Prove Polynomials Form A Vector Space: Degree ≤ 2

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Hey guys! Let's dive into a cool math problem today. We're going to show that the set V of all polynomials with a degree less than or equal to 2 actually forms a vector space. What does this mean? Well, a vector space is a set of objects (in this case, polynomials) that play nicely with addition and scalar multiplication, following certain rules. We need to demonstrate that these rules hold true for our set V.

What is a Vector Space?

Before we get into the nitty-gritty details, let's briefly recap what a vector space is. A vector space is a collection of objects, called vectors, that can be added together and multiplied by scalars (numbers) in a way that satisfies certain axioms. These axioms ensure that the operations behave in a predictable and consistent manner. The eight key axioms for a set V to be a vector space are:

  1. Closure under addition: For any two vectors u and v in V, their sum u + v is also in V.
  2. Commutativity of addition: For any vectors u and v in V, u + v = v + u.
  3. Associativity of addition: For any vectors u, v, and w in V, (u + v) + w = u + (v + w).
  4. Existence of a zero vector: There exists a vector 0 in V such that for any vector u in V, u + 0 = u.
  5. Existence of additive inverses: For any vector u in V, there exists a vector -u in V such that u + (-u) = 0.
  6. Closure under scalar multiplication: For any scalar c and any vector u in V, the scalar product cu* is also in V.
  7. Distributivity of scalar multiplication with respect to vector addition: For any scalar c and any vectors u and v in V, c(u + v) = cu* + cv*.
  8. Distributivity of scalar multiplication with respect to scalar addition: For any scalars c and d and any vector u in V, (c + d)u = cu* + du*.
  9. Compatibility of scalar multiplication with scalar multiplication: For any scalars c and d and any vector u in V, c(du*) = (c d)u.
  10. Identity element of scalar multiplication: For any vector u in V, 1u = u.

To prove that V is a vector space, we must show that all ten of these axioms hold true for polynomials of degree less than or equal to 2. Let's break it down step by step.

Defining Our Set V

First, let's clearly define our set V. A polynomial p(x) in V can be written in the general form:

p(x) = ax² + bx + c

where a, b, and c are real numbers. The highest power of x is 2 (or less), so the degree is at most 2. This is crucial for our proof.

Proving the Vector Space Axioms for V

Now, let's go through each of the vector space axioms and show they hold for our set V of polynomials.

1. Closure Under Addition

This axiom states that if we add two polynomials in V, the result must also be a polynomial in V. Let p(x) = a₁x² + b₁x + c₁ and q(x) = a₂x² + b₂x + c₂ be two polynomials in V. Their sum is:

p(x) + q(x) = (a₁ + a₂)x² + (b₁ + b₂)x + (c₁ + c₂)

Since a₁ + a₂, b₁ + b₂, and c₁ + c₂ are also real numbers, the resulting polynomial is also of degree less than or equal to 2, and thus belongs to V. So, closure under addition holds.

2. Commutativity of Addition

We need to show that p(x) + q(x) = q(x) + p(x) for any polynomials p(x) and q(x) in V. Using the same polynomials as above:

p(x) + q(x) = (a₁ + a₂)x² + (b₁ + b₂)x + (c₁ + c₂)

q(x) + p(x) = (a₂ + a₁)x² + (b₂ + b₁)x + (c₂ + c₁)

Since addition of real numbers is commutative (a₁ + a₂ = a₂ + a₁, etc.), we have p(x) + q(x) = q(x) + p(x). Commutativity holds!

3. Associativity of Addition

We need to show that (p(x) + q(x)) + r(x) = p(x) + (q(x) + r(x)) for any polynomials p(x), q(x), and r(x) in V. Let r(x) = a₃x² + b₃x + c₃. Then:

(p(x) + q(x)) + r(x) = [(a₁ + a₂)x² + (b₁ + b₂)x + (c₁ + c₂)] + [a₃x² + b₃x + c₃] = (a₁ + a₂ + a₃)x² + (b₁ + b₂ + b₃)x + (c₁ + c₂ + c₃)

p(x) + (q(x) + r(x)) = [a₁x² + b₁x + c₁] + [(a₂ + a₃)x² + (b₂ + b₃)x + (c₂ + c₃)] = (a₁ + a₂ + a₃)x² + (b₁ + b₂ + b₃)x + (c₁ + c₂ + c₃)

Since addition of real numbers is associative, we have (p(x) + q(x)) + r(x) = p(x) + (q(x) + r(x)). Associativity holds!

