## Practice Midterm #6 |
carlintou |
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For the first part of question #6, we are given a basis with 3 parts. Is it supposed to be "set of polynomials of degree less than or equal to 2" instead of 3? |

wllmskn: May 3, 2015, 7:37 p.m. |
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Yeah I think it would have to be, especially since none of the elements of the basis have $x^{3}$ in them. As it is, the basis is inadequate to describe the vector space $P_{3}$ |

dyana: May 4, 2015, 9:14 a.m. |
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For the second part of #6, is {(1,1,2,0), (1,0,3,1), (2,1,4,3), (1,-2,4,5)} not a basis because the last value is not linearly independent? I'm still a bit confused as to what makes a basis a basis. Each of the components need to be linearly independent right? |

davidabraham: May 4, 2015, 12:33 p.m. |
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Yes, each component of a basis has to be linearly independent. A basis is the smallest set of vectors with which you can reach any point in the vector space (using scalar multiples and/or adding them together). If one basis is not linearly independent, this means it can be made up of some manipulation of the other bases. In other words, it's not a necessary tool to reach any part of the vector space. The example given in class was giving directions: the basis could be something like (right, forward). Its unnecessary to have (right, left, forward, backward), because left is merely "negative right", and backwards is "negative forward." The basis could also be less simple (right - forward, 4forward), and would still be correct. A basis: (right, left) isn't a basis, because you can't go forward or backward (you can't map to the whole vector space). |

rsehrlich: May 4, 2015, 1:36 p.m. |
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If you want an easy way to check if a basis is made up of linearly independent coordinates, see if you can get it in reduced row echelon form. If one of the columns does not have a leading one, then something must be linearly dependent, and therefore it is not a basis. |

abbyto1607: May 4, 2015, 5:41 p.m. |
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How can we put the basis for the polynomial set into a matrix to row reduce and check for independence? |

abbyto1607: May 4, 2015, 5:47 p.m. |
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To add to my earlier comment, I attempted to apply the book definition of linear independence, where, $c_{1}v_{1}+c_{2}v_{2}+...+c_{n}v_{n}=0$ only when $c_{1},c_{2},...,c_{n}=0$ Anyone else try this? Does it work for this problem? |

Jerry_Zhongyang_Liu: May 4, 2015, 8:53 p.m. |
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It should be 2. I found the following definition on the 115A textbook: In Pn(F) the set { 1, x, x^2, ... x^n } is a basis. We call this basis the standard basis for Pn(F). It also makes sense to change it to "2" since a basis with the highest degree x^2 can only span P2(R). |

brian_li: May 5, 2015, 11:15 p.m. |
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Would someone mind explaining the first part of Problem 6? |