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Answers for “黑板上展示的 R^4 中的具体正交补示例是什么?”

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The example defines a subspace WW spanned by the first two standard basis vectors of R4\mathbb{R}^4, and its orthogonal complement W⊥W^\perp spanned by the last two standard basis vectors. Together, they form a direct sum that equals the entire space R4\mathbb{R}^4.

Conditions: The ambient space is R4\mathbb{R}^4.; The bases are taken as given on the board.

Find a method

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To decompose a vector in R4\mathbb{R}^4, you separate its components based on the bases of WW and W⊥W^\perp. The part corresponding to the basis of WW (the first two coordinates) forms a vector in WW, and the part corresponding to the basis of W⊥W^\perp (the last two coordinates) forms a vector in W⊥W^\perp.

Conditions: The ambient space is R4\mathbb{R}^4.; WW has basis {(1000),(0100)}\left\{\begin{pmatrix}1\\0\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\\0\end{pmatrix}\right\}.; W⊥W^\perp has basis {(0010),(0001)}\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., (43−26)\begin{pmatrix}4\\3\\-2\\6\end{pmatrix}.

Understand why

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The lecturer states that the representation is unique because, in the context of a direct sum W⊕W⊥=R4W \oplus W^\perp = \mathbb{R}^4, there is only one way to split a given vector into a part belonging to WW and a part belonging to W⊥W^\perp. No other pair of vectors from these subspaces can sum to the same original vector.

Conditions: The ambient space is R4\mathbb{R}^4.; WW and W⊥W^\perp are the specific subspaces defined on the board.; The equation W⊕W⊥=R4W \oplus W^\perp = \mathbb{R}^4 holds.

Understand why

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The representation is unique because the direct sum W⊕W⊥=R4W \oplus W^\perp = \mathbb{R}^4 implies that the intersection of WW and W⊥W^\perp contains only the zero vector. Therefore, there is only one way to express any vector in R4\mathbb{R}^4 as the sum of a vector from WW and a vector from W⊥W^\perp.

Conditions: WW and W⊥W^\perp are subspaces of R4\mathbb{R}^4.; W∩W⊥={0}W \cap W^\perp = \{\mathbf{0}\}.; The equation W⊕W⊥=R4W \oplus W^\perp = \mathbb{R}^4 holds.

Find a method

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A general vector in R4\mathbb{R}^4 is decomposed by separating its coordinates into two blocks. The first two coordinates form a vector in WW, and the last two coordinates form a vector in W⊥W^\perp.

Conditions: The ambient space is R4\mathbb{R}^4.; WW has basis {(1000),(0100)}\left\{\begin{pmatrix}1\\0\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\\0\end{pmatrix}\right\}.; W⊥W^\perp has basis {(0010),(0001)}\left\{\begin{pmatrix}0\\0\\1\\0\end{pmatrix},\begin{pmatrix}0\\0\\0\\1\end{pmatrix}\right\}.; The vector is represented as (abcd)\begin{pmatrix}a\\b\\c\\d\end{pmatrix} with a,b,c,d∈Ra,b,c,d \in \mathbb{R}.