Subspace Proof: S In P3 Explained Simply

In linear algebra, proving that a set is a subspace of a vector space is a fundamental concept. Guys, let's break down how to show that a set S is a subspace of P3, the vector space of polynomials with degree at most 3. This involves verifying three essential conditions: the zero vector is in S, S is closed under addition, and S is closed under scalar multiplication. Understanding these conditions and how to apply them is super important for grasping linear algebra. So, let's dive in and make sure we nail this concept!

Understanding Vector Spaces and Subspaces

Before diving into the specifics, let's quickly recap what vector spaces and subspaces are. A vector space is a set of objects (vectors) that satisfy certain axioms, allowing for operations like addition and scalar multiplication. P3, the set of all polynomials of degree at most 3, is a classic example of a vector space. It includes polynomials like ax³ + bx² + cx + d, where a, b, c, and d are real numbers.

A subspace, on the other hand, is a subset of a vector space that itself satisfies the vector space axioms. In simpler terms, a subspace is a vector space contained within another vector space. To prove that a subset S is a subspace of P3, we need to confirm that S meets three critical conditions:

1. Zero Vector: The zero vector (in this case, the zero polynomial) must be in S.
2. Closure Under Addition: If two polynomials p(x) and q(x) are in S, then their sum p(x) + q(x) must also be in S.
3. Closure Under Scalar Multiplication: If p(x) is in S, then for any scalar c, the polynomial c·p(x) must also be in S.

If S satisfies all three of these conditions, then S is indeed a subspace of P3. Failing to meet even one of these conditions means that S is not a subspace. Make sense? Let's move on to a concrete example to illustrate this process.

Defining the Set S

Okay, before we can actually prove anything, we need to define what the set S actually is. Let's say S is the set of all polynomials p(x) in P3 such that p(0) = 0. In other words, S contains all polynomials of degree at most 3 that have a root at x = 0. This is a pretty common type of subspace to work with, so it's a good one to understand.

Mathematically, we can write this as:

S = {p(x) ∈ P3 | p(0) = 0}

So, a polynomial like x³ + 2x² - x would be in S because if you plug in x = 0, you get 0. On the other hand, a polynomial like x² + 1 would not be in S because plugging in x = 0 gives you 1, not 0. Now that we have a clear definition of S, we can move on to verifying the three conditions for it being a subspace of P3.

Condition 1: Zero Vector

The first condition we need to check is whether the zero vector is in S. In the context of polynomials, the zero vector is simply the zero polynomial, p(x) = 0 for all x. To verify this, we need to see if p(0) = 0.

For the zero polynomial, p(x) = 0, when we substitute x = 0, we get p(0) = 0. This satisfies the condition for membership in S. Therefore, the zero vector is indeed in S.

This might seem like a trivial step, but it's an essential one. If the zero vector weren't in S, we could stop right there and conclude that S is not a subspace. But since it is in S, we can move on to the next condition. This is a foundational step, so make sure you've got it down! Next up, we will see if S is closed under addition.

Condition 2: Closure Under Addition

Now, let's tackle the second condition: closure under addition. This means that if we take any two polynomials in S and add them together, the result must also be in S. Let p(x) and q(x) be two arbitrary polynomials in S. By the definition of S, we know that p(0) = 0 and q(0) = 0.

We need to show that their sum, (p + q)(x) = p(x) + q(x), is also in S. To do this, we need to evaluate (p + q)(0) and see if it equals 0:

(p + q)(0) = p(0) + q(0) = 0 + 0 = 0

Since (p + q)(0) = 0, the sum p(x) + q(x) satisfies the condition for membership in S. Therefore, S is closed under addition.

In simpler terms, if you have two polynomials that both equal zero when you plug in x = 0, then their sum will also equal zero when you plug in x = 0. This makes intuitive sense, and it's a key part of proving that S is a subspace. Now we are left with our final step, which will see if S is closed under scalar multiplication!

Condition 3: Closure Under Scalar Multiplication

Finally, we need to check the third condition: closure under scalar multiplication. This means that if we take any polynomial in S and multiply it by a scalar (a real number), the result must also be in S. Let p(x) be a polynomial in S, so p(0) = 0. Let c be any scalar.

We need to show that the scalar multiple, (c · p)(x) = c p(x), is also in S. To do this, we need to evaluate (c · p)(0) and see if it equals 0:

(c · p)(0) = c * p(0) = c * 0 = 0

Since (c · p)(0) = 0, the scalar multiple c p(x) satisfies the condition for membership in S. Therefore, S is closed under scalar multiplication.

In plain English, if a polynomial equals zero when x = 0, then multiplying that polynomial by any number will still result in zero when x = 0. This confirms that S is closed under scalar multiplication, which is our final condition.

Conclusion

Alright, guys! We've successfully verified all three conditions: the zero vector is in S, S is closed under addition, and S is closed under scalar multiplication. Since S satisfies all three conditions, we can confidently conclude that S is a subspace of P3.

Proving that a set is a subspace is a fundamental skill in linear algebra. By methodically checking these three conditions, you can determine whether a subset inherits the vector space structure of its parent space. Understanding these concepts will give you a solid foundation for tackling more advanced topics in linear algebra.

So, to recap, remember these three key conditions:

• Zero Vector: Make sure the zero vector is in the set.
• Closure Under Addition: The sum of any two vectors in the set must also be in the set.
• Closure Under Scalar Multiplication: Multiplying any vector in the set by a scalar must result in a vector that is also in the set.

Keep practicing, and you'll become a pro at identifying subspaces in no time! Good job working through this example – you're one step closer to mastering linear algebra!