24.16. Worked Examples

Today, we'll learn how to determine if a subset of R3 is indeed a subspace. We’ll use W = {(x, y, z) ∈ R3: x + 2y + 3z = 0}. Let's start with the first step: checking if the zero vector is in W. Can anyone tell me what we would need to do?

We need to substitute 0 into the equation, right?

Exactly! So we substitute (0, 0, 0) and see if it satisfies the equation. Now, can someone do that calculation?

If we plug in (0, 0, 0), we get 0 + 2(0) + 3(0) = 0, so it works!

Correct! The zero vector is indeed in W. Now, what do we need to check next for W to be a subspace?

We have to check if it's closed under addition.

Right again! Let’s say we have two vectors in W, (x1, y1, z1) and (x2, y2, z2). When we add these two vectors, what must occur to stay within W?

Their sum must also satisfy the equation x + 2y + 3z = 0.

Exactly! When we add them and plug them into the equation, we should end up showing that it equals zero. So if we manage that, what does it mean for W?

W is closed under addition!

Right! Finally, we need to confirm scalar multiplication. Can someone summarize how that works?

If we take a scalar a and multiply it with any vector in W, it should still satisfy the equation, right?

Perfect! This shows us that W is indeed a subspace. Remember, the key steps were checking the zero vector, addition closure, and scalar multiplication closure.

Now moving on to our second example, let’s find a basis for W = {(x, y, z) ∈ R3: x + y + z = 0}. Who can explain how we might start?

We can express one variable in terms of the others. I think we can set x = -y - z.

Exactly! So if we write vectors in terms of y and z, what does that look like?

It looks like (−y−z, y, z).

Good! Now can you express that as a combination of two different vectors?

Yeah, we can break it down as y(−1, 1, 0) + z(−1, 0, 1).

Perfect! Therefore, which vectors can we consider as a basis for this space?

The basis would be {(−1, 1, 0), (−1, 0, 1)}.

Excellent! Finally, can anyone tell me how we determine the dimension of W based on the basis we've identified?

The dimension is the number of vectors in the basis, which is 2 here.

Correct! Well done, everyone! In summary, we found a basis for W and identified its dimension, which further solidifies our understanding of vector spaces.