Section: 矢量空间实例 | 4.2 矢量空间实例

Section outline

In this lesson we will give some more concrete examples to exactly what vector spaces are and the fields in which they are over. We will show how we deal with vectors in these examples and prove, using the criterion in the last lesson that these are indeed vector spaces.
::在这个教训中,我们将给出一些更具体的例子,说明矢量空间的确切位置和它们过去所在的领域。我们将展示我们如何在这些例子中处理矢量,并用上一个教训中的标准证明这些矢量空间确实是矢量空间。

Example 1: $\begin{align*}\mathbb{F}\end{align*}$
::例1:F

Recall our definition of a field $\begin{align*}\mathbb{F}\end{align*}$ . A field is a set of numbers such that every number has an additive inverse as well as a multiplicative inverse except for the 0 element. Some examples of fields are the rational numbers, the real numbers, the complex numbers and even the set of integers modulo a prime.
::回顾我们对字段 F. A 字段的定义是一组数字,使每个数字都有一个附加数反反反反反反反反反反反反反反反反。一些字段的例子是理性数字、真实数字、复杂数字,甚至一个质数的整数模卢。

First, we have to look at the axioms of a field to show that it is indeed a vector space:
::首先,我们必须看一字段的轴心,以表明它确实是一个矢量空间:

Let $\begin{align*}a,b,c \in \mathbb{F}\end{align*}$ then
::然后,让我们,b,cF

$\begin{align*}a + (b + c) = (a + b) + c\\ (a\cdot b)\cdot c = a\cdot(b\cdot c)\\ a + b = b + a\\ a\cdot b = b\cdot a\\ \exists x \in \mathbb{F} \text{ s.t. } a + x = a, x = 0\\ \exists x' \in \mathbb{F} \text{ s.t. } a\cdot x' = a, x' = 1\\ \forall{a} \in \mathbb{F}, \exists b \in\mathbb{F}\text{ s.t. } a + b = 0, b = -a\\ \forall{a} \in \mathbb{F}, \exists b' \in \mathbb{F}\text{ s.t. } a\cdot b = 1, b = a^{-1} = \frac{1}{a}\\ a\cdot(b + c) = a\cdot b + a\cdot c \end{align*}$
::a+(b+c)=(a+b)+(c)+(c)=(a)+(b)a+b=(b)+(a)a+(b)b=(a)+(a)b=(b)x************************************************************* *****************************************************************************************************************************************************************************************************************************************************************************************************************************************************************************************************

Now, let's see if these axioms add up to show that any field is a vector space (over itself).
::现在,让我们看看这些轴的加起来是否表明任何字段都是矢量空间( 本身) 。

Here, the elements of the vector space as well as the set of scalars are both the field $\begin{align*}\mathbb{F } \end{align*}$ .
::在这里,矢量空间的元素以及星标组都是F字段。

(We are referencing the axioms of a vector space introduced in lesson 4.1).
:我们正在参考第4.1节中介绍的矢量空间的轴心)。

First off, the first axiom of a vector space is true, because addition and multiplication are closed operations in the field.
::首先,矢量空间的第一个轴是真实的,因为添加和乘法是外地的封闭作业。

The second axiom is also true as commutativity holds in a field.
::第二个轴也是正确的,因为一个田地的通货性占据着一个田地。

The third axiom can be verified as we also have associativity from
::第三个轴轴心可以被验证,因为我们与

