How does adding two vectors in the set x₃=x₁+x₂ preserve the defining relation?

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.

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 $\mathbb{R}$.