Problem 18 Let \(U\) and \(V\) denote, resp... [FREE SOLUTION]

Chapter 7: Problem 18

Let \(U\) and \(V\) denote, respectively, the spaces of even and odd polynomials in \(\mathbf{P}_{n}\). Show that \(\operatorname{dim} U+\operatorname{dim} V=n+1 .\) [Hint: Consider \(T: \mathbf{P}_{n} \rightarrow \mathbf{P}_{n}\) where \(T[p(x)]=p(x)-p(-x) .]\)

Short Answer

Expert verified

The dimensions of even and odd polynomial spaces sum to \( n+1 \).

Step by step solution

Understanding the Problem

We want to find the dimensions of the subspaces of even and odd polynomials, denoted as \( U \) and \( V \), respectively, within the space of polynomials \( \mathbf{P}_n \) of degree at most \( n \). Then, we have to show that the sum of these dimensions is \( n + 1 \).

Define the Even and Odd Polynomials

A polynomial \( p(x) \) is even if \( p(x) = p(-x) \) and odd if \( p(x) = -p(-x) \). A general polynomial \( p(x) \in \mathbf{P}_n \) can be written as \( a_nx^n + a_{n-1}x^{n-1} + \, ... \, + a_1x + a_0 \).

Consider Transformation T

Define the transformation \( T: \mathbf{P}_n \rightarrow \mathbf{P}_n \) such that \( T[p(x)] = p(x) - p(-x) \). Due to the properties of even and odd polynomials, \( T[p(x)] \) will isolate the odd component of \( p(x) \).

Properties of Transformation T

For a polynomial \( p(x) \), the even component \( E(x) \) and the odd component \( O(x) \) can be expressed as \[ p(x) = E(x) + O(x) \]. Applying \( T \), we get \( T[p(x)] = O(x) \), because the even component \( E(x) \) cancels out with itself when \( x \) is replaced by \(-x\).

Understanding Image and Kernel of T

The image of \( T \) is the set of odd polynomials because any polynomial returned by \( T \) is purely an odd polynomial. The kernel of \( T \), \( \ker(T) \), is the set of even polynomials since \( T[p(x)] = 0 \) for even polynomials.

Apply the Rank-Nullity Theorem

By the rank-nullity theorem, we know that \[ \dim \ker(T) + \dim \text{Im}(T) = \dim \mathbf{P}_n. \]Since \( \dim \ker(T) = \dim U \) and \( \dim \text{Im}(T) = \dim V \), and \( \dim \mathbf{P}_n = n+1 \), substituting these gives \[ \dim U + \dim V = n+1. \]

Conclude the Solution

Thus, the sum of the dimensions of the spaces of even and odd polynomials in \( \mathbf{P}_n \) is equal to \( n+1 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polynomial Transformation

Polynomial transformations involve altering a polynomial to highlight specific properties or components. The transformation defined in the exercise is a great example. Here, the transformation \( T: \mathbf{P}_{n} \rightarrow \mathbf{P}_{n} \) where \( T[p(x)] = p(x) - p(-x) \) focuses on isolating the odd component of the polynomial. In general:

For any polynomial \( p(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \), the transformation modifies it based on its even and odd nature.
An even polynomial remains unchanged for negative inputs: \( p(x) = p(-x) \).
An odd polynomial changes sign with negative inputs: \( p(x) = -p(-x) \).

The transformation therefore negates the even terms by having them cancel out. This means that after applying \( T \), only the odd terms remain since:

The even component cancels itself out.
The odd component remains unchanged.

Rank-Nullity Theorem

The rank-nullity theorem is a fundamental concept in linear algebra. It connects the dimensions of the image and kernel of a linear transformation. For our transformation \( T \) in the context of polynomials, this theorem is key to proving the exercise statement. Here's how it works:

The theorem states that for any linear map \( T: V \rightarrow W \), the equation \( \dim \ker(T) + \dim \text{Im}(T) = \dim V \) holds true.
In our specific problem, \( V \) represents the space of polynomials \( \mathbf{P}_n \), whose dimension is \( n+1 \).
The kernel of \( T \), \( \ker(T) \), contains all even polynomials, so \( \dim \ker(T) = \dim U \).
The image of \( T \), \( \text{Im}(T) \), contains all odd polynomials, so \( \dim \text{Im}(T) = \dim V \).

Putting it all together, the rank-nullity theorem confirms that the dimensions of even and odd polynomial spaces add up to \( n+1 \):\[ \dim U + \dim V = n+1 \].

Dimension of Polynomial Spaces

Understanding the dimension of polynomial spaces is crucial when dealing with even and odd polynomials. The polynomial space \( \mathbf{P}_n \) is made up of polynomials with degrees ranging from \( 0 \) to \( n \). The dimension here refers to:

How many basis vectors span the entire space of polynomials \( \mathbf{P}_n \).
Each term in the polynomial (\( x^n, x^{n-1}, \ldots, x^0 \)) represents one dimension.

So, \( \mathbf{P}_n \) has a total dimension of \( n+1 \), meaning it has \( n+1 \) independent basis vectors. For this reason:

Even polynomials can be considered half of the basis when viewed over symmetrical powers of \( x \).
Odd polynomials complement this with the alternate powers of \( x \).

When combined, both even and odd polynomial subspaces must fill the entire space \( \mathbf{P}_n \), leading to the conclusion in the exercise:\[ \dim U + \dim V = n+1 \]. This illustrates a perfect alloocation between even and odd spaces in polynomial terms.

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