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Let $W_1=\{(a_1,a_2,...,a_n) \in \mathbb{R}^n : a_1 + a_2 + ... + a_n = 0 \}$. We want to show that $W_1$ is a subspace of $\mathbb{R}^n$.


My attempt:

In order to show that $W_1$ is a subspace of $\mathbb{R}^n$, we must satisfy the following conditions: $W_1 \subset \mathbb{R}^n$, $\exists 0 \in W_1$, $W_1$ is closed under addition, and $W_1$ is closed under scalar multiplication. $W_1$ is clearly a subset of $\mathbb{R}^n$ because any element in $\mathbb{R}^n$ can be defined by a vector $[a_1, a_2, ..., a_n].$ Next, there exists a zero element in $W_1$ as $a_1=a_2=...=a_n=0,$ satisfies our criteria, namely $(0,0,0,...,0)$ is contained in $W_1$. To show closure under addition, take $(a_1, a_2, ..., a_n), (b_1, b_2, ..., b_n) \in W_1$, then $(a_1, a_2, ..., a_n) + (b_1, b_2, ..., b_n) = (a_1+b_1, a_2+b_2, ..., a_n+b_n)$ is also contained within $W_1$. Lastly, checking scalar multiplication, it must be that $c(a_1, a_2, ..., a_n)=(ca_1, ca_2, ..., ca_n) \in W_1$. Hence, $W_1$ is a subspace of $\mathbb{R}^n$.


I feel like I am taking great leaps of faith here, so I just want to double check that I am doing this correctly and understand the Subspace Criterion properly. Thank you for any pointers.

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  • $\begingroup$ It’s not enough for there to be some element of $W_1$ that acts as an additive identity. It must be identical to the zero element of the enclosing space. $\endgroup$ – amd Oct 3 '18 at 18:37
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You just claimed, without any hint of a proof, that if $(a_1, a_2,\ldots, a_n), (b_1, b_2,\ldots, b_n) \in W_1$ and $c\in\mathbb R$, then $(a_1, a_2,\ldots, a_n)+(b_1, b_2,\ldots, b_n) \in W_1$ and $c(a_1, a_2,\ldots, a_n)\in W_1$. It's not hard, though. If $(a_1, a_2,\ldots, a_n), (b_1, b_2,\ldots, b_n) \in W_1$, then $a_1+\cdots+a_n=b_1+\cdots+b_n=0$. But then $(a_1, a_2,\ldots, a_n)+(b_1, b_2,\ldots, b_n) \in W_1$, since$$(a_1+b_1)+\cdots+(a_n+b_n)=a_1+\cdots+a_n+b_1+\cdots+b_n=0+0=0.$$Can you prove now that $c(a_1, a_2,\ldots, a_n)\in W_1$?

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