Question Why is C 1 not the same as R 2 Answer The set is the same namely pairs

# Question why is c 1 not the same as r 2 answer the

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Question: Why is C 1 not the same as R 2 ? Answer: The “set” is the same, namely pairs ( a , b ) of real numbers, but the scalars are di ff erent! Question: Why is V = C [ a , b ], the set of continuous functions from the interval [ a , b ] to C , a complex vector space? Answer: To check closure under both addition and scalar multiplication note: if f , g are continuous functions then ( f + g )( x ) = f ( x ) + g ( x ) for all x defines a continuous function “ f + g ” for any complex numbers , . From this we can easily check that C [0, 1] satisfies the remaining algebraic properties of a vector space. Idea: Think of the x in f ( x ) as a continuous analogue of the i on a i where A = ( a i ) n i =1 . In some sense f is an n –vector in the case n = Œ . Notation: V = C [ a , b ] denotes the vector space of continuous functions on the interval [ a , b ]. Idea: : Many vector spaces W arise as subsets of a larger vector space V . Definition 3. A subset W of a vector space V is called a subspace if it is 1. closed under addition; 2. closed under scalar multiplication. Every subspace is itself a vector space. The point of this is that all the vector space properties for a subset W œ V follow if we check only the two closure conditions at once. Examples 2. 1. V = R 3 . The full list of all subspaces of R 3 consists of: { ˛ 0 } ; all possible lines containing ˛ 0 ; all possible planes containing ˛ 0 ; and finally R 3 itself. 2. R + = { x Ø 0 } µ R . This is closed under addition but NOT closed under scalar multiplication, and is NOT a subspace. 3. Linear ODE with parameter n : L [ y ] = x 2 y ÕÕ + xy Õ + ( x 2 n 2 ) y = 0, for x > 0. (1.1) The solutions y ( x ) of (1.1) are called Bessel functions, and form a subspace of C (0, Œ ) . To check this note that L [ y 1 + y 2 ] = L [ y 1 ] + L [ y 2 ] , so if y 1 , y 2 solve the ODE (1.1) , then y 1 + y 2 also solves the ODE.

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1.3. BASIS 9 Definition 4. Span Given ˛ x 1 , . . . ˛ x m œ V define span ( ˛ x 1 , . . . ˛ x m ) = set of all linear combinations m ÿ n =1 n ˛ x n . Fact: span is always a subspace! Example 1. It will turn out that every solution of (1.1) is of the form y ( x ) = 1 J n ( x ) + 2 Y n ( x ) where J n is the Bessel function of the first kind of order n and Y n is the Bessel function of the second kind of order n . Therefore, the solution space of (1.1) is span ( J n , Y n ) . Typical problem in V = C n (or R n ) : how do you check if a given n –vector X lies in the span ( u 1 , . . . u m ) µ C n ? Idea: Build a matrix U out of the vectors u 1 , . . . u m : U = [ u 1 , u 2 , . . . , u m ] (1.2) U is a matrix with m columns and n rows. Then: X œ span ( u 1 , . . . u m ) X = m ÿ i =1 i u i for some a = ( 1 , . . . , m ) T X = Ua for some a = ( 1 , . . . , m ) T the augmented matrix Ë U - - - X È is consistent. Idea: The trick of writing m ÿ i =1 i u i = Ua (1.3) for a = ( 1 , . . . , m ) T is very helpful! 1.3 Basis Given a set { ˛ u 1 , . . . , ˛ u m } of vectors in V , the span W = span ( ˛ u 1 , . . . , ˛ u m ) is always a subspace of V . This means every ˛ x œ W can be written as ˛ x = m ÿ n =1 n ˛ u n (1.4) for some scalars 1 , m . However, some vector, say ˛ u i , may be redundant in the sense that ˛ u i is itself a linear combination of the remaining vectors in { ˛ u 1 , . . . , ˛ u m } . When this
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