Math Problem Statement

Suppose p,q \in P(\mathbb{C}) are nonconstant polynomials with no zeros in common. Let r \in P_{n-1}(\mathbb{C}) and s \in P_{m-1}(\mathbb{C}). Let m = deg(p) and n = deg(q). Define T : P_{n-1}(\mathbb{C}) \times P_{m-1}(\mathbb{C}) \rightarrow P_{m+n-1}(\mathbb{C}) by T(r,s) = rp + sq. Show that the linear map T is injective and surjective.

Solution

To show that the linear map TT is injective and surjective, we analyze it step by step.


Definitions and Setup

Let:

  • Pn(C)P_n(\mathbb{C}) denote the space of polynomials of degree at most nn with coefficients in C\mathbb{C}.
  • pp and qq are polynomials of degrees m=deg(p)m = \deg(p) and n=deg(q)n = \deg(q), respectively, with no common zeros.
  • T:Pn1(C)×Pm1(C)Pm+n1(C)T: P_{n-1}(\mathbb{C}) \times P_{m-1}(\mathbb{C}) \to P_{m+n-1}(\mathbb{C}) defined by T(r,s)=rp+sqT(r, s) = rp + sq, where rPn1(C)r \in P_{n-1}(\mathbb{C}) and sPm1(C)s \in P_{m-1}(\mathbb{C}).

Step 1: Show that TT is injective

Approach:

To show injectivity, we need to prove that if T(r,s)=0T(r, s) = 0, then r=0r = 0 and s=0s = 0.

  1. Suppose T(r,s)=rp+sq=0T(r, s) = rp + sq = 0.
  2. Rearrange to get rp=sqrp = -sq.
  3. Since pp and qq have no common zeros, the polynomial rprp can only vanish if both r=0r = 0 and s=0s = 0.

Reasoning:

  • The lack of common zeros ensures that pp and qq are linearly independent in the space of polynomials.
  • Therefore, rr and ss must both vanish for the equation rp+sq=0rp + sq = 0 to hold.

Thus, TT is injective.


Step 2: Show that TT is surjective

Approach:

To show surjectivity, we need to prove that for any polynomial hPm+n1(C)h \in P_{m+n-1}(\mathbb{C}), there exist rPn1(C)r \in P_{n-1}(\mathbb{C}) and sPm1(C)s \in P_{m-1}(\mathbb{C}) such that T(r,s)=hT(r, s) = h.

  1. Write hPm+n1(C)h \in P_{m+n-1}(\mathbb{C}). This means hh has degree at most m+n1m+n-1.
  2. Since pp and qq are linearly independent and span Pm+n1(C)P_{m+n-1}(\mathbb{C}), any hh can be written as h=rp+sqh = rp + sq for some rPn1(C)r \in P_{n-1}(\mathbb{C}) and sPm1(C)s \in P_{m-1}(\mathbb{C}).

Constructing rr and ss:

  • Use the Euclidean algorithm to divide hh by pp, yielding h=rp+sh = rp + s, where rPn1(C)r \in P_{n-1}(\mathbb{C}) and sPm1(C)s \in P_{m-1}(\mathbb{C}).
  • Since qq is independent of pp, ss can also be expressed in terms of qq.

Thus, TT is surjective.


Conclusion

Since TT is both injective and surjective, it is a linear isomorphism. This completes the proof.


Would you like additional details or further clarification?

Related Questions

  1. What are the implications of pp and qq having no common zeros?
  2. How does the degree of hh influence the surjectivity of TT?
  3. Can the proof be extended to real polynomials, R\mathbb{R}, instead of C\mathbb{C}?
  4. How does linear independence of pp and qq affect the space they span?
  5. What happens if pp and qq have common zeros?

Tip

To verify injectivity or surjectivity of a linear map, always test the kernel (for injectivity) and image (for surjectivity) explicitly.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Polynomial Functions
Injectivity and Surjectivity of Linear Maps
Vector Spaces

Formulas

T(r, s) = rp + sq
Degree of a polynomial: deg(rp + sq) ≤ max(deg(rp), deg(sq))

Theorems

Linear Independence of Polynomials
Properties of Polynomial Spaces over Complex Numbers

Suitable Grade Level

Undergraduate Mathematics