Why is the zero vector required to be in the set when proving subspace status?

The three standard criteria are: 1) The subset is nonempty (typically verified by showing it contains the zero vector); 2) The subset is closed under vector addition; 3) The subset is closed under scalar multiplication.

Conditions: The subset is part of a larger vector space.; The operations are the inherited vector addition and scalar multiplication.

Scalar multiplication preserves the relation because multiplying the equation $x_3 = x_1 + x_2$ by a scalar $c$ yields $cx_3 = cx_1 + cx_2$. This shows that the third component of the scaled vector $cx$ equals the sum of its first and second components.

Conditions: $x$ is a vector in the set satisfying $x_3 = x_1 + x_2$.; $c$ is a real scalar.; Scalar multiplication is defined componentwise.

Adding two vectors preserves the relation because the sum of the third components equals the sum of the first components plus the sum of the second components. Specifically, if $x_3 = x_1 + x_2$ and $y_3 = y_1 + y_2$, then $(x+y)_3 = x_3 + y_3 = (x_1 + x_2) + (y_1 + y_2) = (x_1 + y_1) + (x_2 + y_2) = (x+y)_1 + (x+y)_2$.

Conditions: $x$ and $y$ are vectors in $\mathbb{R}^3$.; $x$ and $y$ satisfy the equation $x_3 = x_1 + x_2$ and $y_3 = y_1 + y_2$ respectively.; Vector addition is defined componentwise.

To prove that the set $W = \{(x_1, x_2, x_3) \in \mathbb{R}^3 \mid x_3 = x_1 + x_2\}$ is a subspace, you must verify three conditions: nonemptiness (containing the zero vector), closure under vector addition, and closure under scalar multiplication.

Conditions: The set is defined by the linear equation $x_3 = x_1 + x_2$ in $\mathbb{R}^3$.; The field of scalars is $\mathbb{R}$.