prove that a set forms a basis for other set

Question

Prove that $ S:=\{1 + x, x − x^2,2 + 3x^2\}$ forms a basis for $P_2 =\{a+bx+cx^2:a,b,c\inℝ\}$.

Showing that the $P_2=\text{span(S)}$ is trivial. While showing that the elements of S are linearly independent is a bit tricky technically $\exists,c_1,c_2,c_3$ s.t $$c_1*(1+x)+c_2(x-x^2)+c_3(2+3x^2)\neq0$$ One possible solution I think of is bringing it to quadratic equation form and showing that the discriminant<0 i.e. above statement is true. However, it also does not give the desired result. Is there better solution? Also, I don't notice that any of those elements are linear combinations of the other two. Need some hints

Answer 1

$$c_1(1+x)+c_2(x-x^2)+c_3(2+3x^2)=0\iff (3c_3-c_2)x^2+(c_2+c_1)x+(2c_3+c_1)=0$$

But we know that a polynomial is the zero polynomial iff all its coefficients are zero, so we get:

$$\begin{cases}3c_3-c_2=0\\c_2+c_1=0\\2c_3+c_1=0\end{cases}$$

Prove now that the above three equalities force $\;c_1=c_2=c_3=0\;$ and thus the given set of vectors is linearly independent and thus a basis of $\;\Bbb P_2\;$

Answer 2

You want to prove that $$c_1(1+x)+c_2(x-x^2)+c_3(2+3x^2)=0 \ \Rightarrow \ c_1=c_2=c_3=0$$ If you expand, you get $$(3c_3-c_2)x^2+(c_1+c_2)x+(c_1+2c_3)=0$$ and taking into account that a polynomial is $0$ if, and only if, its coefficients are $0$, you end up with the system of equations $$\begin{cases} 3c_3-c_2=0 \\ c_1+c_2=0 \\ c_1+2c_3=0\end{cases}$$ Now, you want to prove that the previous system has only the trivial solution (that is, $c_1=c_2=c_3=0$). That can be done by checking that the determinant of the coefficient matrix is not zero: $$\left|\begin{matrix} 0 & -1 & 3 \\ 1 & 1 & 0 \\1 & 0 & 2 \end{matrix}\right|=3\neq 0$$

Answer 3

$$ \begin{bmatrix} 0 & -x^2 & 3x^2 \\ x & x & 0 \\ 1 & 0 & 2 \\ \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}$$

is the matrix representation of the system of equations we need to have. Here, if we define $A = \begin{bmatrix} 0 & -x^2 & 3x^2 \\ x & x & 0 \\ 1 & 0 & 2 \\ \end{bmatrix}, c = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ \end{bmatrix}$ if $\det(A) = 0$ for all $x$, then we have a non-zero solution for vector $c$ hence the set $S$ is linearly dependent. However, we can find out that $\det(A) = x^2 = 0$ iff $x = 0$, which implies that $c = 0$ and $S$ is a linearly independent set.

Answer 4

Your statement about 'linearly independent', that "there exist $c_1$, $c_2$, $c_3$ such that $c_1(1+ x)+ c_2(x- x^2)+ c_3(2+ 3x^2)\ne 0$" is incorrect. "Linearly independent" means that the only $c_1, c_2, c_3$ such that $c_1(1+ x)+ c_2(x- x^2)+ c_3(2+ 3x^2)= 0$ (for all x) are $c_1= c_2= c_3= 0$. To prove that, do the multiplication to get $c_1+ c_2x+ c_2x- c_2x^2+ 2c_3+ 3c_3x^2= (c_1+ 3c_3)+ c_2x+ (3c_3- c_2)x^2= 0$. Taking x= 0 that becomes $c_1+ 3c_3= 0$. Taking x= 1, $c2+ 3c_3- c_2= 3c_3= 0$ and taking x= -1, $-c_2+ 3c_3- c_2= 3c_3- 2c_2= 0$. From the second equation, $c_3= 0$ and then the first and second equations become $c_1= 0$ and $-c_2= 0$ so that $c_2= 0$.