02-28-03 - Lecture 12 Andrei Antonenko February 28, 2003 1...

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Lecture 12 Andrei Antonenko February 28, 2003 1 Examples of bases and dimensions Last lecture we stated the result that each basis has the same number of vectors. From this result very important corollary follows. Corollary 1.1. If the dimension of the vector space V is equal to n : dim V = n , then any n linearly independent vectors form a basis. Corollary 1.2. If the dimension of the vector space V is equal to n : dim V = n , then any n vectors which span all the vector space form a basis. Now we’ll consider some examples. Examples of bases in R 2 . Example 1.3. The easiest basis in R 2 is a basis u 1 = (1 , 0) , and u 2 = (0 , 1) . These vectors are linearly independent, and span R 2 . So, we have 2 vectors in the basis of R 2 , and thus dim R 2 = 2 . This particular basis is called standard basis . Now we can take any pair of linearly independent vectors and it will be a basis. Example 1.4 (Slight modification of standard basis). Let u 1 = (0 , 5) and u 2 = (1 , 0) . This slight modification of the standard basis is a basis itself, since it contains 2 vectors, and is linearly independent. Moreover, we can see that any vector ( a,b ) can be represented as a linear combination of u 1 and u 2 in the following way: ˆ a b ! = b 5 u 1 + au 1 = b 5 ˆ 0 5 ! + a ˆ 1 0 ! Example 1.5 (Less trivial example of basis). Let u 1 = (1 , 1) and u 2 = (0 , 1) . This is a basis since there are 2 vectors and they are linearly independent. We can check that these 1
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vectors are linearly independent by forming a linear combination which is equal to 0 and proving that this combination is trivial. x ˆ 1 1 ! + y ˆ 0 1 ! = ˆ 0 0 ! . This is equivalent to the following system which has unique zero solution: ( x = 0 x + y = 0 So, the linear combination should be trivial, and vectors are independent. Example 1.6 (Even more nontrivial example of basis for R 2 ). Consider the following pair of vectors u 1 = (1 , 2) , and u 2 = (3 , 5) . To check that this is a basis we have to prove that these 2 vectors are linearly independent. So, again we form a linear combination: x ˆ 1 2 ! + y ˆ 3 5 ! = ˆ 0 0 ! . and try to find x ’s and y ’s —coefficients of linear combination. In order for this set to be linearly dependent, coefficients should not be zeros simultaneously. The expression above is equivalent to the following system of linear equation.
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This note was uploaded on 05/28/2011 for the course AMS 2010 taught by Professor Andant during the Spring '03 term at SUNY Stony Brook.

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02-28-03 - Lecture 12 Andrei Antonenko February 28, 2003 1...

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