Title Linear dependence and independence_02.pdf ✅

Page 4 of 8 Linear Dependency If * + is non-empty set of vectors, then the vector equation or has at least one solution, namely .  If there is a unique solution, then is called a linearly independent set.  If there are many solutions, then is called linearly dependent set Gaussian Elimination on Linearly Dependent Column To test for linear dependence/independence we can write the corresponding matrix in echelon form. Example: Are the following vectors linearly dependent/ Independent? a) {( ) ( ) ( )} b) {( ) ( ) ( )} Solution: a) We have ( ) ( ) ( ) ( ) The augmented matrix: ( | ) ( | ) ( | ) ( ⁄ | )
Page 5 of 8 The pivot in the last column is missing. Therefore is a dependent column. Therefore is a free variable and we are having many solution of the system. Let any arbitrary value (real number) In we have: ⁄ ⁄ In we have: ( ⁄ ) ( ) . Solution of the system is ( ) ( ⁄ ) while (any real number).  Non trivial solution  The set of vectors is linearly dependent. Trivial solution: The only solution to is . For example has at least one solution, namely . Non-trivial solution: There exists x for which where .
Page 6 of 8 b) We have ( ) ( ) ( ) ( ) The augmented matrix: ( | ) ( | ) ( | ) ( | ) We will omit in our next step since it is entirely “0” ( | ) ( | ) Hence by back substitution we have  Trivial solution  The set of vectors is linearly independent.
Page 7 of 8 Linear Dependency in Square and non-Square matrices We have observed linear dependency in a square matrix while solving Example (a) from Gaussian Elimination on Linearly Dependent Column. Let’s consider non-Square matrices or Rectangular matrices: Few Facts about Rectangular Matrices  Consider the following system of linear Equation: Converting the system to augmented matrix ( | ) It is a tall matrix with dimension 3×2 {Number of variables are less than the number of equations, i.e. variables and equations}. After applying Gaussian Elimination Method we have ( | ) ( | )  If then the system has unique solution. By back substitution we have: ( ) ( ) ( )