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# 数学代写|线性代数代写Linear algebra代考|Orthogonal Complements

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## 数学代写|线性代数代写Linear algebra代考|Orthogonal Complements

If $S$ is a subset of a metric vector space $V$, then $S$ inherits the metric structure from $\mathrm{V}$. With this structure, we refer to $\mathrm{S}$ as a subspace of $V$.

Definition Two subspaces $S$ and $T$ of a metric vector space $\mathrm{V}$ are orthogonal, denoted by $S \perp T$, if $\langle\mathbf{s}, \mathbf{t}\rangle=0$ for all $s \in S$ and $t \in T$. The orthogonal complement of $S$, denoted by $S^{\perp}$, is the set
$$S^{\perp}={\mathbf{v} \in \mathbf{V} \mid \mathbf{v} \perp \mathrm{S}}$$
Definition If $\mathrm{V}$ is a metric vector space, then $\mathrm{V}^{\perp}$ is called the radical of $\mathrm{V}$, and denoted by $\operatorname{Rad}(\mathrm{V})$.

Thus, $\mathrm{V}$ is nonsingular if and only if $\operatorname{Rad}(\mathrm{V})={0}$. Note that if $S$ is a subspace of $V$, then the radical of $S$ is $\operatorname{Rad}(\mathrm{S})=\mathrm{S} \cap \mathrm{S}^{\perp}$.
It should be emphasized that the properties of orthogonality can be quite different for arbitrary base fields than for the familiar case of the real base field. For instance, in the case of real metric vector spaces, we have $\mathrm{S} \cap \mathrm{S}^{\perp}={0}$, whereas in the case of metric vector spaces over finite fields, for instance, we may even have $S=S^{\perp}$, as the next example shows.
Example 11.3 It is easy to see that the subspace
$$\mathrm{S}={0000,1100,0011,1111}$$
of $\mathrm{V}(4,2)$ has the property that $\mathrm{S}=\mathrm{S}^{\perp}$. Note that $\mathrm{V}(4,2)$ is nonsingular, and yet the subspace $S$ is quite singular.

The previous example notwithstanding, we do have the following important result concerning dimensions.

## 数学代写|线性代数代写Linear algebra代考|Orthogonal Direct Sums

Definition Let $\mathrm{V}$ be a metric vector space. If $\mathrm{S}$ and $T$ are subspaces of $V$ with the property that $V=S \oplus T$ and $S \perp T$, then we say that $V$ is the orthogonal direct sum of $S$ and $T$ and write $\mathrm{V}=\mathrm{S} \oplus \mathrm{T}$.

In view of Example 11.3, it is reasonable to ask under what conditions on a subspace $S$ is it true that $V=S \oplus S^{\perp}$. The answer is given by the following theorem.

Theorem 11.8 Let $\mathrm{S}$ be a subspace of a nonsingular metric vector space V. The following statements are equivalent.
1) $S$ is nonsingular
2) $S^{\perp}$ is nonsingular
3) $\mathrm{S} \cap \mathrm{S}^{\perp}={0}$
4) $\mathrm{V}=\mathrm{S}+\mathrm{S}^{\perp}$
5) $\mathrm{V}=\mathrm{S} \oplus \mathrm{S}^{\perp}$
Proof. According to Theorem 11.7, statements 1, 2 and 3 are equivalent. For any subspaces $S$ and $T$ of a vector space $\mathrm{V}$,
$$\operatorname{dim}(\mathrm{S}+\mathrm{T})=\operatorname{dim}(\mathrm{S})+\operatorname{dim}(\mathrm{T})-\operatorname{dim}(\mathrm{S} \cap \mathrm{T})$$
and so
\begin{aligned} \operatorname{dim}\left(\mathrm{S}+\mathrm{S}^{\perp}\right) & =\operatorname{dim}(\mathrm{S})+\operatorname{dim}\left(\mathrm{S}^{\perp}\right)-\operatorname{dim}\left(\mathrm{S} \cap \mathrm{S}^{\perp}\right) \ & =\operatorname{dim}(\mathrm{V})-\operatorname{dim}\left(\mathrm{S} \cap \mathrm{S}^{\perp}\right) \end{aligned}
which shows that 3 is equivalent to 4 , and that 4 implies 5 . Since 5 clearly implies 4 , the proof is complete.

Most of the important results that we have established so far require that the space be nonsingular. Fortunately, the following theorem says that we may restrict attention to such spaces, without loosing any important structure.

## 数学代写|线性代数代写Linear algebra代考|Orthogonal Complements

$$S^{\perp}=\mathbf{v} \in \mathbf{V} \mid \mathbf{v} \perp \mathrm{S}$$

$$\mathrm{S}=0000,1100,0011,1111$$

## 数学代写|线性代数代写Linear algebra代考|Orthogonal Direct Sums

1) $S$ 是非奇异的
2) $S^{\perp}$ 是非奇异的
3) $\mathrm{S} \cap \mathrm{S}^{\perp}=0$
4) $\mathrm{V}=\mathrm{S}+\mathrm{S}^{\perp}$
5) $\mathrm{V}=\mathrm{S} \oplus \mathrm{S}^{\perp}$

$$\operatorname{dim}(S+T)=\operatorname{dim}(S)+\operatorname{dim}(T)-\operatorname{dim}(S \cap T)$$

$$\operatorname{dim}\left(\mathrm{S}+\mathrm{S}^{\perp}\right)=\operatorname{dim}(\mathrm{S})+\operatorname{dim}\left(\mathrm{S}^{\perp}\right)-\operatorname{dim}\left(\mathrm{S} \cap \mathrm{S}^{\perp}\right) \quad=\operatorname{dim}(\mathrm{V})-\operatorname{dim}\left(\mathrm{S} \cap \mathrm{S}^{\perp}\right)$$

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