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If \(U\) and \(W\) denote the subspaces of even and odd polynomials in \(\mathbf{P}_{n}\), respectively, show that \(\mathbf{P}_{n}=U \oplus W\). (See Exercise 6.3.36.)

Short Answer

Expert verified
Polynomials in \(\mathbf{P}_n\) can be uniquely written as a sum of even and odd polynomials; hence, \(\mathbf{P}_n = U \oplus W\).

Step by step solution

01

Understand Subspaces U and W

Subspaces of polynomials are given: \(U\) consists of even polynomials and \(W\) of odd polynomials. A polynomial \(p(x)\) is even if \(p(x) = p(-x)\) and odd if \(p(x) = -p(-x)\). We want to prove that any polynomial in \(\mathbf{P}_{n}\) can be written uniquely as a sum of an even and an odd polynomial.
02

Consider any polynomial in \(\mathbf{P}_{n}\)

Let \(p(x)\) be any polynomial in \(\mathbf{P}_{n}\). We can always express \(p(x)\) as a sum of its even and odd parts: \(p(x) = e(x) + o(x)\) where \(e(x) = \frac{p(x) + p(-x)}{2}\) and \(o(x) = \frac{p(x) - p(-x)}{2}\).
03

Verify even and odd components

Check that \(e(x)\) is an even polynomial: \(e(x) = \frac{p(x) + p(-x)}{2}\). It follows that \(e(-x) = \frac{p(-x) + p(x)}{2} = \frac{p(x) + p(-x)}{2} = e(x)\), confirming \(e(x)\) is even. Similarly, check that \(o(x)\) is odd: \(o(x) = \frac{p(x) - p(-x)}{2}\). Then \(o(-x) = \frac{p(-x) - p(x)}{2} = -\frac{p(x) - p(-x)}{2} = -o(x)\), confirming \(o(x)\) is odd.
04

Uniqueness of Decomposition

Suppose \(p(x) = e_1(x) + o_1(x) = e_2(x) + o_2(x)\) with \(e_1, e_2\) even and \(o_1, o_2\) odd. Then, \(e_1(x) - e_2(x) = o_2(x) - o_1(x)\). An even function equals an odd function only if both are zero. Therefore, \(e_1(x) = e_2(x)\) and \(o_1(x) = o_2(x)\), proving uniqueness.
05

Conclusion

We have expressed any polynomial \(p(x)\) in \(\mathbf{P}_{n}\) as a sum \(p(x) = e(x) + o(x)\), where \(e(x)\) is even and \(o(x)\) is odd, with this representation being unique. Thus, \(\mathbf{P}_{n} = U \oplus W\).

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Key Concepts

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

Even Polynomials
Polynomials can be much easier to understand when we classify them into types based on their behavior. One type is the *even polynomial*. An even polynomial is a function where the same value results when you substitute a number and its negative into the polynomial. Mathematically, we describe it as:
  • For an even polynomial, if we have a function \( p(x) \), \( p(-x) = p(x) \).
This property makes even polynomials symmetric with respect to the y-axis on a graph.
This symmetry implies that it "mirrors" itself perfectly across the vertical axis.
To get the even part of a polynomial \( p(x) \), you calculate \( e(x) = \frac{p(x) + p(-x)}{2} \).
  • Why this works: By adding \( p(x) \) and \( p(-x) \) and then dividing by 2, we cancel out any terms that change signs, leaving us with an even function.
Odd Polynomials
Unlike their even counterparts, *odd polynomials* behave differently when swapping a number for its negative. Their defining feature is:
  • For an odd polynomial \( p(x) \), \( p(-x) = -p(x) \).
This distinctive property means they are symmetric with respect to the origin, resembling a 180-degree rotation.
On the graph, they appear to flip over the x-axis and y-axis simultaneously to land back on the same position.
To extract the odd part of a polynomial \( p(x) \), calculate \( o(x) = \frac{p(x) - p(-x)}{2} \).
  • Calculating in this way nullifies elements that remain the same, emphasizing those that flip in sign.
Subspaces
In mathematics, *subspaces* are a critical concept when discussing structures such as polynomials. A subspace is essentially a subset of a larger space that is itself a space, satisfying certain conditions.
  • A subspace must contain the zero vector (or zero polynomial in polynomial terms).
  • It must be closed under addition, meaning adding any two elements in the subspace results in another element in this subspace.
  • The subspace must be closed under scalar multiplication; multiplying any element in the subspace by a scalar results in another element in the subspace.
When talking about polynomials, the subspaces of interest often involve even and odd polynomials. Both even and odd polynomials form valid subspaces of the set of all polynomials.
Understanding subspaces implies recognizing how these distinct parts (even and odd) contribute to the entirety of the polynomial structure.
Polynomial Uniqueness
*Polynomial uniqueness* refers to the interesting property that any polynomial can be uniquely divided into its even and odd components. In other words, each polynomial has a distinct "fingerprint" made up of an even part and an odd part.
This concept means:
  • If you express a polynomial \( p(x) \) as \( p(x) = e(x) + o(x) \), where \( e(x) \) is an even polynomial and \( o(x) \) is an odd polynomial, this representation is one-of-a-kind.
  • If another form \( e_1(x) \) and \( o_1(x) \) claims to break down the same polynomial but does not match \( e(x) \) and \( o(x) \), it implies \( e_1(x) = e(x) \) and \( o_1(x) = o(x) \).
The uniqueness arises from the fact that the only way for an even function to equal an odd one is if both are zero.
This principle assures that the decomposition of any polynomial into even and odd parts is definitive and conclusive, showing Polynomial substructure's beauty and precision.

