Answer to Question #87108 in Linear Algebra for rahul bisht

Question

Let P superscript (e) ={p(x)∈R[x]|p(x) = p(−x)} P superscript(o) ={p(x)∈R[x]|p(x) =−p(−x)} a) Check that P superscript (e) and P superscript (o) are subspace of R[x]. b) Show that P superscript (e) =(∑ i a subscript i x superscript i ∈R[x]



 a subscript = 0 if i is odd.) P superscript(o) =(∑ i a subscript i x superscript i ∈R[x]



 a subscript = 0 if i is even.) Deduce that P superscript (o)∩P superscript (e) ={0}. ( c) Check p(x)+p(−x)∈P superscript(e) for every p(x)∈R(x). Check that the map ψ: R[x]→P superscript(e) given byψ(p(x)) = p(x)+p(−x)/ 2 is a linear map. Further, check that ψ superscript 2 =ψ. Determine the kernel of ψ

Answer 1

a) If $p_1 (x) \in P^{(e)}$ and $p_2 (x) \in P^{(e)}$ and if $\alpha, \beta \in \R$, then

$$p (x) \equiv \alpha p_1 (x) + \beta p_2 (x) = \alpha p_1 (- x) + \beta p_2 (- x) = p (-x) ,$$

so that $p (x) \in P^{(e)}$. Similarly, if $p_1 (x) \in P^{(o)}$ and $p_2 (x) \in P^{(o)}$, then

$$p (x) \equiv \alpha p_1 (x) + \beta p_2 (x) \\ {} = - \alpha p_1 (- x) - \beta p_2 (- x) = - p (-x) \, ,$$

so that $p (x) \in P^{(o)}$. This means that $P^{(e)}$ and $P^{(o)}$ are linear subspaces of $\R [x]$.

b) Let $p (x) = \sum_i a_i x^i \in P^{(e)}$. Then $p(x) = p(-x)$, so that

$$\sum_i a_i x^i = p(x) = \frac12 \left[ p(x) + p (-x) \right] \\ {} = \frac12 \left[ \sum_i a_i x^i + \sum_i a_i (- 1)^i x^i \right] \\ {} = \frac12 \sum_i a_i \left[ 1 + (-1)^i \right] x^i \, .$$

Equating the coefficients on the left-hand side and on the right-hand side, we have

$$a_i = \frac12 a_i \left[ 1 + (-1)^i \right] \, ,$$

placing no restriction on $a_i$ if $i$ is even, and implying $a_i = 0$ if $i$ is odd. Hence,

$$P^{(e)} = \left\{ \sum_i a_i x^i \in \R [x] \, , a_i = 0 ~ \text{if}~ i ~ \text{is odd} \right\} \, .$$

Let $p (x) = \sum_i a_i x^i \in P^{(o)}$. Then $p (x) = - p (-x)$, so that

$$\sum_i a_i x^i = p(x) = \frac12 \left[ p(x) - p (-x) \right] \\ {} = \frac12 \left[ \sum_i a_i x^i - \sum_i a_i (- 1)^i x^i \right] \\ {} = \frac12 \sum_i a_i \left[ 1 - (-1)^i \right] x^i \, .$$

Equating the coefficients on the left-hand side and on the right-hand side, we have

$$a_i = \frac12 a_i \left[ 1 - (-1)^i \right] \, ,$$

placing no restriction on $a_i$ if $i$ is odd, and implying $a_i = 0$ if $i$ is even. Hence,

$$P^{(o)} = \left\{ \sum_i a_i x^i \in \R [x] \, , a_i = 0 ~ \text{if}~ i ~ \text{is even} \right\} \, .$$

The space $P^{(e)} \cap P^{(o)}$ consists of $p (x) = \sum_i a_i x^i \in \R[x]$ such that $a_i = 0$ for $i$ both odd and even, hence, for all $i$. Thus, $P^{(e)} \cap P^{(o)} = \{ 0 \}$.

c) We have $q (x) = p (x) + p (-x) = p (-x) + p(x) = q (-x) \in P^{(e)}$ for every $p (x) \in \R [x]$.

If $p_1 (x), p_2 (x) \in \R [x]$ and $\alpha , \beta \in \R$, then

$$\psi \left( \alpha p_1 (x) + \beta p_2 (x) \right) \\ {} = \frac12 \left( \alpha p_1 (x) + \beta p_2 (x) + \alpha p_1 (-x) + \beta p_2 (-x) \right) \\ {} = \frac12 \left( \alpha p_1 (x) + \alpha p_1 (-x) \right) + \frac12 \left( \beta p_2 (x) + \beta p_2 (-x) \right) \\ {} = \alpha \frac12 \left( p_1 (x) + p_1 (-x) \right) + \beta \frac12 \left( p_2 (x) + p_2 (-x) \right) \\ {} = \alpha \psi \left( p_1 (x) \right) + \beta \psi \left( p_2 (x) \right) \, .$$

This demonstrates that $\psi \left( p (x) \right)$ is a linear map.

Further, we have

$$\psi^2 \left( p(x) \right) = \psi \left( \frac12 \left( p(x) + p(-x) \right) \right) \\ {} = \frac12 \left( \frac12 \left( p(x) + p(-x) \right) + \frac12 \left( p(-x) + p(x) \right) \right) \\ {} = \frac12 \left( p(x) + p(-x) \right) = \psi \left( p(x) \right) \, .$$

Thus, $\psi^2 = \psi$.

The kernel of $\psi$ is the subspace formed by $p (x) \in \R[x]$ such that $\psi \left( p (x) \right) = 0.$ The last condition is equivalent to $p (x) + p (-x) = 0$, or $p (-x) = - p (x)$. Hence, the kernel of $\psi$ is the subspace $P^{(o)}$, ${\rm ker}\, \psi = P^{(o)}$.