如何证明满足6x-2y-4z=0的ℝ³向量集合对加法封闭？

Let's walk through this proof clearly, leveraging the given simplification that the set only contains vectors where (x=y=z):

Step 1: Define the Set Clearly #

First, let's formalize our set (V):

(V = { (x, y, z) \in \mathbb{R}^3 \mid 6x - 2y - 4z = 0 })
And we know this simplifies to (V = { (t, t, t) \mid t \in \mathbb{R} }) (since substituting (x=y=z=t) satisfies the original equation, and any solution to the equation must have (x=y=z)).

Step 2: Pick Arbitrary Elements from (V) #

To prove closure under addition, we need to show that any two vectors in (V), when added together, still land in (V).

• Let (\mathbf{v}_1) be an arbitrary vector in (V). By the simplified condition, we can write (\mathbf{v}_1 = (a, a, a)) where (a) is some real number.
• Let (\mathbf{v}_2) be another arbitrary vector in (V). Similarly, (\mathbf{v}_2 = (b, b, b)) where (b) is some real number.

Step 3: Compute the Sum and Verify Membership #

Calculate the sum of (\mathbf{v}_1) and (\mathbf{v}_2):

v₁ + v₂ = (a + b, a + b, a + b)

Now check if this sum is in (V):

• The sum has all three components equal to (a+b), so it fits the form ((t, t, t)) where (t = a+b) (and (a+b) is a real number, since real numbers are closed under addition).
• Alternatively, substitute into the original equation to confirm:
  (6(a+b) - 2(a+b) - 4(a+b) = (6 - 2 - 4)(a+b) = 0 \times (a+b) = 0), which satisfies the original equation defining (V).

Conclusion #

Since the sum of any two vectors in (V) is also in (V), the set is closed under vector addition.

内容的提问来源于stack exchange，提问作者Is12Prime