{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

s08week04 - Math 205 Spring 2008 B Dodson Week 4 1.6...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 205, Spring 2008 B. Dodson Week 4: 1.6 Properties of Determinants 2.1 R n and Vector Spaces 2.2 Subspaces/Spanning ———– 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.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 ) . ———- V any collection, elements v V called vectors. Formula for vector plus and scalar mult. Usually scalars are real k R ; may also have k C , complex. ( V , + , · ) is a Vector Space if the vector + and scalar mult satisfy 10 rules: 2 closure rules 4 rules for + (including 0; and - x ) 2 rules for · (including 1 · v = v ) 2 distributive laws. (2+4+2+2 = 10 or 2+8.) ——- Examples: R n ; V = M 2 × 2 ( R ); F ( a, b ) = functions: f+g, k · f, 0, -f. P k = polynomials with real coef. degree < k. Problem Is P = polynomials of degree exactly 2 a vector space? Solution. (closure?), zero vector?
Background image of page 2
3 (try p 1 ( x ) + p 2 ( x ) = ( x 2 + 3 x ) + ( - x 2 - 2).) 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).
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}