5 Standard R n the set of real n 1 matrices X ie column n vectors is likewise

5 standard r n the set of real n 1 matrices x ie

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5. “Standard” R n , the set of real [ n , 1] matrices X (i.e. column n -vectors), is likewise the first example of a real vector space , i.e a vector space taken with real scalars.
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8 CHAPTER 1. LINEAR ALGEBRA (APPROX. 9 LECTURES) The central theme here is that many important examples exist which are not naturally a C n or R n . 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 different! 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 ( αf + βg )( x ) = αf ( x ) + βg ( x ) for all x defines a continuous function “ αf + βg ” for any complex numbers α , β . 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 = . 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 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 0 } ⊂ 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 + xy + ( 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, ) . To check this note that L [ αy 1 + βy 2 ] = α L [ y 1 ] + β L [ y 2 ] , so if y 1 , y 2 solve the ODE (1.1) , then αy 1 + βy 2 also solves the ODE. Definition 4. Span Given x 1 , . . . x m V define span ( x 1 , . . . x m ) = set of all linear combinations m n =1 α n x n .
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1.3. BASIS 9 Fact: span is always a subspace! Example 1. It will turn out that every solution of (1.1) is of the form y ( x ) = α 1 J n ( x ) + α 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 ) 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 span ( u 1 , . . . u m ) X = m i =1 α i u i for some a = ( α 1 , . . . , α m ) T X = Ua for some a = ( α 1 , . . . , α m ) T the augmented matrix U X is consistent.
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