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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.

The representation is unique because the direct sum $W \oplus W^\perp = \mathbb{R}^4$ implies that the intersection of $W$ and $W^\perp$ contains only the zero vector. Therefore, there is only one way to express any vector in $\mathbb{R}^4$ as the sum of a vector from $W$ and a vector from $W^\perp$.

Conditions: $W$ and $W^\perp$ are subspaces of $\mathbb{R}^4$.; $W \cap W^\perp = \{\mathbf{0}\}$.; The equation $W \oplus W^\perp = \mathbb{R}^4$ holds.

The representation is unique because the direct sum $W \oplus W^\perp = \mathbb{R}^4$ implies that the intersection of $W$ and $W^\perp$ contains only the zero vector. Therefore, there is only one way to express any vector in $\mathbb{R}^4$ as the sum of a vector from $W$ and a vector from $W^\perp$.

Conditions: $W$ and $W^\perp$ are subspaces of $\mathbb{R}^4$.; $W \cap W^\perp = \{\mathbf{0}\}$.; The equation $W \oplus W^\perp = \mathbb{R}^4$ holds.