Subspace Of P3: Find A Basis For W

Hey guys! Let's dive into the fascinating world of linear algebra, specifically, subspaces and bases. Today, we're tackling a classic problem: proving that a set W is a subspace of P3 (the vector space of polynomials of degree at most 3) and then figuring out its basis. This is super important because understanding subspaces and bases is like having a key to unlock many concepts in linear algebra. You'll use these ideas everywhere, from machine learning to physics. So, buckle up! We'll break down the concepts, making sure everything clicks. Ready?

Understanding the Basics: Subspaces and P3

Alright, before we get our hands dirty, let's make sure we're on the same page. A subspace is a subset of a vector space that has all the characteristics of a vector space itself. Think of it like a mini-universe within a bigger universe, following the same rules. To prove something's a subspace, we need to show three things: It contains the zero vector, it's closed under vector addition, and it's closed under scalar multiplication. We'll get into the details soon.

Now, what about P3? P3 is the set of all polynomials of the form ax³ + bx² + cx + d, where a, b, c, and d are real numbers. This includes constants, linear, quadratic, and cubic polynomials, along with the zero polynomial (where a = b = c = d = 0). This is our big universe, and our W is going to be a special part of it.

We'll show you how to demonstrate that W is indeed a subspace and how to pinpoint its basis. Having a firm grasp of these principles will make your future linear algebra adventures much smoother. This is the cornerstone for understanding more complex topics like linear transformations and eigenvalues.

The Importance of the Zero Vector, Closure Under Addition, and Scalar Multiplication

Why are those three conditions – zero vector, closure under addition, and closure under scalar multiplication – so critical? Because they guarantee that our subset W behaves like a vector space. The zero vector is the identity element for addition. Without it, you can't guarantee that the space has a starting point. Closure under addition ensures that when you add two vectors in W, the result also stays within W. This is critical for internal consistency. Scalar multiplication closure means if you multiply any vector in W by a real number, the result stays in W. It makes sure that you can stretch, shrink, or flip vectors within W without escaping its boundaries. These three conditions form the fundamental requirements for W to possess the structure of a vector space.

Defining W and Setting the Stage

Let's assume W is defined as the set of all polynomials in P3 such that p(1) = 0. In simpler words, W is composed of all polynomials of degree at most 3 where, when you plug in x = 1, the polynomial evaluates to zero. This could include polynomials like (x - 1), (x - 1)², or even more complex ones like (x - 1)(x² + 2). Remember, we need to confirm that W adheres to all the vector space rules, specifically, that it's a subspace of P3.

Examining p(1) = 0 and Its Implications

This condition, p(1) = 0, is what defines W. It means that every polynomial in W has a root at x = 1. This might seem like a simple constraint, but it significantly shapes the structure of W. When x = 1 is a root, the polynomial can be factored as (x - 1) times another polynomial. This particular detail is crucial to understanding the behavior of elements within W.

Proving W is a Subspace

Now, let's formally verify that W is indeed a subspace. We'll use the three conditions mentioned earlier. It is very important to show each condition.

Condition 1: The Zero Vector

Does W contain the zero vector? Yep! The zero vector in P3 is the zero polynomial, p(x) = 0. If we evaluate this at x = 1, we get p(1) = 0. Therefore, the zero polynomial is in W.

Condition 2: Closure Under Addition

If we take two polynomials, p(x) and q(x), both in W, does their sum, (p + q)(x), also belong to W? Let's see. Since p(x) and q(x) are in W, we know p(1) = 0 and q(1) = 0. Now, consider (p + q)(1) = p(1) + q(1) = 0 + 0 = 0. This shows that (p + q)(x) also has a value of zero at x = 1, meaning it's in W. Closure under addition holds.

Condition 3: Closure Under Scalar Multiplication

Let's take a polynomial p(x) from W and a scalar c (a real number). Does the polynomial cp(x) stay within W? Because p(x) is in W, we know p(1) = 0. If we evaluate cp(x) at x = 1, we get (cp)(1) = c * p(1) = c * 0 = 0. This tells us that cp(x) also equals zero at x = 1, placing it in W. This demonstrates the closure under scalar multiplication.

