6.2 Review: Linear Independence and Systems of Equations

 A. Solving Systems of Linear Equations

Solving systems of linear equations is a fundamental aspect of linear algebra. To solve these systems efficiently, we often express them in matrix form. Consider a system of linear equations

 [asciimath]{(a_11x_1+a_12x_2+...+a_(1n)x_n=b_1),(a_21x_1+a_22x_2+...+a_(2n)x_n=b_2 ),(\ \ \ \...),(\ \ \ \...),(\ \ \ \...),(a_(m1)x_1+a_(m2)x_2+...+a_(m\n)x_n=b_m ) :}[/asciimath] 

Such a system can be represented in matrix form as

 [asciimath][[a_11, a_12 ,..., a_(1n)],[a_21,a_22, ..., a_(2n)], [vdots,vdots ,ddots, vdots] ,[a_(m1), a_(m2) , ..., a_(mn)] ][/asciimath] [asciimath][[x_1],[x_2], [vdots] ,[x_n] ]=[/asciimath] [asciimath][[b_1],[b_2], [vdots] ,[b_m] ][/asciimath]

which is simply denoted as

 [asciimath]A bb(x) = bb(b)[/asciimath]

Here, [asciimath]A[/asciimath] is the coefficient matrix, containing the coefficients of the variables in the system, [asciimath]bb(x)[/asciimath] is the vector (column matrix) representing the variables, and [asciimath]bb(b)[/asciimath]  is the vector (column matrix) representing the constants on the right side of each equation. If all the constant terms in the vector [asciimath]bb(b)[/asciimath]  are zero, then the system of linear equations is referred to as a homogeneous system. Conversely, if any of the constants in [asciimath]bb(b)[/asciimath] are non-zero, the system is classified as a nonhomogeneous system.

To simplify this system, we use an augmented matrix, which combines the coefficient matrix [asciimath]A[/asciimath] and the constant vector [asciimath]bb(b)[/asciimath]  into a single matrix. This is done by appending [asciimath]bb(b)[/asciimath] as an additional column to [asciimath]A[/asciimath].

 [asciimath][[a_11, a_12 ,..., a_(1n),|,b_1],[a_21,a_22, ..., a_(2n),|,b_2], [vdots,vdots ,ddots, vdots,|,vdots] ,[a_(m1), a_(m2) , ..., a_(mn),|,b_m] ][/asciimath]

Row operations are then used to systematically simplify this augmented matrix, maintaining the equivalence of the system. The goal is to achieve either row echelon form (REF) or reduced row echelon form (RREF). REF, achieved through the Gaussian elimination method, simplifies the matrix into an upper triangle form where all non-zero rows are above rows of all zeros, and each leading coefficient (first non-zero number in a row) is to the right of the leading coefficient of the row above it. RREF achieved through the Gauss-Jordan elimination method, further simplifies REF so that each leading coefficient is the only non-zero number in its column and is equal to 1, making it easier to read the solutions directly from the matrix.

 Solution Possibilities

The solution of the system depends on the final form of the augmented matrix after applying row operations:

• Unique Solution: If the augmented matrix can be reduced to row echelon form where each variable has a leading 1 and there are no inconsistent equations (like  [asciimath]0 = 1[/asciimath]), the system is consistent and has a unique solution.
  Watch Video: Unique Solution

   
• No Solution: If the matrix yields a contradiction (such as  [asciimath]0 = 1[/asciimath]), it indicates that the system is inconsistent and has no solution.
  Watch Video: Possible Solutions

   

• Infinite Solutions: If the system has at least one row where all coefficients are zero, but the system is consistent (like  [asciimath]0 = 0[/asciimath]), the system has an infinite number of solutions. In such cases, the solution is typically expressed in a parametric form. This case typically happens when there are fewer independent equations than variables.
  Watch Video: Infinite Solutions

   

B. Linear Independence

Understanding linear independence is also important for solving systems of linear equations and differential equations. It helps in determining whether a set of solutions forms a valid basis for the solution space and whether the solutions are unique and span the entire solution space.

A set of vectors in a vector space is said to be linearly independent if no vector in the set can be written as a linear combination of the others. Consider a set of vectors

 [asciimath]{ bb(v)_1, bb(v)_2, ..., bb(v)_n }.[/asciimath]

The vectors are linearly independent if the only solution to the equation

[asciimath]c_1 bb{v}_1 + c_2 bb{v}_2 + ... + c_n bb{v}_n = bb{0}[/asciimath]

is  [asciimath]c_1 = c_2 = ... = c_n = 0[/asciimath] , where [asciimath]bb{0}[/asciimath] is the zero vector and [asciimath]c_1, c_2, ..., c_n[/asciimath] are constants. In other words, none of the vectors can be expressed as a linear combination of the others. If there exists at least one non-trivial solution (where not all [asciimath]c_i[/asciimath] are zero) to this equation, then the vectors are linearly dependent. This means at least one of the vectors in the set can be written as a linear combination of the others.

To test for linear independence or dependence, we can represent this system in matrix form as

 Testing for Linear Independence

• Using a Matrix: Form a matrix [asciimath]V[/asciimath] with these vectors as columns. The set of vectors is linearly independent if the determinant of [asciimath]V[/asciimath] is non-zero. If the determinant is zero, the vectors are linearly dependent.
• Row Reduction: Alternatively, use row reduction to bring the matrix [asciimath]V[/asciimath] into row echelon form (REF) or reduced row echelon form (RREF). If any column in [asciimath]V[/asciimath] lacks a leading 1 (pivot), the vectors are linearly dependent.

If the vectors are found to be linearly dependent, the specific relationship among them can be found by solving the system [asciimath]V bb(c)=bb(0)[/asciimath] for the constants [asciimath]c_1,c_2,...,c_n[/asciimath].

Section 6.2 Exercises