vault backup: 2022-08-18 19:14:44

content/notes/HW2.md

@@ -23,4 +23,4 @@ Since $P_2(\mathbb{R})$ has dimension 3, by default all bases of $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 $P_2(\mathbb{R})$
+To show linear independence, if $a(0 +2x+0x^2)+b(4+2x-x^2)+c(-4-6x+x^2)=(0+0x+0x^2)$, we have $4b-4c=0$, $2a+2b-6c=0$ and $-b+c = 0$. So $c=b$ which implies $b+4b = 0$ so $b=c=0$ which then implies $a=0$. 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 $P_2(\mathbb{R})$