Summarizing the Subspace Proof

Great job! We've successfully shown that W includes the zero vector and is closed under both vector addition and scalar multiplication. Thus, W meets all the requirements to be a subspace of P3. This is a critical result, and it unlocks the potential for us to further explore the properties of W, such as finding its basis.

Finding a Basis for W

Now, the fun part: finding a basis for W. A basis is a set of linearly independent vectors that span the subspace. This means that any polynomial in W can be written as a linear combination of the basis vectors, and no vector in the basis can be expressed as a linear combination of the others. We're looking for the simplest building blocks that generate the whole space W.

Understanding Linear Independence and Spanning Sets

What do we mean by linear independence? Vectors are linearly independent if no vector in the set can be written as a linear combination of the others. In simpler terms, none of the vectors are redundant. A spanning set is a set of vectors where every vector in the space can be written as a linear combination of these vectors. Combining linear independence and spanning sets leads us to the basis, which is the perfect set of vectors needed to define the subspace.

Constructing a Basis for W

Since every polynomial in W has x = 1 as a root, it must be divisible by (x - 1). So, any polynomial in W can be written as (x - 1) times another polynomial of degree at most 2. Let's think about this: We can write a general polynomial in W as: p(x) = (x - 1)(ax² + bx + c). Expanding this, we get p(x) = ax³ + (b - a)x² + (c - b)x - c. This is where the magic happens!

We can rewrite this in terms of the basis vectors by factoring out a, b, and c. This simplifies our expression to a(x³ - x²) + b(x² - x) + c(x - 1). So, the vectors (x³ - x²), (x² - x), and (x - 1) are likely candidates for a basis. They are linearly independent and span W. Let's quickly verify that these polynomials are linearly independent. Consider α(x³ - x²) + β(x² - x) + γ(x - 1) = 0. For this to be zero for all x, the coefficients must all be zero. This gives us α = 0, β = 0, and γ = 0. These vectors are therefore linearly independent.

Let's prove the spanning property. Any polynomial in W can be written as a linear combination of those three vectors: a(x³ - x²) + b(x² - x) + c(x - 1). The resulting polynomials will always satisfy the p(1) = 0 condition, therefore, they are the right bases.

The Basis Revealed

Thus, a basis for W is the set: {(x - 1), (x² - x), (x³ - x²)}. Any polynomial in W can be created using a combination of these three. The basis gives us a perfect and minimal set of building blocks for the subspace. The dimension of W, which is the number of vectors in the basis, is 3. This means that W is a three-dimensional subspace within the four-dimensional space P3.

Confirming Linear Independence and Spanning

We've already shown linear independence. But for good measure, here's a recap. A set of vectors is linearly independent if the only way to get the zero vector as a linear combination is by using zero coefficients. Now, to confirm the spanning property, we've essentially constructed any general polynomial within W by using linear combinations of our proposed basis vectors. This proves that any polynomial in W can be represented by our suggested vectors.

Conclusion: Wrapping Up the Subspace and Basis Quest

Excellent work, everyone! We've successfully proven that W is a subspace of P3 and identified its basis. This exercise is foundational in linear algebra, and you can now confidently approach similar problems. Remember, understanding subspaces and bases is crucial. They are not just abstract concepts. They are tools that help us understand and model real-world problems.

Key Takeaways and Next Steps

Let's recap: W is a subspace of P3 because it contains the zero vector and is closed under addition and scalar multiplication. The basis for W is the set {(x - 1), (x² - x), (x³ - x²)}. The dimension of W is 3. For further exploration, try changing the definition of W. What if W was defined by p(2) = 0? Or p(1) = p(0)? How would the basis change? Keep experimenting and pushing your understanding of these concepts. Keep practicing with different subspace definitions and see how the basis changes. Good luck, and keep learning! You are doing great!