1
$\begingroup$

Let $p_{1}(x)=-2x+2$, $p_{2}(x)=x+2$, $p_{3}(x)=x^{2}+2x+3$, $p_{4}(x)=x^{2}-x+3$.

a) From the above four polynomials, determine a linearly independent subset that spans the polynomials.

b) What is the dimension of the (sub)space that this basis spans?

c) Does this basis span the space of the second degree polynomials? If not, describe the subspace it spans in terms of the coefficients of a polynomial that belongs to this subspace?

I said since the Wronskian is 0, the functions are linearly dependent. We see three non-zero rows there. This means three of the functions form a linearly independent set of functions. So, {$p_{1}(x),p_{2}(x),p_{3}(x)$} is subset that spans the space of second degree polynomials and is a basis for this space. The dimension is three since there are three functions that form the basis. But parts a and c yelling at me: "Wrong!". Can somebody help me?

$\endgroup$
  • $\begingroup$ You determined that the 4 are linearly dependent. Then you listed 3 and proposed they are a basis. You still need to show these 3 are independent. (They are.) THEN you conclude the dimension is 3. $\endgroup$ – GEdgar Dec 26 '14 at 0:08
1
$\begingroup$

(a) Let $\vec p_i$ be the coefficient vector of $p_i(x)$. Then, $$M= \begin{bmatrix} \vec p_1 & \vec p_2 & \vec p_3 & \vec p_4 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 & 1 \\ -2 & 1 & 2 & -1 \\ 2 & 2 & 3 & 3 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 \end{bmatrix} .$$

Columns $1$, $2,$ and $3$ are pivot columns $\Rightarrow T =\{p_1, p_2, p_3 \}$ is a linearly independent subset that spans $\{p_1, p_2, p_3, p_4 \} \Rightarrow T$ forms a basis for $\operatorname{Span} \{p_1, p_2, p_3, p_4 \}$.

(b) $\dim (\operatorname{Span} T) = \operatorname{Rank} M = 3.$

(c) Let $p \in \mathbb{P}^2$. Then, $p$ is of the form $ax^2+bx+c$. The corresponding system to solve is $$ \begin{bmatrix} \vec p_1 & \vec p_2 & \vec p_3 & \vec p \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 & a \\ -2 & 1 & 2 & b \\ 2 & 2 & 3 & c \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 & \frac{1}{6} a + \frac{1}{6}c - \frac{1}{3} b \\ 0 & 1 & 0 & - \frac{5}{3} a + \frac{1}{3}c + \frac{1}{3} b \\ 0 & 0 & 1 & a \end{bmatrix}. $$

We can see that the system is consistent, and $$p = \left( \frac{1}{6} a + \frac{1}{6}c - \frac{1}{3} b \right)p_1 + \left( -\frac{5}{3} a + \frac{1}{3}c + \frac{1}{3} b \right) p_2 + a p_3. $$ Hence, $\operatorname{Span} T = \mathbb{P}^2$. But, we showed that $T$ was linearly independent $\Rightarrow T$ forms a basis for $\mathbb{P}^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.