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The polynomials are $(t-a)(t-b), \space(t-b)(t-c),\space (t-a)(t-c)$. I know that in order for them to be LI there need to be such coefficients $d1,\space d2, \space d3$ so that $\sum_{i=0}^{3} d_ip(x) = 0$ if at least one $d_i \neq 0$.

This leads me to the following equation: $$(d_1 +d_2 +d_3)t^2 - (d_1(a+b) +d_2(b+c) + d_3(a+c))t + (d_1ab + d_2bc + d_3ac) = 0$$

How do I proceed from here? I know the solution is $(a-b)(a-c)(b-c) \neq 0$, I just don't know how to get there.

Kris
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1 Answers1

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These quadratic polynomials are said to be linearly independent if and only if any quadratic polynomials $f(t)$ can be expressed as a linear combination of the three quadratic polynomials.

$f(t)=k_{1}(t-a)(t-b)+k_{2}(t-b)(t-c)+k_{3}(t-a)(t-c)$

Now consider a quadratic polynomial of which roots are not $a$ , $b$ , or $c$.

$f(a)=k_{2}(a-b)(a-c) \neq0$

$f(b)=k_{3}(b-a)(b-c) \neq0$

$f(c)=k_{1}(c-a)(c-b) \neq0$

Combine these and we obtain $(a-b)(b-c)(a-c)\neq 0$

acat3
  • 12,197