Subspace Of P3: Proving S Is A Subspace

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Show That S is a Subspace of P3

In linear algebra, proving that a set is a subspace of a vector space is a fundamental concept. Guys, today, let's dive into showing that a set S is a subspace of P3, the vector space of polynomials with degree at most 3. This involves verifying that S satisfies the three conditions required for it to be considered a subspace. So, grab your thinking caps, and let's get started!

Understanding Vector Spaces and Subspaces

Before diving into the specifics, let's 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. These operations must result in vectors that are still within the same set. A subspace, on the other hand, is a subset of a vector space that is itself a vector space under the same operations. In simpler terms, a subspace is a smaller vector space contained within a larger one. To prove that a set is a subspace, we need to verify three essential conditions:

  1. The zero vector must be in the set.
  2. The set must be closed under addition. This means that if you add any two vectors in the set, the result must also be in the set.
  3. The set must be closed under scalar multiplication. This means that if you multiply any vector in the set by a scalar, the result must also be in the set.

If a set satisfies all three of these conditions, it is indeed a subspace of the larger vector space. Now that we have a clear understanding of the basics, let's proceed with showing that S is a subspace of P3.

Defining S and P3

First, let's define P3. P3 is the vector space of all polynomials with degree at most 3. A general element in P3 can be written as:

p(x) = ax³ + bx² + cx + d

where a, b, c, and d are real numbers. The operations in P3 are standard polynomial addition and scalar multiplication.

Now, let's define S. S is a subset of P3 with a specific condition. For example, 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 pass through the origin (0,0).

Verifying the Subspace Conditions

To prove that S is a subspace of P3, we need to verify the three conditions mentioned earlier.

1. The Zero Vector

The zero vector in P3 is the zero polynomial, z(x) = 0 for all x. We need to check if z(x) is in S. Since z(0) = 0, the zero polynomial satisfies the condition p(0) = 0. Therefore, the zero vector is in S.

2. Closure Under Addition

Let p(x) and q(x) be two polynomials in S. This means 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 evaluate the sum at x = 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 being in S. Therefore, S is closed under addition.

3. Closure Under Scalar Multiplication

Let p(x) be a polynomial in S, so p(0) = 0. Let c be any scalar (real number). We need to show that the scalar multiple, (cp)(x) = c * p(x), is also in S. To do this, we evaluate the scalar multiple at x = 0:

(cp)(0) = c * p(0) = c * 0 = 0

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

Conclusion

We have shown that S satisfies all three conditions for being a subspace:

  1. The zero vector is in S.
  2. S is closed under addition.
  3. S is closed under scalar multiplication.

Therefore, S is a subspace of P3. Nice job, guys! Understanding these concepts is crucial for further studies in linear algebra. Keep practicing, and you'll master it in no time!

Let's consider another example to solidify our understanding. Suppose S is the set of all polynomials p(x) in P3 such that p(1) = 0. We can follow the same steps to verify if S is a subspace of P3.

1. The Zero Vector

The zero polynomial z(x) = 0 is in S because z(1) = 0.

2. Closure Under Addition

Let p(x) and q(x) be in S, so p(1) = 0 and q(1) = 0. Then (p + q)(1) = p(1) + q(1) = 0 + 0 = 0. Thus, p(x) + q(x) is in S.

3. Closure Under Scalar Multiplication

Let p(x) be in S, so p(1) = 0. Let c be a scalar. Then (cp)(1) = c * p(1) = c * 0 = 0. Thus, c * p(x) is in S.

Again, S satisfies all three conditions, so S is a subspace of P3.

However, consider a different set S defined as the set of all polynomials p(x) in P3 such that p(0) = 1. Let's check if this S is a subspace of P3.

1. The Zero Vector

The zero polynomial z(x) = 0 is not in S because z(0) = 0 ≠ 1. Since S does not contain the zero vector, S is not a subspace of P3. We don't need to check the other conditions.

This example illustrates that not every subset of a vector space is a subspace. The conditions must be strictly satisfied. Remember, the zero vector must always be in the set for it to be a subspace. Closure under addition and scalar multiplication are also crucial. Always verify these conditions carefully when determining if a set is a subspace of a vector space.

In summary, to show that a set S is a subspace of P3, you must demonstrate that S contains the zero vector, and is closed under both addition and scalar multiplication. By carefully applying these conditions, you can confidently determine whether a given subset is indeed a subspace. Keep up the great work, and you'll become a pro at linear algebra in no time!

Now you know how to prove that S is a subspace of P3. Keep practicing and exploring different sets S to deepen your understanding. Good luck, and have fun with linear algebra!