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**Unformatted text preview: **Week 4: (2.1 R n and Vector Spaces - finish) 2.2 Subspaces/Spanning 2.3 Independence/Bases 2.4 Nullspaces 1. Vector addition; scalar multiplication in R 2 . Vectors are x = ( x, y ) , with x, y real numbers. ( x 1 , y 1 ) + ( x 2 , y 2 ) = ( x 1 + x 2 , y 1 + y 2 ). parallelogram law: (2 , 1) + (1 , 4) = (3 , 5) (picture!) k ( x, y ) = ( kx, ky ) , scaling factor 3(2 , 1) = (6 , 3);- 1 2 (2 , 1) = (- 1 ,- 1 2 ) . (sketch) zero vector: 0 = (0 , 0) . additive inverse:- x =- ( x, y ) = (- x,- y ) . distributive rules (2) - standard unit vectors: i = (1 , 0) , j = (0 , 1) . Linear combination property: x = ( x, y ) = ( x, 0) + (0 , y ) = xi + yj. R 3 : x + y, k x, 0 = (0 , , 0);- x =- ( x, y, z ) = (- x,- y,- z ) . standard unit vectors i, j, k. x = ( x, y, z ) = ( x, , 0) + (0 , y, 0) + (0 , , z ) = xi + yj + zk. 2 R n : x + y = = ( x 1 , x 2 , . . . , x n ) + ( y 1 , y 2 , . . . , y n ) ( x 1 + y 1 , . . . , x n + y n ) . k x = k ( x 1 , x 2 , . . . , x n ) = ( kx 1 , . . . , kx n ) 0 = (0 , , . . . , 0) ,- x =- ( x 1 , x 2 , . . . , x n ) = (- x 1 ,- x 2 , . . . ,- x n ) . Problem Express the solutions of Ax = 0 as a subset of R 4 for A = 1 3- 2 1 3 10- 4 6 2 5- 6- 1 Solution: A 1 3- 2 1 1 2 3- 1- 2- 3 1- 8- 8 1 2 3 = A R in Reduced Row Echelon Form (RREF). - Last Steps: for A R = 1- 8- 8 1 2 3 x 1 and x 2 are the bound variables (since there are leading 1s in col 1 and in col 2); so x 3 = s and x 4 = t are the free variables. Use the i th row of A R to solve for the i th bound variable: 3 x 1- 8 x 3- 8 x 4 = 0 , so x 1 = 8 s + 8 t, and x 2 + 2 x 3 + 3 x 4 = 0 , so x 2 =- 2 s- 3 t ; so ( x 1 , x 2 , x 3 , x 4 ) = (8 s + 8 t,- 2 s- 3 t, s, t ) , for s, t R . Note 1: We reduce A rather than A # = ( A | 0) , but the equations correspond to the augmented matrix ( A R | 0) . Note 2: (Pre-view of 2.4) Using vector addition and scalar mult in R 4 , we say that the solutions (Nullspace) are spanned by the coefficient vectors (8 ,- 2 , 1 , 0) and (8 ,- 3 , , 1) and that these two vectors are a spanning set for the solution space. Properties of the free variables x 3 and x 4 will show that these two vectors are independent, so we say that these two vectors are a basis for the solutions....

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