The development of Linear Algebra

          

       Linear Algebra is the most useful and powerful tool in Mathematics. In the last blog "Linear Algebra - Purpose to Application" we saw how a higher-order system of linear equations has been solved by transforming the equation into the matrix.

                 Although it's good to mention that it only works if in the AX=B [A is m*n, X is n*1, B is m*1, so it has m equations and n unknowns ] our A matrix is invertible, if it is not invertible i.e, in the sense of matrix it is rank of A is less than n and rank of (A;B) = rank of A < n, so the system of linear equation AX=B has infinitely many solutions, we can find the solution by converting the A matrix into row-echelon form, simply you will get the solution from bottom.

        Note:- 1. The rank of a matrix - It is defined as the largest order square matrix whose determinant is non-zero.

                 2. Row-echelon form - We said a matrix is in a row echelon form if in the matrix following property is satisfied, 

      a) The first nonzero element in each row is 1,

      b) in each column containing the leading 1 of some row, the leading is the only non-zero element,

     c) all the non-zero rows should appear first then zero rows, 

     d) the i th non-zero row occurs in the k(i) th column of A then k(i) < k(i+1).

        You can see when you convert the matrix in row-echelon form, if the rank is less than n let's say r < n then the matrix has r non-zero rows so you will get r unknowns value and in other (n-r) unknown boxes you can put any number it doesn't matter so the conclusion we will get an infinite number of solutions.

 

  Q. What will happen if in AX=B we have B=0 i.e., AX=0 ? What can you tell about its solution, a) Always exist and unique, b) Always exist and but not unique, c) May not exist, d) never exist, e) other?

- Comment your answer below, you can explain why you think so.

        Till now I have tried to convince you why converting a system of linear equations into a matrix form is useful. But one thing you may notice is that although converting to a matrix may be reducing my work but still if you go for a much higher order, computing a matrix and working on it is not a tall joke. So here is the truth, although converting my problem to a matrix gives me a unified structure but it does not solve our problem. That's the primary reason for the huge development of Linear Algebra that leads us to real-life applications in various places.

        The way of approaching this problem is numerical analysis, there are several methods like Gauss Jordan elimination, LU decomposition and different iterative methods like Gauss-Seidel, and Jacobi method.

        So what we do is make computer programs in a programming language and use the above method algorithm to solve the problem. It's quite useful and using the iterative method we can get very less error-free solutions up to the required approximation. 

        Now we more or less know how we can solve the system of linear equations but our many questions are not resolved because only solving a linear equation doesn't give us many things. 

       Let me give you a scenario," Think you run a parcel service in your country. For simplicity think you only run this business in your country not outside and your country has a total 5 states, each state is different in its size, different geography, and different numbers of people use your service so you have different approaches so people use your service and your goal is to make your profit high, give best services at most minimum cost and also manage your employees and their salary and other transportation cost, packaging cost, other various cost."

       This kind of problem with given data can be converted into a system of linear equations and can be solved, although particularly this kind of problem is called LPP (linear programming problem) there are various ways to solve but you think you convert it into a system of a linear equation and solve. But only solving is not enough we have to understand in a business where we are spending much more and where less for that we have to analyze the solution in the given scenario, so we can make much more predictions about the system and make a more effective business model. 

 

          To analyze the solution we have to understand what it means by solution.

           So let again the same problem, AX=B [A is m*n, X is n*1, B is m*1, so it has m equations and n unknowns ] since one solution exists it has n unknowns so we can write in form (x(1),x(2),.....,x(n)), where each x(i) is a number belongs to space  (let's say for simplicity its ⁠, you can take other things also) that space with a following structure called vector space (V) and (x(1),x(2),.....,x(n)) called a vector (v).

           Let V is a nonempty set and (F,+,*) is a field, (+) is a binary composition on V [(+): V ✕ V -> V] and (.) be an external composition of F with V [(.): F ✕ V -> V]

     a) With respect to (+) it's a commutative group.

     b) c(.)v ∈ V

     c) c(.)(d(.)v)= (c*d)(.)v

    d) c(.)(u(+)v)=c(.)u(+)c(.)v

    e) (c+d)(.)v=c(.)v+d(.)v

    f) there is 1 in V such that 1(.)v =v

Above all true for all c,d ∈ F and u,v ∈ V

Then we called (V,F,+,*,(+),(.)) is a vector space, we called V is a vector space over F.

Example - (⁠, ⁠ ,+,*,(+),(.)) called ⁠ over ⁠ where +, (+) is addition of two real numbers and * is multiplication of real numbers and (.) scalar multiply with real number, you can check it satisfy all above condition.

(This example part was included on 21-08-2024)

So we get a well-defined powerful structure called vector space, in the next blog, we will see how useful it is and how linear algebra develops around it.