this implies that A does not have full column rank, and therefore, it cannot have n pivot positions.

In both cases, because the equation Ax = 0 has a nontrivial solution, we can conclude that A has fewer than n pivot positions.

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