Question

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

Answer 1

ANSWER on Question #82560 – Math – Linear Algebra

QUESTION

Let

$$P^{(e)} = \{p(x) \in \mathbb{R}[x] \mid p(x) = p(-x)\}$$

Find $W = P^{(e)} \cap P_3$. Find a basis for $W$ and find the dimension of $W$.

SOLUTION

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

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

Since $p(x)\in W\to \left\{ \begin{array}{l} p(x)\in P_3 \\ p(x)\in P^{(e)} \end{array} \right.$. Then,

$$\begin{array}{l} p(x) \in P^{(e)} \to p(x) = p(-x) \to \\ a x^3 + b x^2 + c x + d = a \cdot (-x)^3 + b \cdot (-x)^2 + c \cdot (-x) + d \to \\ a x^3 + b x^2 + c x + d = -a x^3 + b x^2 - c x + d \to \\ a x^3 + b x^2 + c x + d + a x^3 - b x^2 + c x - d = 0 \to \\ 2 a x^3 + 2 c x = 0 \to \left\{ \begin{array}{l} 2 a = 0 \\ 2 c = 0 \end{array} \right. \to \boxed{\left\{ \begin{array}{l} a = 0 \\ c = 0 \end{array} \right.} \end{array}$$

Conclusion,

$$\boxed{W = \{p(x) = b \cdot x^2 + d \cdot 1, \forall b, d \in \mathbb{R}\}}$$ $$\boxed{A \text{ basis for } W \text{ is } \{1, x^2\}}$$

Indeed, these two elements are linearly independent since

$$a \cdot 1 + b \cdot x^2 = 0 \quad \text{for all } x \in \mathbb{R} \to \left\{ \begin{array}{l} a = 0 \\ b = 0 \end{array} \right.$$

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

Hence, the dimension of $W$ is $\dim(W) = 2$.

ANSWER:

$$W = \{p(x) = b \cdot x^2 + d \cdot 1, \forall b, d \in \mathbb{R}\}$$

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

$$\dim(W) = 2$$

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