Show That S is a Subspace of P3

Hey guys! Today, we’re diving into a fun little problem in linear algebra: proving that a set S is a subspace of P3 . What’s P3 , you ask? Well, it’s the vector space of all polynomials with a degree of 3 or less. So, things like x^3 + 2x^2 - x + 5 or even just the number 7 are elements of P3 . Now, what does it mean for S to be a subspace of P3 ? Buckle up, because we’re about to find out!

Understanding Subspaces

Before we jump into the proof, let’s quickly recap what a subspace is. In simple terms, a subspace is a subset of a vector space that itself satisfies the properties of a vector space. This means that for S to be a subspace of P3 , it must meet three crucial conditions:

1. The zero vector must be in S: In our case, the zero vector is the zero polynomial, 0 . So, S must contain the polynomial that is simply zero.
2. S must be closed under addition: If we take any two polynomials in S and add them together, the result must also be in S . This means that adding two polynomials from S won’t magically kick you out of S .
3. S must be closed under scalar multiplication: If we take any polynomial in S and multiply it by any scalar (a real number), the result must also be in S . So, scaling a polynomial from S keeps you safely within S .

If S satisfies all three of these conditions, then congrats! It’s a subspace of P3 . Let’s look at how to prove a set S is a subspace of P3 with an example.

Example: Proving S is a Subspace of P3

Let’s say S is defined as the set of all polynomials in P3 that have the form ax^2 + bx , where a and b are real numbers. In other words, S contains all polynomials of degree 2 or less that have no constant term. Our mission, should we choose to accept it, is to prove that S is a subspace of P3 .

1. The Zero Vector

First, we need to check if the zero vector (the zero polynomial) is in S . Can we write the zero polynomial in the form ax^2 + bx ? Absolutely! Just set a = 0 and b = 0 , and you get 0x^2 + 0x = 0 . So, the zero vector is indeed in S . That’s one down, two to go!

2. Closure Under Addition

Next, we need to show that S is closed under addition. Let’s take two arbitrary polynomials from S : p(x) = a1x^2 + b1x and q(x) = a2x^2 + b2x , where a1 , b1 , a2 , and b2 are real numbers. Now, let’s add them together:

p(x) + q(x) = (a1x^2 + b1x) + (a2x^2 + b2x) = (a1 + a2)x^2 + (b1 + b2)x

Notice that the resulting polynomial is also in the form ax^2 + bx , where a = a1 + a2 and b = b1 + b2 . Since a1 + a2 and b1 + b2 are also real numbers, the sum p(x) + q(x) is an element of S . This proves that S is closed under addition. Awesome!

3. Closure Under Scalar Multiplication

Finally, we need to show that S is closed under scalar multiplication. Let’s take an arbitrary polynomial from S : p(x) = ax^2 + bx , where a and b are real numbers. Let’s also take an arbitrary scalar c , which is a real number. Now, let’s multiply p(x) by c :

*c * p(x) = c * (ax^2 + bx) = (ca)x^2 + (cb)x*

Again, notice that the resulting polynomial is in the form ax^2 + bx , where a = ca and b = cb . Since ca and cb are also real numbers, the scalar multiple *c * p(x)* is an element of S . This proves that S is closed under scalar multiplication. Fantastic!

Conclusion

We’ve shown that S satisfies all three conditions for being a subspace: it contains the zero vector, it’s closed under addition, and it’s closed under scalar multiplication. Therefore, we can confidently conclude that S is a subspace of P3 . High five!

Another Example with a Different Set

Okay, let’s solidify this concept with another example. This time, let’s consider a set T defined as all polynomials in P3 that have the form ax^3 + c , where a and c are real numbers. In other words, T contains cubic polynomials where the x^2 and x terms are missing. Let’s see if T is a subspace of P3 .

1. The Zero Vector

Is the zero polynomial in T ? To find out, we need to see if we can write 0 in the form ax^3 + c . If we set a = 0 and c = 0 , we get 0x^3 + 0 = 0 . So, the zero vector is in T . Good start!

2. Closure Under Addition

Now, let’s check for closure under addition. Let’s take two arbitrary polynomials from T : p(x) = a1x^3 + c1 and q(x) = a2x^3 + c2 , where a1 , c1 , a2 , and c2 are real numbers. Let’s add them:

p(x) + q(x) = (a1x^3 + c1) + (a2x^3 + c2) = (a1 + a2)x^3 + (c1 + c2)

The result is in the form ax^3 + c , where a = a1 + a2 and c = c1 + c2 . Since a1 + a2 and c1 + c2 are real numbers, the sum p(x) + q(x) is in T . So far, so good!

3. Closure Under Scalar Multiplication

Finally, let’s check for closure under scalar multiplication. Let’s take an arbitrary polynomial from T : p(x) = ax^3 + c , where a and c are real numbers. Let’s also take an arbitrary scalar k , which is a real number. Now, let’s multiply p(x) by k :

*k * p(x) = k * (ax^3 + c) = (ka)x^3 + (kc)*

The result is in the form ax^3 + c , where a = ka and c = kc . Since ka and kc are real numbers, the scalar multiple *k * p(x)* is in T . Excellent!

Conclusion

Since T satisfies all three conditions – containing the zero vector, closure under addition, and closure under scalar multiplication – we can conclude that T is also a subspace of P3 .

When a Set is NOT a Subspace

Now that we’ve seen two examples of sets that are subspaces of P3 , let’s consider a set that isn’t . This will help us understand the conditions even better.

Let’s define a set U as all polynomials in P3 that have the form x^3 + ax^2 + bx + c , where a , b , and c are real numbers. Notice that every polynomial in U must have an x^3 term with a coefficient of 1. Let’s see if U is a subspace of P3 .

1. The Zero Vector

Is the zero polynomial in U ? To be in U , a polynomial must have an x^3 term. The zero polynomial, 0 , does not have an x^3 term. Therefore, the zero vector is not in U . Right away, we know that U cannot be a subspace of P3 . We don’t even need to check the other conditions!

Conclusion

Since U does not contain the zero vector, it fails the first condition for being a subspace. Therefore, U is not a subspace of P3 . This example highlights the importance of checking all three conditions. If even one condition fails, the set is not a subspace.

Key Takeaways

• Subspace Definition: A subspace S of a vector space V must contain the zero vector, be closed under addition, and be closed under scalar multiplication.
• Proving a Subspace: To prove that a set S is a subspace of P3 , you must demonstrate that all three conditions are met.
• Counterexamples: To prove that a set is not a subspace, you only need to show that one of the three conditions fails.
• P3 Elements: Elements of P3 are polynomials with a degree of 3 or less.

Alright, that’s a wrap on proving whether a set is a subspace of P3 ! I hope these examples and explanations have helped you understand the concept a little better. Keep practicing, and you’ll be a subspace pro in no time! Happy learning, guys!