Row space | Column Space | Null space | Rank | Nullity

Row space | Column Space | Null space | Rank | Nullity

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  Gandhinagar Institute of Technology(012) Subject: VCLA (2110015) Active Learning Assignment Branch : Computer DIV. : AG2 Prepared by : - Vishvesh jasani (160120107042) Guided By: Prof. Priya Jani Topic Name: Row Space,Column Space & Null space,Rank & Nullity
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  3  Row space: Therow space of A is the subspace of Rn spanned by the row vectors of A.  Column space: The column space of A is the subspace of Rm spanned by the column vectors of A.   },,{ 21 )((2) 2 (1) 1 RAAAACS n n n    }|{)( 0xx  ARANS n  Null space: The null space of A is the set of all solutions of Ax=0 and it is a subspace of Rn. Let A be an m×n matrix Raw space ,column space and null space The null space of A is also called the solution space of the homogeneous system Ax = 0.
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                                  mmnmm n n A A A aaa aaa aaa A      2 1 21 22221 11211 )( (2) (1) ],,,[ ],,,[ ],,,[ nmnm2m1 2n2221 1n1211 Aaaa Aaaa Aaaa        Row vectors of A Row Vectors:        n mnmm n n AAA aaa aaa aaa A      21 21 22221 11211                                                   mn n n mm a a a a a a a a a    2 1 2 22 12 1 21 11 Column vectors of A Column Vectors: || || || A (1) A (2) A (n)
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  5 THEOREM 1 Elementary rowoperations do not change the null space of a matrix. THEOREM 2 The row space of a matrix is not changed by elementary row operations. RS(r(A)) = RS(A) r: elementary row operations THEOREM 3 If a matrix R is in row echelon form, then the row vectors with the leading 1′s (the nonzero row vectors) form a basis for the row space of R, and the column vectors with the leading 1′s of the row vectors form a basis for the column space of R.
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  6 Find a basisof row space of A =                    2402 1243 1603 0110 3131 Sol:                    2402 1243 1603 0110 3131 A=                 0000 0000 1000 0110 3131 3 2 1 w w w B = .E.G bbbbaaaa 43214321 Finding a basis for a row space EXAMPLE A basis for RS(A) = {the nonzero row vectors of B} (Thm 3) = {w1, w2, w3} = {(1, 3, 1, 3), (0, 1, 1, 0), (0, 0, 0, 1)}
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  7 THEOREM 4 If Aand B are row equivalent matrices, then: (a) A given set of column vectors of A is linearly independent if and only if the corresponding column vectors of B are linearly independent. (b) A given set of column vectors of A forms a basis for the column space of A if and only if the corresponding column vectors of B form a basis for the column space of B. If a matrix A is row equivalent to a matrix B in row-echelon form, then the nonzero row vectors of B form a basis for the row space of A. THEOREM 5
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  8 Finding a basisfor the column space of a matrix                    2402 1243 1603 0110 3131 A Sol: 3 2 1 .. 00000 11100 65910 23301 21103 42611 04013 23301 w w w                                 BA EGT EXAMPLE
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  9 CS(A)=RS(AT) (a basis forthe column space of A) A Basis For CS(A) = a basis for RS(AT) = {the nonzero vectors of B} = {w1, w2, w3}                                                                        1 1 1 0 0 , 6 5 9 1 0 , 2 3 3 0 1
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  10 Finding the solutionspace of a homogeneous system Ex: Find the null space of the matrix A. Sol: The null space of A is the solution space of Ax = 0.              3021 4563 1221 A                          0000 1100 3021 3021 4563 1221 .. EJG A x1 = –2s – 3t, x2 = s, x3 = –t, x4 = t 21 vvx tsts t t s ts x x x x                                                          1 1 0 3 0 0 1 232 4 3 2 1 },|{)( 21 RtstsANS  vv
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  11 Rank and Nullity IfA is an mn matrix, then the row space and the column space of A have the same dimension. dim(RS(A)) = dim(CS(A)) THEOREM The dimension of the row (or column) space of a matrix A is called the rank of A and is denoted by rank(A). rank(A) = dim(RS(A)) = dim(CS(A)) Rank: Nullity: The dimension of the null space of A is called the nullity of A. nullity(A) = dim(NS(A))
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  12 THEOREM If A isan mn matrix of rank r, then the dimension of the solution space of Ax = 0 is n – r. That is n = rank(A) + nullity(A) • Notes: (1) rank(A): The number of leading variables in the solution of Ax=0. (The number of nonzero rows in the row-echelon form of A) (2) nullity (A): The number of free variables in the solution of Ax = 0.
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  13 Rank and nullityof a matrix Let the column vectors of the matrix A be denoted by a1, a2, a3, a4, and a5.                  120930 31112 31310 01201 A a1 a2 a3 a4 a5 EXAMPLE Find the Rank and nullity of the matrix. Sol: Let B be the reduced row-echelon form of A.                                  00000 11000 40310 10201 120930 31112 31310 01201 BA a1 a2 a3 a4 a5 b1 b2 b3 b4 b5 235)(rank)(nuillity  AnA G.E. rank(A) = 3 (the number of nonzero rows in B) 235)(rank)(nuillity  AnA