How does scalar multiplication preserve the relation x₃=x₁+x₂?

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.

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.

The zero vector must be in the set to ensure the set is nonempty. A subspace is a vector space in its own right, and every vector space must contain an additive identity, which is the zero vector.

Conditions: The set is a candidate for being a subspace of a vector space.; The definition of a subspace requires it to be a vector space under the inherited operations.

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 $\mathbb{R}$.