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The set is considered a subspace because vectors of the form (x,0,−x) stay in the same form after addition or scalar multiplication. This set is the span of (1,0,−1), a line through the origin.

Conditions: The vectors are of the form (x,0,-x).; The operation is vector addition or scalar multiplication.

The set is considered a subspace because vectors of the form (x,0,−x) stay in the same form after addition or scalar multiplication. This set is the span of (1,0,−1), a line through the origin.

Conditions: The vectors are of the form (x,0,-x).; The operation is vector addition or scalar multiplication.

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 $\mathbb{R}$.

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 $\mathbb{R}$.

The lecturer states that the representation is unique because, in the context of a direct sum $W \oplus W^\perp = \mathbb{R}^4$, there is only one way to split a given vector into a part belonging to $W$ and a part belonging to $W^\perp$. No other pair of vectors from these subspaces can sum to the same original vector.

Conditions: The ambient space is $\mathbb{R}^4$.; $W$ and $W^\perp$ are the specific subspaces defined on the board.; The equation $W \oplus W^\perp = \mathbb{R}^4$ holds.

The lecturer states that the representation is unique because, in the context of a direct sum $W \oplus W^\perp = \mathbb{R}^4$, there is only one way to split a given vector into a part belonging to $W$ and a part belonging to $W^\perp$. No other pair of vectors from these subspaces can sum to the same original vector.

Conditions: The ambient space is $\mathbb{R}^4$.; $W$ and $W^\perp$ are the specific subspaces defined on the board.; The equation $W \oplus W^\perp = \mathbb{R}^4$ holds.