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Chapter 4: Vector Spaces - Part 4/Slides By Pearson

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Let Then A and B are row equivalent (You don\'t have to check this). For the following problems a, b, c, d, and e, justify your answers. Find rank A and dim Nul A. Find a basis for Col A. Find a basis for Col A. Find a basis for Row A Find a basis for Nul A. Solution (a) A and B are row equivalent.so their ranks are same. Thus rank(A) = rank(B) = 3 (first second and fifth columns of B are independent and other columns are in the span of these vectors) dim (nullA) = no of columns - rank(A) = 6 - 3 = 3 (b) a basis of column space of A are first second and fifth columns of A (since A and B are row equivalent and B has first second and fifth columns as basis of its column space) (c)basis of row space of A must contain three vectors(since rank = 3) Easy to see first and second rows are independent and third row is in their span(R3 = R1-R2) Hence R1,R2,R4 is a basis (d)A basis of null space of B is (3,0,0,0,0,-1) , (0,1,0,-1,0,0) , (0,0,1,0,0,1). Since these are in null space of B,are independent,and dimension of nullspace of B is 3 Now A and B are row equivalent So a basis of A is (3,0,0,0,0,-1) , (0,1,0,-1,0,0) , (0,0,1,0,0,1) .

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- Let Then A and B are row equivalent (You don't have to check this). For the following problems a, b, c, d, and e, justify your answers. Find rank A and dim Nul A. Find a basis for Col A. Find a basis for Col A. Find a basis for Row A Find a basis for Nul A. Solution (a) A and B are row equivalent.so their ranks are same. Thus rank(A) = rank(B) = 3 (first second and fifth columns of B are independent and other columns are in the span of these vectors) dim (nullA) = no of columns - rank(A) = 6 - 3 = 3 (b) a basis of column space of A are first second and fifth columns of A (since A and B are row equivalent and B has first second and fifth columns as basis of its column space) (c)basis of row space of A must contain three vectors(since rank = 3) Easy to see first and second rows are independent and third row is in their span(R3 = R1-R2) Hence R1,R2,R4 is a basis (d)A basis of null space of B is (3,0,0,0,0,-1) , (0,1,0,-1,0,0) , (0,0,1,0,0,1). Since these are in null space of B,are independent,and dimension of nullspace of B is 3 Now A and B are row equivalent So a basis of A is (3,0,0,0,0,-1) , (0,1,0,-1,0,0) , (0,0,1,0,0,1)