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Let $\mathbb{R}_2[t]$ be the space of polynomials of one variable with real coefficients and degree less than or equal to two. Show that $p_1(t) = 1−t, p_2(t) = 1−t^2, p_3(t) = t$ form a basis for $\mathbb{R}_2[t]$.

Also, compute the coordinates of $q_1(t) = 2−5t + t^2$ and $q_2(t) = 3−6t + t^2$ with respect to such basis.

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closed as off-topic by B. Goddard, Parcly Taxel, Strants, Fabio Somenzi, Dietrich Burde Feb 14 '18 at 16:02

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Show that $p_1(t) = 1−t, p_2(t) = 1−t^2, p_3(t) = t$ form a basis for $\mathbb{R}_2[t]$

Depending on what you know and what you are allowed to use, this can be shown easily or it can require a bit more effort. If you know $\mathbb{R}_2[t]$ is three-dimensional, it suffices to show that $\left\{ p_1,p_2,p_3\right\}$ is linearly independent, so: $$\forall t : \alpha \left( 1−t \right)+\beta \left(1−t^2\right)+\gamma t = 0 \implies \alpha = \beta = \gamma = 0$$ If you explicitly need to show that $\left\{ p_1,p_2,p_3\right\}$ spans $\mathbb{R}_2[t]$, then you need to show that you can solve the following system for $\alpha,\beta,\gamma$ in terms of $a,b,c$ so that the equality holds for all $t$: $$at^2+bt+c = \alpha \left( 1−t \right)+\beta \left(1−t^2\right)+\gamma t$$

Also, compute the coordinates of $q_1(t) = 2−5t + t^2$ and $q_2(t) = 3−6t + t^2$ with respect to such basis.

For example, for $q_1$, solve for $\alpha,\beta,\gamma$: $$\forall t : \alpha \left( 1−t \right)+\beta \left(1−t^2\right)+\gamma t = 2−5t + t^2 \iff \left\{\begin{array}{rcl} \alpha + \beta & = & 2 \\ -\alpha+\gamma & = & -5 \\ -\beta & = & 1 \end{array}\right. \iff \ldots$$ Likewise for $q_2$.

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