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Consider the $\mathbb{R}$-vector space $P_3(\mathbb{R})$ of real polynomials $$a_0+a_1X+a_2X^2,$$ of degree $\leq 2$. Which of the following collections of elements span $P_3(\mathbb{R})$?

  1. $1,X,X^2$

  2. $1,X,X^2-1$

  3. $1-X,X-X^2,X^2-1$

  4. $1,X^2-X$

My book doesn't explain span very well when it comes to polynomials, so I'm having some trouble here.. How do I solve this?

EDIT: $P_3(\mathbb{R})$, not $P_2(\mathbb{R})$ because that is the convention that our professor uses.

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    $\begingroup$ Denote the polynomial $a_0+a_1X+a_2X^2$ by the vector $(a_0,a_1,a_2)$. For example, $1=(1,0,0),X=(0,1,0),X-1=(-1,1,0)$ and so on. Then all you have to do is find out which set of vectors spans $\Bbb R^3$. $\endgroup$ – Shubham Johri May 20 '19 at 9:23
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You can simply study the linear systems that need to be solved in order to get the coefficients on each base.

(1) obviously works and (4) obviously doesn't work. For instance for (2) what you want to know is if every polynomial $a_0+a_1 X + a_2 X^2$ can be written in the form $b_0+ b_1 X+ b_2(X^2-1)$, which amounts to solve the linear system $$ \begin{cases} b_0 - b_2 = a_0\\ b_1 = a_1 \\b_2 = a_2\end{cases}\Leftrightarrow \begin{cases} b_0 = a_0 + a_2\\ b_1 = a_1 \\b_2 = a_2\end{cases}. $$

Since the system has a single solution, you conclude that every second degree polynomial can be uniquely written as a linear combination of $1, X, X^2 -1$, which means that it forms a basis.

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Hint: Since $\{1, x, x^2\}$ is a basis of $P_2(\mathbb{R})$, write your elements in that basis (they already are). Then, form a matrix using the coefficients. They span $P_2(\mathbb{R})$ iff the matrix formed is non-singular. Since option $4$ doesn't have $3$ elements, you can rule it out.

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You have to remember the dimension of a vector space is the minimal number of elements in a spanning set. $P_2(\mathbf R)$ has dimension $3$, and is obviously generated by the set $\{1,X,X^2\}$. This eliminates 4, which has only two elements.

As to 2 and 3, you can calculate their determinants: if they're a spanning set with $3$ elements, they're a basis, and their determinant must be nonzero.

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$1,X^{2}-X$ cannot span a three dimensional space.

$1-X,X-X^{2},X^{2}-1$ are linearly dependent (because if you add the first and the third and multiply by $-1$ you get the second), so they cannot span the spacee.

The first and the second sets do span the space. Note that $a_0+a_1X+a_2X^{2}= (a_0+a_2)+a_1X+a_2(X^{2}-1)$.

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Hint: A polynomial $a+bX+cX^2$ in $P_2({\Bbb R})$ can be canonically viewed as a vector $$\left(\begin{array}{c}a\\b\\c\end{array}\right)$$ in ${\Bbb R}^3$. Moreover, addition of polynomials is term-wise which corresponds to addition of vectors component-wise. In this way, the space $P_2({\Bbb R})$ is an ${\Bbb R}$-vector space in a very natural sense. All notions such as ''span'' carry over. With this correspondence, the exercise should be doable.

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