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02-26-03 - Lecture 11 Andrei Antonenko 1 Examples of bases...

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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....
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02-26-03 - Lecture 11 Andrei Antonenko 1 Examples of bases...

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