By Prabhat Choudhary

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33 Systems of Linear Equations From the above analysis of pivots we get several very important corollaries. The main observation. In echelon form, any row and any column have no more than 1 pivot in it (it can have 0 pivots) Corollaries about Linear Independence and Bases Questions as to when a system of vectors in ~n is a basis, a linearly independent or a span'1ing system, can be easily answered by the row reduction. Proposition. • Vm E ~n. • vm. Then 1. The system vI' v2, ... , vm is linearly independent i echelonform of A has a pivot in every column; ~n 2.

Let A be an an m x n matrix. Then the equation Ax = b has a solution for every b ATx=O E lR m if and only if the dual equation 46 Systems of Linear Equations has a unique (only the trivial) solution. (Note, that in the second equation we have AT, not A). Proof The proof follows immediately from Theorem by counting the dimensions. There is a very nice geometric interpretation of the second rank theorem. Namely, statement 1 of the theorem says, that if a transformation a: IR n ~ IRtn has trivial kernel (KerA = {O}), then the dimensions of the domain Rn and of the range Ran A coincide.

Systems of Linear Equations 43 The null space KerA. , found all the solutions, then any vector in Ker A is a linear combination of the vectors we obtained. Thus, the vectors we obtained form a spanning system in Ker A. To see that the system is linearly independent, let us multiply each vector by the corresponding free variable and add everything. Then for each free variable x k , the entry number k of the resulting vector is exactly x k' so the only way this vector (the linear combination) can be 0 is when all free variables are O.