Consider the following problem:

Show that the following three polynomials in $\mathbb{R}[x]$: $${{x^2}/2}-x/2, 1-x^2,{{x^2}/2}+x/2$$ are linearly independent in $\mathbb{R}[x]$. Write the monomials $1, x, x^2$ as a linear combination of the polynomials.

My attempt:

Let $P_1(x)={{x^2}/2}-x/2,P_2(x)=1-x^2,$ and $P_3(x)={{x^2}/2}+x/2$.

Let $a,b,c$ be scalars. I'm trying to show that there isn't a collection of scalars $a,b,c$ (not all $0$) such that $$aP_1(x)+bP_2(x)+cP_3(x)=0$$ I'm don't know how to proceed after this, any help is welcome

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    $\begingroup$ Expand out the polynomials and then group together like terms in terms of $t.$ Two polynomials are equal iff their coefficients are equal, so each coefficient of your polynomial on the left-hand side must be zero. Now solve the resulting system. $\endgroup$ – Stephen Donovan May 12 at 19:15
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    $\begingroup$ Follow the suggestion in the question, presumably you know that $1,t,t^2$ are linearly independent? $\endgroup$ – copper.hat May 12 at 19:17
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    $\begingroup$ Hint: Let $\alpha=\frac x2$, $\beta=x-1$, and $\gamma=x+1$. Then, $P_1=\alpha\beta$, $P_2=-\beta\gamma$, $P_3=\alpha\gamma$. Why does that help? $\endgroup$ – Don Thousand May 12 at 19:23
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    $\begingroup$ In your equation $a P_1(x)+... = 0$, if you set $x=0$ what can you say about $b$? $\endgroup$ – copper.hat May 12 at 19:25
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    $\begingroup$ @user926356 Now differentiate $aP_1(x)+cP_3(x)$ and set $x=0$, this will give an equation for $a,c$. Repeat with one more differentiation. $\endgroup$ – copper.hat May 12 at 19:57


$a\left(\dfrac{x^2}2-\dfrac x2\right)+b\big(1-x^2\big)+c\left(\dfrac{x^2}2+\dfrac x2\right)=0$

$\left(\dfrac a2-b+\dfrac c2\right)x^2+\left(-\dfrac a2+\dfrac c2\right)x+b=0$

By applying the Identity Theorem for polynomials, we get that

$\begin{cases} \dfrac a2-b+\dfrac c2=0\\ -\dfrac a2+\dfrac c2=0\\ b=0 \end{cases}$

$\begin{cases} \dfrac a2+\dfrac c2=0\\ -\dfrac a2+\dfrac c2=0\\ b=0 \end{cases}$

$\begin{cases} a+c=0\\ -a+c=0\\ b=0 \end{cases}$

$\begin{cases} a+c=0\\ c=a\\ b=0 \end{cases}$

$\begin{cases} 2a=0\\ c=a\\ b=0 \end{cases}$

$\begin{cases} a=0\\ c=0\\ b=0 \end{cases}$



consequently the three polynomials $\;P_1(x)\;,\;P_2(x)\;,\;P_3(x)\;$ are linearly independent.



Note that $$\begin{pmatrix}1/2&-1/2&0\\-1&0&1\\1/2&1/2&0\end{pmatrix}\begin{pmatrix}x^2\\x\\1\end{pmatrix}=\begin{pmatrix}P_1(x)\\P_2(x)\\P_3(x)\end{pmatrix}$$

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    $\begingroup$ This is assuming that OP understands why the monomial basis is a basis, which, from the comments, is not clear. $\endgroup$ – Don Thousand May 12 at 19:25
  • $\begingroup$ Well, I have considered that possibility. If my answer is not useful, I'll just delete it. Nevertheless, it can be useful for someone else. +1 to your comment. $\endgroup$ – ajotatxe May 12 at 19:26
  • $\begingroup$ @ajotatxe I like that approach as well, it is efficient. Though the OP might not proceed that way for now, it might be interesting for the OP to keep that in mind. +1 $\endgroup$ – Axel May 12 at 19:36

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