Given the following polynomials, determine whether the set $\{p_1, p_2, p_3\}$ forms a basis for the vector space $V_3$ of all polynomials of degree less than or equal to 3:

$$p_1(X) = X^3 - X +1$$ $$p_2(X) = X^3 - 1$$ $$p_3(X) = X^2 - X$$

So I need to determine whether $p_1, p_2, p_3$ are linearly independent and span $V_3$.

$p_1, p_2, p_3$ are linearly independent if for $\alpha_1p_1 + \alpha_2p_2 + \alpha_3p_3 = 0$ it follows that $\alpha_1,\alpha_2, \alpha_3 = 0$ is the only solution.

By plugging in the specific polynomials and setting the sum of $p_1, p_2, p_3$ equal to 0 I get: $$(\alpha_1 + \alpha_2)X^3 + \alpha_3X^2 + (-\alpha_1 - \alpha_3)X + (\alpha_1 - \alpha_2)X^0 = 0$$

I know that the set $\{X^3, X^2, X, X^0\}$ is a basis of $V_3$ and $X^3, X^2, X, X^0$ therefore are linearly independent, it follows that we have: $$\alpha_1 + \alpha_2= 0$$ $$\alpha_3 = 0$$ $$ -\alpha_1 - \alpha_3 = 0$$ $$ \alpha_1 - \alpha_2 = 0$$

Solving this system of linear equations with the Gauss algorithm, the only solution is the one where $\alpha_1, \alpha_2, \alpha_3 = 0$, therefore $p_1(X), p_2(X)$ and $p_3(X)$ are linearly independent.

To show that $(p_1, p_2, p_3)$ spans $V_3$, I let $p\in V_3$ be an arbitrary polynomial of $V_3$, then it follows that $$p = \lambda_3X^3 + \lambda_2X^2 + \lambda_1X + \lambda_0$$

Setting $\lambda_3 = (\alpha_1 + \alpha_2)$, $\lambda_2 = \alpha_3$, $\lambda_1 = (-\alpha_1 - \alpha_3)$ and $\lambda_0 = (\alpha_1 - \alpha_2)$, it follows that $p\in span(p_1,p_2,p_3)$. Since p was arbitrary $(p_1, p_2, p_3)$ spans $V_3$.

Could anybody please give me some brief feedback on my solution. I just started learning linear algebra and I'm not sure at all whether my approach was correct and whether my solution is valid.

  • 2
    $\begingroup$ $V_3$ is dimension $4$ and you only have three polynomials.... $\endgroup$ – JSchlather Nov 24 '13 at 17:11
  • $\begingroup$ We haven't covered dimensions yet. What does it mean for $V_3$ to have dimension 4? $\endgroup$ – eager2learn Nov 24 '13 at 17:14
  • $\begingroup$ What is a basis of $V_3$? $\endgroup$ – eager2learn Nov 24 '13 at 17:22

The space of polynomials of degree 3 and less is actually a $4$ dimensional vector space. Meaning, you require a minimum of 4 vectors to span your space. Since you only have 3 polynomials, you can't span the whole space. But, if they're linearly independent, you will span a 3-dimensional subspace. The error you made in your proof is when you let $$p = \lambda_3X^3 + \lambda_2X^2 + \lambda_1X + \lambda_0.$$ You've implicitly included a fourth vector, namely $X^0$ or, $1$--the $\lambda_0$ term. This allowed you to complete your basis, in particular, it shows that $p_4=X^0=1$ is independent from $p_1,p_2,p_3$ above. In whole, your three vectors do not span $V_3$.

In your solution of the problem, you even mention a basis for $V_3$, namely: $\{X^3,X^2,X,X^0\}=\{X^3,X^2,X,1\}$. Notice, any other basis for $V_3$ will have four elements, as will any other 4-dimensional vector space (and in general any $n$ dimensional vector space will have $n$ elements in it's basis, for any $n \in \mathbf{N}$.)

  • $\begingroup$ Thanks a lot for your answer. Could you comment on whether my solution as to whether $p_1, p_2, p_3$ are linearly independent was correct? $\endgroup$ – eager2learn Nov 24 '13 at 17:32
  • 1
    $\begingroup$ By showing that $p_1,p_2,p_3,p_4$ (with my $p_4=1$ above) form a basis you also shows that $\{p_1,p_2,p_3\}$ were linearly independent. You have $4$ vectors that form a basis for a 4-dimensional vector space, so in particular they are linearly independent. Remove one, and the others are still linearly independent. I've been slightly vague on purpose in this statement, I leave it to you to formalize. Linear algebra isn't terribly tricky, and once you see it, you realize how interconnected most of the terms are--basis, dimension, linear independence, and so on. Good luck! $\endgroup$ – mojambo Nov 24 '13 at 17:38

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