This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Lecture 11 Andrei Antonenko February 26, 2003 1 Examples of bases Last time we studied bases of vector spaces. Today we’re going to give some examples of bases. Example 1.1. Consider the vector space P 2 — the space of polynomials with degree less than or equal to 2. Let’s consider the following 3 vectors in this vector space: u 1 = t 2 + 1 , u 2 = t + 1 , u 3 = t 1 . Let’s determine whether it is a basis or not. We have to check 2 conditions: Spanning set To check that these vectors form a spanning set for P 2 we should take arbitrary vector from P 2 and try to express it as a linear combination of the vectors from the basis. Let’s take arbitrary polynomial at 2 + bt + c : at 2 + bt + c = x ( t 2 + 1) + y ( t + 1) + z ( t 1) = xt 2 + ( y + z ) t + ( x + y z ) . So, we can see that this is equivalent to the following system of linear equations, which we will try to solve: x = a y + z = b x + y z = c subtract the 1st eq. from the 3rd one ˆ x = a y + z = b y z = c a subtract the 2nd eq. from the 3rd one ˆ x = a y + z = b 2 z = c a b 1 So, we see that z = 1 2 ( a + b c ) , y = 1 2 ( b + c a ) , and x = a . So, we got the expression for arbitrary polynomial as a linear combination of given: at 2 + bt + c = a ( t 2 + 1) + 1 2 ( b + c a )( t + 1) + 1 2 ( a + b c )( t 1) . So, this system is a spanning set. Linear independence To check that these vectors are linearly independent we form a linear combination which is equal to 0: x ( t 2 + 1) + y ( t + 1) + z ( t 1) = 0 ⇔ xt 2 + ( y + z ) t + ( x + y z ) = 0 . This is equivalent to the following linear system: x = 0 y + z = 0 x + y z = 0 subtract the 1st eq. from the 3rd one ˆ x = 0 y + z = 0 y z = 0 subtract the 2nd eq. from the 3rd onesubtract the 2nd eq....
View
Full Document
 Spring '03
 Andant
 Linear Algebra, Vectors, Vector Space, basis, Linear combination

Click to edit the document details