1.8 KiB
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Assignment 2
Jet Hughes 9474308
1.
Let V = P_2(\mathbb{R}) with the ususal vector addition and scalar multiplication. For each of the following subsets of V, either prove that it is a basis of V or explain why it is not a basis of V . You may use any result from class.
(a) \{759, 20+2x+43x^2\}
(b) \{1-x, 2x^{2},3+x^2\}
(c) \{2x, 4+2x-x^{2}, -4-6x+x^2\}
(d) \{-1+3x, 1+x^{2,}x-3x^{2,}4+4x-11x^2\}
Since P_2(\mathbb{R}) has dimension 3, by default all bases of P_2(\mathbb{R}) have three elements. Hence (a) and (d) cannot possibly be bases of P_2(\mathbb{R})
(b) Since this set has 3 vectors and P_2(\mathbb{R}) has dimension 3, it is enough to check either that is is linearly independent or that it spans P_2(\mathbb{R}).
To show linear independence, if a(1-x)+b(2x^2)+c(3+x^2)=(0x^2 + 0x + 0), we have 2b+c=0, -a=0 and a+3c = 0. So a=0 which implies c=0 which then implies b=0. So the only linear combination equal to the zero vector is the one where a=b=c=0, hence this set in linearly independent. Since it is linearly independent and has three vectors, its span is a subspace of P_2(\mathbb{R}) of dimension 3, i.e., all of P_2(\mathbb{R})
(c) Since this set has 3 vectors and P_2(\mathbb{R}) has dimension 3, it is enough to check either that is is linearly independent or that it spans P_2(\mathbb{R}).
To show linear independence, if a(2x)+b(4+2x-x^2)+c(-4+6x+x^2)=(0x^2 + 0x + 0), we have 2b+c=0, -a=0 and a+3c = 0. So a=0 which implies c=0 which then implies b=0. So the only linear combination equal to the zero vector is the one where a=b=c=0, hence this set in linearly independent. Since it is linearly independent and has three vectors, its span is a subspace of P_2(\mathbb{R}) of dimension 3, i.e., all of P_2(\mathbb{R})