Consider the polynomials $p_1(t) = 1 + t^2$ and $p_2(t) = 1 − t^2$. Is ${p1, p2}$ a linearly independent set in $\mathbb{P_3}$? Why or why not?

Solution: Recall $\mathbb{P_3}$ is the vector space of polynomials with degree $ \leq 3$. Neither polynomial is a multiple of the other, so the answer is no.

I thought the answer would be "no" because there is no $t^3$ in the polynomials. I don't see how that "neither polynomial is a multiple of the other" relates because wouldn't the fact that $p_1$ and $p_2$ are not multiples of each other make it linearly independent?

  • $\begingroup$ $t$ and $t^2$ are linearly independent even though there is not $t^3$ term. $\endgroup$ – user223391 Apr 13 '17 at 18:26

The set $\{1,t,t^2,t^3\}$ is a basis for $P_3$. So, we can write $p_1$ and $p_2$ in coordinates:

$$p_1=(1,0,1,0) , \qquad p_2=(1,0,-1,0)$$

Now, $(1,0,1,0)$ and $(1,0,-1,0)$ are linearly independent since they are not proportional.

You can approach the problem without using a basis, but I think it's simpler to visualize the space $P_3$ this way. Suppose $p_1$ and $p_2$ are proportional, that is $p_1(x)=ap_2(x)$ for $a\in \mathbb R$. Then $p_1(x)=0$ if and only if $p_2(x)=0$, which clearly doesn't occur, since $p_1$ has no real roots.

  • $\begingroup$ Thank you! In your solution using the basis, since we see that $(1,0,1,0)$ and $(1,0,1,0)$ are linearly independent and sets in $\mathbb{P_3}$ why is the ultimate answer "no" for the problem? $\endgroup$ – stumped Apr 13 '17 at 18:34
  • $\begingroup$ I think that should be a mistake. Maybe they were asking if $\{p_1, p_2\}$ is a linearly dependent set. Anyway, the reasoning of the solution you provided leads to conclude they are linearly independent. $\endgroup$ – A. Salguero-Alarcón Apr 13 '17 at 18:37

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