4. Existence of a Zero Vector

We need to find a polynomial 0(x) in V such that p(x) + 0(x) = p(x) for any polynomial p(x) in V. The zero vector is the zero polynomial:

0(x) = 0x² + 0x + 0 = 0

Clearly, p(x) + 0(x) = (a₁x² + b₁x + c₁) + (0x² + 0x + 0) = a₁x² + b₁x + c₁ = p(x). The zero vector exists!

5. Existence of Additive Inverses

For any polynomial p(x) in V, we need to find a polynomial -p(x) in V such that p(x) + (-p(x)) = 0(x). If p(x) = a₁x² + b₁x + c₁, then the additive inverse is:

-p(x) = -a₁x² - b₁x - c₁

Now, p(x) + (-p(x)) = (a₁x² + b₁x + c₁) + (-a₁x² - b₁x - c₁) = (a₁ - a₁)x² + (b₁ - b₁)x + (c₁ - c₁) = 0x² + 0x + 0 = 0(x). Additive inverses exist!

6. Closure Under Scalar Multiplication

This axiom states that if we multiply a polynomial in V by a scalar, the result must also be a polynomial in V. Let p(x) = a₁x² + b₁x + c₁ be a polynomial in V, and let k be any real number (scalar). Then:

k p(x) = k(a₁x² + b₁x + c₁) = (ka₁)x² + (kb₁)x + (kc₁)

Since ka₁, kb₁, and kc₁ are also real numbers, the resulting polynomial is also of degree less than or equal to 2, and thus belongs to V. So, closure under scalar multiplication holds.

7. Distributivity of Scalar Multiplication with Respect to Vector Addition

For any scalar k and polynomials p(x) and q(x) in V, we need to show that k(p(x) + q(x)) = kp(x) + kq(x)*. Using our previous definitions:

k(p(x) + q(x)) = k[(a₁ + a₂)x² + (b₁ + b₂)x + (c₁ + c₂)] = [k(a₁ + a₂)]x² + [k(b₁ + b₂)]x + [k(c₁ + c₂)] = (ka₁ + ka₂)x² + (kb₁ + kb₂)x + (kc₁ + kc₂)*

kp(x) + kq(x) = k(a₁x² + b₁x + c₁) + k(a₂x² + b₂x + c₂) = (ka₁)x² + (kb₁)x + (kc₁) + (ka₂)x² + (kb₂)x + (kc₂) = (ka₁ + ka₂)x² + (kb₁ + kb₂)x + (kc₁ + kc₂)*

Since k(a₁ + a₂) = ka₁ + ka₂, distributivity holds!

8. Distributivity of Scalar Multiplication with Respect to Scalar Addition

For any scalars k and l and polynomial p(x) in V, we need to show that (k + l)p(x) = kp(x) + lp(x)*. Using our previous definitions:

(k + l)p(x) = (k + l)(a₁x² + b₁x + c₁) = [(k + l)a₁]x² + [(k + l)b₁]x + [(k + l)c₁] = (ka₁ + la₁)x² + (kb₁ + lb₁)x + (kc₁ + lc₁)*

kp(x) + lp(x) = k(a₁x² + b₁x + c₁) + l(a₁x² + b₁x + c₁) = (ka₁)x² + (kb₁)x + (kc₁) + (la₁)x² + (lb₁)x + (lc₁) = (ka₁ + la₁)x² + (kb₁ + lb₁)x + (kc₁ + lc₁)*

Since (k + l)a₁ = ka₁ + la₁, distributivity holds!

9. Compatibility of Scalar Multiplication with Scalar Multiplication

For any scalars k and l and polynomial p(x) in V, we need to show that k(lp(x)) = (k l)p(x)*. Using our previous definitions:

k(lp(x)) = k[l(a₁x² + b₁x + c₁)] = k((la₁)x² + (lb₁)x + (lc₁)) = (k(la₁))x² + (k(lb₁))x + (k(lc₁))

(k l)p(x) = (k l)(a₁x² + b₁x + c₁) = ((k l)a₁)x² + ((k l)b₁)x + ((k l)c₁)*

Since k(la₁) = (k l)a₁, compatibility holds!

10. Identity Element of Scalar Multiplication

For any polynomial p(x) in V, we need to show that 1p(x) = p(x). This is straightforward:

1p(x) = 1(a₁x² + b₁x + c₁) = (1a₁)x² + (1b₁)x + (1c₁) = a₁x² + b₁x + c₁ = p(x)

The identity element exists!

Conclusion

We've successfully shown that all ten axioms of a vector space hold true for the set V of all polynomials of degree less than or equal to 2. Therefore, V is indeed a vector space! Great job, everyone! This is a fundamental concept in linear algebra, and understanding it will help you tackle more advanced topics later on. Keep up the great work!