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# week04 - Week 4(2.1 Rn and Vector Spaces finish 2.2...

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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 , 0); - x = - ( x, y, z ) = ( - x, - y, - z ) . standard unit vectors i, j, k. x = ( x, y, z ) = ( x, 0 , 0) + (0 , y, 0) + (0 , 0 , z ) = xi + yj + zk.

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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 , . . . , 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 0 1 2 3 0 - 1 - 2 - 3 1 0 - 8 - 8 0 1 2 3 0 0 0 0 = A R in Reduced Row Echelon Form (RREF). ————- Last Steps: for A R = 1 0 - 8 - 8 0 1 2 3 0 0 0 0 x 1 and x 2 are the bound variables (since there are leading 1’s 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 , 0 , 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. Postscript. We are asked two other questions related to the homogeneous system in 2.4. First, as we have already discussed, the Rank is the number of non-zero rows in the RREF. We also feature the non-zero rows themselves, which are called the basis of the Row Space of A, for reasons described later (shortly!). ———-

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4 V any collection, elements v V called vectors.
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