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# 数学代写|线性代数代写Linear algebra代考|Examples of vector subspaces

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## 数学代写|线性代数代写Linear algebra代考|Examples of vector subspaces

Let $V$ be a vector space and $S$ be a non-empty subset of $V$. If the set $S$ satisfies all 10 axioms of a vector space with respect to the same vector addition and scalar multiplication as $V$ then $S$ is also a vector space. We say $S$ is a subspace of $V$.

Definition (3.4). A non-empty subset $S$ of a vector space $V$ is called a subspace of $V$ if it also a vector space with respect to the same vector addition and scalar multiplication as $V$.
We illustrate this in Fig. 3.3.

Note the difference between subspace and subset. A subset is merely a specific set of elements chosen from $V$. A subset must also satisfy the 10 axioms of vector space to be called a subspace.

## 数学代写|线性代数代写Linear algebra代考|How do we prove this proposition?

$(\Rightarrow)$. We first assume that $S$ is a subspace of the vector space $V$, and from this we deduce conditions (a) and (b) [show that we have closure under vector addition and scalar multiplication]. $(\Leftarrow)$. Then we assume conditions (a) and (b) are satisfied, and from this we deduce that the set $S$ is a subspace of the vector space $V$.
Proof.
Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in the set $S$.
$(\Rightarrow)$. Let $S$ be a subspace of the vector space $V$. By the above definition (3.4) we have closure under vector addition and scalar multiplication because the set $S$ is itself a vector space. [Remember, axioms 1 and 6 state that we have closure under vector addition and scalar multiplication]. Hence conditions (a) and (b) hold.
$(\Leftarrow)$. Assume conditions (a) and (b) are satisfied, that is we have closure under vector addition and scalar multiplication.
Required to prove that all 10 axioms of the last section are satisfied.
We have closure, therefore axioms 1 and 6 are satisfied. Axioms 2, 3, 7, 8, 9 and 10 are satisfied because these axioms are true for all vectors in the vector space $V$ and vectors $\mathbf{u}$ and $\mathbf{v}$ are vectors in the vector space $V$.
For the set $S$ we need to prove axioms 4 and 5 which are:

There is a zero vector $\mathbf{O}$ in $V$ which satisfies
$$\mathbf{u}+\mathbf{O}=\mathbf{u} \quad \text { for every vector } \mathbf{u} \text { in } V$$

For every vector $\mathbf{u}$ there is a vector $-\mathbf{u}$ which satisfies the following:
$$\mathbf{u}+(-\mathbf{u})=\mathbf{O}$$
We have to show (Fig. 3.6) that the zero vector, $\mathbf{O}$, and $-\mathbf{u}$ are also in $S$ for any $\mathbf{u}$ in $S$.

## 数学代写|线性代数代写Linear algebra代考|Examples of vector subspaces

Let $V$ be a vector space and $S$ be a non-empty subset of $V$. If the set $S$ satisfies all 10 axioms of a vector space with respect to the same vector addition and scalar multiplication as $V$ then $S$ is also a vector space. We say $S$ is a subspace of $V$.

Definition (3.4). A non-empty subset $S$ of a vector space $V$ is called a subspace of $V$ if it also a vector space with respect to the same vector addition and scalar multiplication as $V$.
We illustrate this in Fig. 3.3.

Note the difference between subspace and subset. A subset is merely a specific set of elements chosen from $V$. A subset must also satisfy the 10 axioms of vector space to be called a subspace.

## 数学代写|线性代数代写Linear algebra代考|How do we prove this proposition?

$(\Rightarrow)$. We first assume that $S$ is a subspace of the vector space $V$, and from this we deduce conditions (a) and (b) [show that we have closure under vector addition and scalar multiplication]. $(\Leftarrow)$. Then we assume conditions (a) and (b) are satisfied, and from this we deduce that the set $S$ is a subspace of the vector space $V$.
Proof.
Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in the set $S$.
$(\Rightarrow)$. Let $S$ be a subspace of the vector space $V$. By the above definition (3.4) we have closure under vector addition and scalar multiplication because the set $S$ is itself a vector space. [Remember, axioms 1 and 6 state that we have closure under vector addition and scalar multiplication]. Hence conditions (a) and (b) hold.
$(\Leftarrow)$. Assume conditions (a) and (b) are satisfied, that is we have closure under vector addition and scalar multiplication.
Required to prove that all 10 axioms of the last section are satisfied.
We have closure, therefore axioms 1 and 6 are satisfied. Axioms 2, 3, 7, 8, 9 and 10 are satisfied because these axioms are true for all vectors in the vector space $V$ and vectors $\mathbf{u}$ and $\mathbf{v}$ are vectors in the vector space $V$.
For the set $S$ we need to prove axioms 4 and 5 which are:

There is a zero vector $\mathbf{O}$ in $V$ which satisfies
$$\mathbf{u}+\mathbf{O}=\mathbf{u} \quad \text { for every vector } \mathbf{u} \text { in } V$$

For every vector $\mathbf{u}$ there is a vector $-\mathbf{u}$ which satisfies the following:
$$\mathbf{u}+(-\mathbf{u})=\mathbf{O}$$
We have to show (Fig. 3.6) that the zero vector, $\mathbf{O}$, and $-\mathbf{u}$ are also in $S$ for any $\mathbf{u}$ in $S$.

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