Is W a Subspace of V3(F)? Let’s Prove It!

Hey math lovers, gather ‘round! Today, we’re diving deep into the awesome world of vector spaces and tackling a super important concept: subspaces . Specifically, we’re going to show that a certain set, which we’ll call ‘W’, is a subspace of a larger vector space, V3(F). You know, those V3(F) spaces are like the VIP lounges of vector spaces – they’re all about vectors with three components over a field F. Now, proving something is a subspace might sound a bit intimidating, but trust me, guys, it’s all about following a few simple rules. Think of it like a checklist; if you tick all the boxes, then BAM! You’ve got yourself a subspace. So, let’s get down to business and break down exactly what this set W is and why it fits the bill as a subspace of V3(F). Get ready to flex those linear algebra muscles!

So, what exactly is this set W we’re talking about? It’s defined as all vectors of the form (a, b, 8712f) where ‘a’ and ‘b’ can be any numbers from our field F. Now, V3(F) is just the space of all possible vectors with three components, where each component is from that field F. So, V3(F) looks like (x, y, z) where x, y, and z are all in F. Our set W is a subset of V3(F), meaning all its elements are also elements of V3(F). But just being a subset isn’t enough to be a subspace. We need to make sure it behaves nicely when we do vector operations within it. To prove that W is a subspace of V3(F), we need to verify three fundamental conditions. These conditions are like the golden rules of subspaces, and if W satisfies all of them, then we’re golden. First off, the zero vector must be in W. Second, if you take any two vectors that are already in W, their sum must also be in W. This is called closure under addition. And third, if you take any vector in W and multiply it by any scalar (a number from our field F), the result must still be in W. This is called closure under scalar multiplication. If all three of these conditions hold true, then congratulations, you’ve officially proven that W is a subspace of V3(F)! It’s like showing that a smaller club fits perfectly within a larger club, following all the same rules.

Let’s kick things off with the first crucial condition: the zero vector . For V3(F), the zero vector is always (0, 0, 0) . Now, we need to check if this (0, 0, 0) vector can be represented in the form of our set W, which is (a, b, 8712f) . Remember, in the definition of W, the first component ‘a’ can be any number from F, the second component ‘b’ can be any number from F, and the third component is fixed at 8712f . To get the zero vector (0, 0, 0) , we need to see if we can plug in values for ‘a’ and ‘b’ that make the components match. So, we can set a = 0 and b = 0 . This gives us the vector (0, 0, 8712f) . Now, here’s the snag, guys. For (0, 0, 8712f) to be equal to (0, 0, 0) , the third component must be zero. This means we’d need 8712f = 0 . But is that always true? Not necessarily! The field F could be any field, and 8712f is a specific value. Unless 8712f happens to be zero in the field F, the zero vector (0, 0, 0) is not in our set W. This is a pretty big deal! If the very first condition fails – if the zero vector isn’t chilling in W – then W cannot be a subspace of V3(F). So, this initial check is super important, and it looks like our set W might be in trouble right from the start, depending on the value of 8712f in the field F. It’s a crucial first step, and sometimes it’s the step that shows us something isn’t a subspace!

Now, let’s say, for the sake of argument and to explore the other conditions, that 8712f is zero in our field F. So, our set W actually consists of vectors of the form (a, b, 0) , where ‘a’ and ‘b’ are any elements from F. In this hypothetical scenario, our first condition (the zero vector) would be satisfied because we can set a=0 and b=0 to get (0, 0, 0) , which is indeed in W. Okay, so assuming this, let’s move on to the second condition: closure under addition . This means if we take any two vectors that are in W, their sum must also be in W. Let’s pick two arbitrary vectors from W. Let v1 = (a1, b1, 0) and v2 = (a2, b2, 0) , where a1, b1, a2, b2 are all elements of F. Now, we add them together: v1 + v2 = (a1 + a2, b1 + b2, 0 + 0) . Simplifying this, we get v1 + v2 = (a1 + a2, b1 + b2, 0) . Look at the result, guys! The resulting vector has the form (some number, some number, 0) . Since a1 and a2 are in F, their sum a1 + a2 is also in F (because F is a field, and fields are closed under addition). Similarly, b1 + b2 is in F. The third component is 0, which matches the required form of vectors in W. Therefore, the sum v1 + v2 is indeed in W. So, W is closed under addition ! This is a big win and means our hypothetical W is looking more and more like a subspace.

We’re on the home stretch, guys! The third and final condition for W to be a subspace is closure under scalar multiplication . This means if we take any vector in W and multiply it by any scalar (which is just a number from our field F), the resulting vector must also be in W. Let’s take our hypothetical vector v = (a, b, 0) from W, where ‘a’ and ‘b’ are in F. Let ‘c’ be any scalar from our field F. We multiply v by c: c * v = c * (a, b, 0) . Performing the scalar multiplication, we get c * v = (c*a, c*b, c*0) . Simplifying, this becomes c * v = (c*a, c*b, 0) . Now, let’s inspect this resulting vector. Since ‘c’ and ‘a’ are both in F, their product c*a is also in F (because F is a field and is closed under multiplication). Similarly, c*b is in F. The third component is c*0 , which is always 0, regardless of what ‘c’ is. So, the resulting vector (c*a, c*b, 0) has the form (some number, some number, 0) , which is exactly the form of vectors in our set W. This means that W is closed under scalar multiplication ! With this, we’ve ticked off the second and third boxes in our subspace checklist. If 8712f were 0, our set W would satisfy all the conditions to be a subspace.

However, we must return to the original definition of W: vectors of the form (a, b, 8712f) . As we identified in the first step, the crucial point is whether the zero vector (0, 0, 0) is present in W. For (0, 0, 0) to be in W, we need to be able to find ‘a’ and ‘b’ from F such that (a, b, 8712f) = (0, 0, 0) . This requires a=0 , b=0 , and critically, 8712f = 0 . If 8712f is not equal to 0 in the field F, then the zero vector is not an element of W. Since one of the fundamental requirements for a subspace is that it must contain the zero vector, the set W = {(a, b, 8712f) | a, b in F} is generally NOT a subspace of V3(F) , unless the specific value 8712f happens to be zero within the field F. It’s a classic example where failing just one condition means it’s not a subspace. So, while it’s fun to check closure under addition and scalar multiplication, the zero vector test is often the quickest way to disqualify a set from being a subspace. Always remember that the zero vector must be present! It’s the bedrock of any subspace, guys. Keep this in mind for your next linear algebra challenge!