Question

Let P^(e)(x) = {p(x) ∈ R[x]|p(x) = p(−x)} Find W =P(e)∩P3. Find a basis for W and the dimension of W.

Answer 1

Answer on Question #80890 – Math – Linear Algebra

Question

Let $P^{\wedge}(e)(x) = \{p(x) \in R[x] \mid p(x) = p(-x)\}$ Find $W = P(e) \cap P3$. Find a basis for $W$ and the dimension of $W$.

Solution

$P_3$ consists of all polynomials of degree 3, i.e. of all functions

$$y = a x^3 + b x^2 + c x + d$$

$y \in P(e)$ means

$$a x^3 + b x^2 + c x + d = a(-x)^3 + b(-x)^2 + c(-x) + d$$

from which

$$a x^3 + c x = 0$$

This holds for all $x$ only if $a = c = 0$.

So

$$W = \{y = b x^2 + d, b \in \mathbb{R}, d \in \mathbb{R}\}.$$

A basis for $W$ is $\{1, x^2\}$.

Indeed, these two elements are linearly independent since

$$\alpha + \beta x^2 = 0 \text{ for all } x \text{ yields } \alpha = \beta = 0$$

and all elements of $W$ are linear combinations of 1 and $x^2$.

Hence the dimension of $W$ is 2.

**Answer**: $W = \{y = b x^2 + d, b \in \mathbb{R}, d \in \mathbb{R}\}$; a basis for $W$ is $\{1, x^2\}$; $\dim(W) = 2$.

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