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Lecturenotes2March2011

Lecturenotes2March2011 - Theorem 4.4.1(Uniqueness of Basis...

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Theorem 4.4.1 (Uniqueness of Basis Representation) If S = { v 1 , v 2 , . . . , v n } is a basis for a vector space V, then every vector v in V can be expressed in the form v = c 1 v 1 + c 2 v 2 + . . . + c n v n in exactly one way. Proof: Assume there is another expression that is v = d 1 v 1 + d 2 v 2 + . . . + d n v n . Then d 1 v 1 + d 2 v 2 + . . . + d n v n = c 1 v 1 + c 2 v 2 + . . . + c n v n ( d 1 - c 1 ) v 1 +( d 2 - c 2 ) v 2 + . . . +( d n - c n ) v n = 0 Since { v 1 , . . . , v n } is linearly independent set, d i = c i for i = 1 , . . . , n Definition: If S = { v 1 , . . . , v n } is a basis for a vector space V, and v = c 1 v 1 + c 2 v 2 + . . . + c n v n is the expression for a vector v in terms of the basis S, then the scalars c 1 , . . . , c n are called the coordinates of v relative to the basis S. The vector ( c 1 , c 2 , . . . , c n ) in R n constructed from these coordinates is called the coordinate vector v relative to S, denoted as ( v ) S = ( c 1 , c 2 , . . . , c n ) Example: Find the coordinate vector of v = (2 , - 1 , 3) relative to the basis S = { (1 , 0 , 0) , (2 , 2 , 0) , (3 , 3 , 3) } . Solution: We want to find ( c 1 , c 2 , c 3 ) such that (2 , - 1 , 3) = c 1 (1 , 0 , 0)+ c 2 (2 , 2 , 0)+ c 3 (3 , 3 , 3) From this equation we see that c 1 = 3 , c 2 = - 2 , c 3 = 1 . Hence ( v ) S = (3 , - 2 , 1) . Definition: A nonzero vector space V is called finite dimensional if it contains a finite set of vectors { v 1 , v 2 , . . . , v n } that forms a basis. If no such set exists, V is called infinite dimensional. Example: 1. R n is finite dimensional 2. A line passing through origin in R n is finite dimensional 3. M 22 is finite dimensional 4. P vector space of all polynomials with real coefficients is infinite dimensional 5. Vector space of real-valued functions is infinite dimensional 1
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Section 4.5 Dimension To define dimension of a vector space we need the following two theorems: Theorem 4.5.2:
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