What is the concrete orthogonal-complement example shown on the board in $R^4$?

To decompose a vector in $\mathbb{R}^4$, you separate its components based on the bases of $W$ and $W^\perp$. The part corresponding to the basis of $W$ (the first two coordinates) forms a vector in $W$, and the part corresponding to the basis of $W^\perp$ (the last two coordinates) forms a vector in $W^\perp$.

Conditions: The ambient space is $\mathbb{R}^4$.; $W$ has basis $\left\{\begin{pmatrix}1\\0\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\\0\end{pmatrix}\right\}$.; $W^\perp$ has basis $\left\{\begin{pmatrix}0\\0\\1\\0\end{pmatrix},\begin{pmatrix}0\\0\\0\\1\end{pmatrix}\right\}$.; The target vector is given, e.g., $\begin{pmatrix}4\\3\\-2\\6\end{pmatrix}$.

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.

A general vector in $\mathbb{R}^4$ is decomposed by separating its coordinates into two blocks. The first two coordinates form a vector in $W$, and the last two coordinates form a vector in $W^\perp$.

Conditions: The ambient space is $\mathbb{R}^4$.; $W$ has basis $\left\{\begin{pmatrix}1\\0\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\\0\end{pmatrix}\right\}$.; $W^\perp$ has basis $\left\{\begin{pmatrix}0\\0\\1\\0\end{pmatrix},\begin{pmatrix}0\\0\\0\\1\end{pmatrix}\right\}$.; The vector is represented as $\begin{pmatrix}a\\b\\c\\d\end{pmatrix}$ with $a,b,c,d \in \mathbb{R}$.