Lecturenotes16-18March2011

Lecturenotes16-18March2011 - Relation of row space and...

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Relation of row space and column space of A, to Ax=b Theorem 4.7.1: A system of linear equations Ax=b is consistent if and only if b is in the column space of A. Idea: Consider Ax = b where x = x 1 . . . x n , A = ± c 1 ... c n ² . c 1 ,...,c n denotes the columns of A. Then we can write Ax = x 1 c 1 + x 2 c 2 + ... + x n c n = b so if Ax=b is consistent this means there are x 1 ,x 2 ,...,x n satisfying the above equation. This means b is in the column space of A. On the other hand if b is in the column space of A then it means we can write b = c 1 x 1 + ... + c n x n which means Ax=b has a solution. Theorem 4.8.5: If Ax=b is a consistent linear system of m equations in n unknowns, and if A has rank r, then the general solution of the system contains n-r parameters. Idea: rank(A)=r=number of leading variables, hence there will be n-r free variables. Definition: If W is a subspace of R n , then the set of all vectors in R n that are orthog- onal to every vector in W is called the orthogonal complement of W and is denoted by W . Example: (a) Consider the plane x + 2 y + 3 z = 0 . This plane is a vector space. Its orthogonal complement is the line represented by the vector equation x = t (1 , 2 , 3) where t R . (b) L1: x=t(1,2) and L2: t(-2,1) are vector spaces orthogonal complement of each other. Example: Find a basis and dimension of W where W is a plane in R 3 passing through origin and parallel to the vectors v 1 = (1 , 2 , 3) and v 2 = (4 , 0 , 2) . Solution: W = { x = t 1 v 1 + t 2 v 2 | t 1 ,t 2 R } , then W = { v R 3 | v · x = 0 , for all x W } . 1st way: We want
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Lecturenotes16-18March2011 - Relation of row space and...

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