10-dimension

10-dimension - DIMENSION Homework Section 3.3...

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DIMENSION Math 21b, O. Knill Homework: Section 3.3: 22,24,32,38,52,36*,56* LINEAR SPACE. X is a linear space if ~ 0 X and if X is closed under addition and scalar multiplica- tion. Examples: R n , X = ker( A ) , X = im( A ) are linear spaces. REVIEW BASIS. B = { ~v 1 , . . . , ~v n } ⊂ X B linear independent : c 1 ~v 1 + ... + c n ~v n = 0 implies c 1 = ... = c n = 0. B span X : ~v X then ~v = a 1 ~v 1 + ... + a n ~v n . B basis : both linear independent and span. BASIS: ENOUGH BUT NOT TOO MUCH. The spanning condition for a basis assures that there are enough vectors to represent any other vector, the linear independence condi- tion assures that there are not too many vectors. A basis is, where J.Lo meets A.Hi: Left: J.Lopez in ”Enough” , right ”The man who new too much ” by A.Hitchcock AN UNUSUAL EXAMPLE. Let X be the space of polynomials up to degree 4. For example p ( x ) = 3 x 4 + 2 x 3 + x + 5 is an element in this space. It is straightforward to check that X is a linear space. The ”zero vector” is the function f ( x ) = 0 which is zero everywhere. We claim that e 1 ( x ) = 1 , e 2 ( x ) = x, e 3 ( x ) = x 2 , e 4 ( x ) = x 3 and e 5 ( x ) = x 4 form a basis in X . PROOF. The vectors span the space: every polynomial f ( x ) = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + c 4 x 4 is a sum f = c 0 e 1 + c 1 e 2 + c 2 e 3 + c 3 e 4 + c 4 e 5 of basis elements. The vectors are linearly independent: a nontrivial relation 0 = c 0 e 1 + c 1 e 2 + c 2 e 3 + c 3 e 4 + c 4 e 5 would mean that c 0 + c 1 x + c 2 x 2 + c 3 x 3 + c 4 x 4 = 0 for all x which is not possible unless all c j are zero. DIMENSION. The number of elements in a basis of
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