## Practice Midterm Solution #6 |
aleksanderjanczewski |
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To show that some some basis is a basis for set of polynomials of degree less than or equal to 2. The given basis {x+1,x-2,$x^{2}$+x}. There is an easy way to solve this problem. The given basis for $P_{2}$ can be converted to a basis in $R^{3}$. Say you assume a standard basis s={1,x,$x^{2}$}. Now express the basis in the problem in terms of a standard basis: $[x+1]_s$=[1,1,0] $[x-2]_s$=[-2,1,0] $[x^{2}+x]_s$=[0,1,1] Now, since we may assume that if the basis consists of linearly independent vectors, in any other basis (in this case in standard basis) it will also consist of linearly independent vectors. To check that, create a matrix M with a column vectors [$[x+1]_s$, $[x-2]_s$, $[x^{2}+x]_s$]. $M= \begin{bmatrix} 1 & -2 & 0 \\ 1 & 1 & 1 \\ 0 & 0 & 1 \\ \end{bmatrix}$ Now, row reduce to echelon form $rref(M)= \begin{bmatrix} 1 & -2 & 0 \\ 0 & 1 & 1/3 \\ 0 & 0 & 1 \\ \end{bmatrix}$ Since there is a pivot in each of the column vectors, the set is linearly independent. Since there are three independent vectors in the basis, assuming that we talk about basis for polynomial space, the basis must span a space $P_{2}$ - basis for set of polynomials of degree less than or equal to 2. |

danielkim: May 5, 2015, 8:24 p.m. |
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excellent explanation! |

aleksanderjanczewski: May 5, 2015, 9:33 p.m. |
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Thank you, I forgot to mention that this also applies to the second part of the question, all you have to do is create a matrix using column vectors and then row reduce to echelon form. Again if the number of pivots in the matrix is equal to the number of columns then all vectors are independent. In case the number of pivots is less than the number of columns, you choose columns with pivots, and the column vectors associated with row reduced column vectors containing pivots will constitute the basis of the vector space spanned. The number of the independent vectors determines the dimension of the vector space spanned. |

ShengyueHuo: May 5, 2015, 9:40 p.m. |
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I do not understand why you choose s={1,x,x2} as a standard basis? Is that mean if I have to prove a set of polynomials of degree less that 3, I have to choose a standard basis like s = {1,x,x2,x3} ? |

aleksanderjanczewski: May 5, 2015, 10:03 p.m. |
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There was a mistake in this question, professor sent a correction, it was supposed to be a set of polynomials of degree less or equal to 2. |