关于‘X、Y为V的子空间时X+Y也为V的子空间’的证明及自证逻辑合理性验证请求

Hey there! I noticed you're working through a subspace proof and want to verify if your logic is fully sound. Let's start by laying out your proposition and proof clearly, then I'll walk through where things are solid and where there's room to tighten up the reasoning.

Proposition #

If X is a subspace of V and Y is a subspace of V, then X + Y is a subspace of V.

Your Submitted Proof #

Let $\vec{a},\vec{b} \in X$, and $l,k,\phi,\lambda \in \mathbb{R}$

• Since X is a subspace, $ l\vec{a} + k\vec{b} \in X $
• Since Y is a subspace, $ \phi\vec{p} + \lambda\vec{q} \in Y $ (note: $\vec{p},\vec{q} \in Y$ wasn't explicitly defined here)

X + Y implies $ (l\vec{a} + k\vec{b}) + (\phi\vec{p} + \lambda\vec{q}) \in X + Y $

Thus, X + Y is closed under vector addition and scalar multiplication.

• Since X is a subspace, $ \vec{0} \in X $
• Since Y is a subspace, $ \vec{0} \in Y $

X + Y implies $ \vec{0} + \vec{0} = \vec{0} $

Thus, X + Y is nonempty.

Hence, X + Y is a subspace of V.

Logic Soundness Analysis #

Your core intuition is correct, but there are a few gaps and missteps in the reasoning that need fixing to make the proof rigorous:

1. Undefined variables: You mention $\vec{p},\vec{q} \in Y$ but never explicitly state they're elements of Y at the start. This is a small oversight but important for clarity.
2. Confusion between linear combinations and subspace closure: To properly prove closure for X+Y, you should start with arbitrary elements of X+Y, not linear combinations of elements from X and Y directly. Here's the standard, tighter approach:
   • Closure under addition: Let $\vec{u} = \vec{x_1} + \vec{y_1}$ and $\vec{v} = \vec{x_2} + \vec{y_2}$ where $\vec{x_1},\vec{x_2} \in X$ and $\vec{y_1},\vec{y_2} \in Y$. Then $\vec{u} + \vec{v} = (\vec{x_1} + \vec{x_2}) + (\vec{y_1} + \vec{y_2})$. Since X is a subspace, $\vec{x_1} + \vec{x_2} \in X$; since Y is a subspace, $\vec{y_1} + \vec{y_2} \in Y$. Therefore $\vec{u} + \vec{v} \in X + Y$.
   • Closure under scalar multiplication: Let $\vec{u} = \vec{x} + \vec{y}$ where $\vec{x} \in X$, $\vec{y} \in Y$, and $c \in \mathbb{R}$. Then $c\vec{u} = c\vec{x} + c\vec{y}$. Since X is a subspace, $c\vec{x} \in X$; since Y is a subspace, $c\vec{y} \in Y$. Therefore $c\vec{u} \in X + Y$.
3. Nonemptiness: Your proof of the zero vector being in X+Y is solid—since $\vec{0} \in X$ and $\vec{0} \in Y$, $\vec{0} + \vec{0} = \vec{0} \in X + Y$, which also confirms nonemptiness.

With those adjustments, your proof will be fully logically sound!

备注：内容来源于stack exchange，提问作者Colton Fitzjarrald