Is W A Subspace Of R3? Let's Find Out!
Hey guys! Today, we're diving deep into the world of linear algebra to tackle a super common question: how do you show that a set W is a subspace of R3? This might sound a bit intimidating at first, but trust me, it's all about following a few key steps. Think of R3 as our big playground, and W is a smaller section within that playground. We need to prove that this section W follows some specific rules to be considered a valid 'subspace'. It's like making sure a specific park area is officially designated for, say, picnics – it needs to meet certain criteria. We'll break down these criteria and walk through an example so you can see exactly what's going on. So grab your favorite study snack, get comfy, and let's get this linear algebra party started!
The Three Crucial Tests for a Subspace
Alright, so to prove that a set W is a subspace of R3 (or any vector space, really!), we need to put it through three important tests. These aren't just random hoops to jump through; they're fundamental properties that define what a subspace is. If W passes all three, then congratulations, it's officially a subspace! If it fails even one, then it's not.
Test 1: Does it Contain the Zero Vector?
The first and often the easiest test is to check if the zero vector is in W. In R3, the zero vector is (0, 0, 0). Why is this important? Well, think about it – a subspace needs to be a sort of 'self-contained' little universe within the larger vector space. If it doesn't even have the origin (the zero vector), it's like a galaxy without a central star, which just doesn't feel right in the linear algebra cosmos. So, the very first thing you do is plug in 0 for all your variables in the definition of W and see if you get (0, 0, 0). If you do, it passes this test with flying colors! If not, you can stop right there – W is not a subspace. Easy peasy, right?
Test 2: Is it Closed Under Addition?
This test is about what happens when you take two elements (vectors) from W and add them together. Closure under addition means that if you pick any two vectors, let's call them u and v, that are inside W, their sum, u + v, must also be inside W. Imagine W is a set of points on a specific plane. If you take any two points on that plane (within W) and draw a line segment between them, every point on that segment (their sum, in a sense) should also land on that same plane. It's like saying that if you're playing within a designated zone, adding any two things within that zone should keep you within that zone. This is super important because it ensures that W 'behaves' nicely when you perform the basic operation of vector addition. We'll usually do this by picking generic vectors u and v that satisfy the conditions of W, adding them, and then checking if the resulting vector also satisfies those conditions. If it does, W passes this crucial test.
Test 3: Is it Closed Under Scalar Multiplication?
The final test is closure under scalar multiplication. This one is pretty similar in spirit to the addition test. It means that if you take any vector v that is inside W, and you multiply it by any scalar (just a regular number, like 2, -5, or 1/3), the resulting vector, cv, must also be inside W. Think of it like stretching or shrinking a vector that's already in W. If you stretch it (multiply by a scalar greater than 1) or shrink it (multiply by a scalar between 0 and 1) or even flip it (multiply by a negative scalar), the new, transformed vector has to stay within the boundaries of W. It's like saying that if you have a certain 'direction' and 'magnitude' defined by a vector in W, scaling that vector should keep you within the same 'space' W occupies. Again, we'll typically prove this by taking a generic vector v from W, multiplying it by a general scalar 'c', and then verifying that the outcome cv satisfies the defining conditions of W. If it does, W nails this final test!
Putting it all Together: An Example in R3
Okay, theory is great, but let's get our hands dirty with a concrete example. Suppose we are asked to determine if the set W defined as follows is a subspace of R3:
W = { (x, y, z) in R3 | 2x - y + 3z = 0 }
This definition says that W is the set of all vectors (x, y, z) in R3 where the coordinates satisfy the equation 2x - y + 3z = 0. This equation represents a plane passing through the origin in R3. Now, let's put W through our three tests:
Test 1: The Zero Vector Check
First, we need to see if the zero vector (0, 0, 0) is in W. We plug x=0, y=0, and z=0 into the equation defining W:
2(0) - (0) + 3(0) = 0 - 0 + 0 = 0
Since the equation holds true (0 = 0), the zero vector is in W. So, W passes Test 1! Phew, one down!
Test 2: Closure Under Addition
Next, we need to check if W is closed under addition. Let's pick two arbitrary vectors from W. Let u = (x1, y1, z1) and v = (x2, y2, z2) be two vectors in W. Because they are in W, they must satisfy the equation:
For u: 2x1 - y1 + 3z1 = 0 For v: 2x2 - y2 + 3z2 = 0
Now, let's find their sum, u + v:
u + v = (x1 + x2, y1 + y2, z1 + z2)
To see if u + v is in W, we need to check if it satisfies the defining equation. Let's substitute the components of u + v into the equation 2x - y + 3z = 0:
2(x1 + x2) - (y1 + y2) + 3(z1 + z2) = 2x1 + 2x2 - y1 - y2 + 3z1 + 3z2 = (2x1 - y1 + 3z1) + (2x2 - y2 + 3z2)
We know from our initial assumptions that 2x1 - y1 + 3z1 = 0 and 2x2 - y2 + 3z2 = 0. So, we can substitute these values back in:
= 0 + 0 = 0
Since the sum u + v satisfies the equation 2x - y + 3z = 0, the vector u + v is indeed in W. Therefore, W is closed under addition and passes Test 2! Awesome!
Test 3: Closure Under Scalar Multiplication
Finally, let's check for closure under scalar multiplication. Let v = (x, y, z) be any vector in W. This means that 2x - y + 3z = 0. Now, let 'c' be any scalar (any real number). We want to see if the vector cv is also in W.
cv = c(x, y, z) = (cx, cy, cz)
We substitute the components of cv into the defining equation 2x - y + 3z = 0:
2(cx) - (cy) + 3(cz) = c(2x) - c(y) + c(3z) = c(2x - y + 3z)
Since we know that v is in W, we know 2x - y + 3z = 0. Substituting this back in:
= c(0) = 0
Because the vector cv satisfies the equation 2x - y + 3z = 0, it means that cv is in W. So, W is closed under scalar multiplication and passes Test 3! High five!
Conclusion: W is Definitely a Subspace!
Since W passed all three tests – it contains the zero vector, it's closed under addition, and it's closed under scalar multiplication – we can confidently conclude that W is a subspace of R3. 🎉
See? It's not so scary after all! The key is to understand the three essential properties and apply them systematically to the definition of your set W. Keep practicing with different examples, and you'll become a subspace-proving pro in no time. You've got this, guys!