Find $a_1,a_2,a_3\in\mathbb{R}$ such that vectors $e_i=(x-a_i)^2,i=1,2,3$ form a basis for $\mathcal{P_2}$ (space of polynomials)


$e_1= \begin{bmatrix} 1 \\ -2a_1 \\ a_1^{2} \\ \end{bmatrix}$

$e_2= \begin{bmatrix} 1 \\ -2a_2 \\ a_2^{2} \\ \end{bmatrix}$

$e_3= \begin{bmatrix} 1 \\ -2a_3 \\ a_3^{2} \\ \end{bmatrix}$

Matrix of column vectors $e_i$ is $A= \begin{bmatrix} 1 & 1 & 1 \\ -2a_1 & -2a_2^2 & -2a_3^2 \\ a_1^2 & a_2^2 & a_3^2 \\ \end{bmatrix}$

$A$ must be invertible $\Rightarrow \det(A)\neq 0$


Question: How to factorize this function to get the conditions for $a_1,a_2,a_3$?

Every vector $e_i$ can be written as a linear comination of $e_i\Rightarrow e_i$ span $\mathcal{P_2}$

If $A$ is invertible, $e_i$ are linearly independent.

From the span and linear independence, set of $e_i$ form a basis for $\mathcal{P_2}$.

  • 2
    $\begingroup$ What is $x$? If it's a real, how can the square of the difference of two reals be a vector? If it's a vector - how do you define the difference of a vector and a real and the square of this? $\endgroup$
    – Roland
    Dec 23 '15 at 16:31
  • 1
    $\begingroup$ @user300044 Squaring a vector doesn't make sense $\endgroup$ Dec 23 '15 at 16:33
  • $\begingroup$ Look at $e_i$ as polynomials with real variable $x$. $\endgroup$
    – user300048
    Dec 23 '15 at 16:35
  • $\begingroup$ The determinant should be divisible by $a_2-a_3$ and $a_1-a_2$ and $a_3-a_1$ since it's obviously zero if any of those are. $\endgroup$ Dec 23 '15 at 16:35
  • $\begingroup$ @Milo Brandt Thanks, but how to factorize $\det(A)$? $\endgroup$
    – user300048
    Dec 23 '15 at 16:37


Write $\det (A)=0$ as an equation in $a_1$ : $$ a_1^2(a_3-a_2)+a_1(a_2^2-a_3^2)+a_2a_3(a_3-a_2)=0 $$

Solve for $a_1$ this second degree equation, with $a_2$ and $a_3$ as parameter.

We have the discriminant $\Delta=(a_2-a_3)^4$ and the solutions, for $a_2 \ne a_3$, are $a_1=a_2$ or $a_1=a_3$ . Now you can find when the three column vectors are linearly independent and, as you have noted, they are a basis in$\mathbb{R}^3$.

Note that this result is the same as suggested by Milo Brandt in his comment (now I see it !).


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