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Most popular questions from this chapter

Let \(U\) and \(W\) denote subspaces of a vector space \(V\). a. If \(V=U \oplus W,\) define \(T: V \rightarrow V\) by \(T(\mathbf{v})=\mathbf{w}\) where \(\mathbf{v}\) is written (uniquely) as \(\mathbf{v}=\mathbf{u}+\mathbf{w}\) with \(\mathbf{u}\) in \(U\) and \(\mathbf{w}\) in \(W\). Show that \(T\) is a linear transformation, \(U=\operatorname{ker} T, W=\operatorname{im} T,\) and \(T^{2}=T\). b. Conversely, if \(T: V \rightarrow V\) is a linear transformation such that \(T^{2}=T,\) show that \(V=\) ker \(T \oplus \operatorname{im} T\). [Hint: \(\mathbf{v}-T(\mathbf{v})\) lies in ker \(T\) for all \(\mathbf{v}\) in \(V\).]

In each case, show that \(U\) is \(T\) -invariant, use it to find a block upper triangular matrix for \(T,\) and use that to compute \(c_{T}(x)\). a. \(T: \mathbf{P}_{2} \rightarrow \mathbf{P}_{2}\) $$\begin{aligned} T(&\left.a+b x+c x^{2}\right) \\ &=(-a+2 b+c)+(a+3 b+c) x+(a+4 b) x^{2} \\ U &=\operatorname{span}\left\\{1, x+x^{2}\right\\} \end{aligned}$$ b. \(T: \mathbf{P}_{2} \rightarrow \mathbf{P}_{2}\) $$\begin{aligned} T &\left(a+b x+c x^{2}\right) \\ &=(5 a-2 b+c)+(5 a-b+c) x+(a+2 c) x^{2} \\ U&=\operatorname{span}\left\\{1-2 x^{2}, x+x^{2}\right\\} \end{aligned}$$

Let \(U\) and \(W\) be subspaces of \(V,\) let \(\operatorname{dim} V=n,\) and assume that \(\operatorname{dim} U+\operatorname{dim} W=n\) a. If \(U \cap W=\\{\mathbf{0}\\}\), show that \(V=U \oplus W\). b. If \(U+W=V\), show that \(V=U \oplus W\).

Let \(B\) be an ordered basis of the \(n\) dimensional space \(V\) and let \(C_{B}: V \rightarrow \mathbb{R}^{n}\) be the coordinate transformation. If \(D\) is the standard basis of \(\mathbb{R}^{n}\), show that \(M_{D B}\left(C_{B}\right)=I_{n}\).

Use Theorem 9.1 .4 to show that \(T: V \rightarrow V\) is not an isomorphism if \(\mathrm{ker} T \neq 0\) (assume \(\operatorname{dim} V=n\) ).

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