Example 2: $\begin{align*}\mathbb{R}^{n}\end{align*}$
::实例2:Rn

$\begin{align*}\mathbb{R}^{n}\end{align*}$ is a vector space over the real numbers. We define $\begin{align*}\mathbb{R}^{n} = \big\{(x_{1},...,x_{n})\mid x_{1},...,x_{n} \in \mathbb{R}\big\}\end{align*}$ . Hence, it is the set of n-tuples which are all real and the scalars are just the reals. Addition in $\begin{align*}\mathbb{R}^{n}\end{align*}$ is defined component-wise, so for $\begin{align*}\vec{x} = \begin{bmatrix}x_{1}\\\cdot\\\cdot\\\cdot\\x_{n}\end{bmatrix}, \vec{y} = \begin{bmatrix}y_{1}\\\cdot\\\cdot\\\cdot\\y_{n}\end{bmatrix} \in \mathbb{R}^{n} \text{ we get } \vec{x} + \vec{y} = \vec{x+y} = \begin{bmatrix}x_{1} + y_{1}\\\cdot\\\cdot\\\cdot\\x_{n} + y_{n}\end{bmatrix} \end{align*}$ . Likewise, scalar multiplication is defined component-wise, so for   $\begin{align*}\vec{x} = \begin{bmatrix}x_{1}\\\cdot\\\cdot\\\cdot\\x_{n}\end{bmatrix} \in \mathbb{R}^{n},\text{ and } c \in \mathbb{R}, c\vec{x} = \begin{bmatrix}c\cdot x_{1}\\\cdot\\\cdot\\\cdot\\c\cdot x_{n}\end{bmatrix}\end{align*}$ . Now, because addition in the reals is commutative and addition is closed, the first down axioms are satisfied trivially. Similarly addition is associative in the reals, so it is associative in every component implying it is associative in the entirety of the reals in n-dimensions. Also, scalar multiplication is closed in the n-dimensional reals, because it is also closed in the 1 dimensional reals. We can also see that it is closed, because addition in $\begin{align*}\mathbb{R}^{n}\end{align*}$ is closed so repeated addition which is just multiplication must be closed as well. Now, our zero vector is just the n-dimensional vector all of whose entries are zeros.
::Rn 是真实数字的矢量空间。我们定义了 Rn 的矢量。我们定义了 Rn 的矢量( x1,..., xn) xxxxx1, xxxxx1, xn) xx1, xxxxxxxxxx1, ..., xn{R} 。因此, 一组 n- tuples 都是真实的, 而 n- tuples 是真实的, c_x=nxxxxxn, 和 c_xxxxslax 仅仅是真实的。添加在 Rn- 中是封闭的。现在, 因为在 xxx=[x1]xxxxxxxxxxxxxxxxn, y=[y1] {y_xxxxxxxxxxxxxxxyn+yn] 中, 添加的附加是微不足道的。同样, 添加是真实的 n- dismlxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Next we need to show that every element has an inverse. This follows from every real number having an inverse, so the additive inverse in $\begin{align*}\mathbb{R}^{n}\end{align*}$ is defined component-wise.
::接下来,我们需要显示每个元素都有反向的。它来自每个真实数字都有反向的, 所以Rn的反向添加剂是定义成份的。

We also satisfy the 1 axiom, because 1 times any real equals that real and then putting that into a n dimensional vector which is equal to the regular vector. Now for the scalar axioms we know that distributivity and associativity of scalars holds in the reals so it is trivial that $\begin{align*}c\cdot d\vec{v} = d\cdot c \vec{v} = cd\vec{v}\end{align*}$ and that $\begin{align*}c(\vec{u} + \vec{v}) = c\vec{u}+c\vec{v}\end{align*}$ . Similarly the opposite identity holds where $\begin{align*}\vec{v}(c+d) = c\vec{v} + d\vec{v}\end{align*}$ where vectors are distributive over scalars which is basically just saying that the sum of two scalars which is another scalar times a vector exists and so we are done and know that $\begin{align*}\mathbb{R}^{n}\end{align*}$ is a vector space over $\begin{align*}\mathbb{R}\end{align*}$ .
::我们还满足了 1 轴值, 因为任何真实值的乘以 1 等于真实值, 然后将该值置入与正向量相等的正维矢量中。现在我们知道, 弧值的分布性和关联性存在于真实值中, 因此, cdv=dcv=cdv; c( uv) =cu+cv。同样, 正好相反的身份所在的 v( c+d) =cv+dv 的正维矢量分布于正负值的正维量矢量中。这基本上只是说, 两个斜度的和两个斜度之和是另一个矢量的斜度乘数存在, 所以我们知道 Rn 是R 的矢量空间。

Example 3: $\begin{align*}\{\vec{0}\}\end{align*}$
::例 3: 0}

This is a very trivial vector space. Obviously additive closure holds, because $\begin{align*}\vec{0} + \vec{0} = \vec{0}\end{align*}$ in any dimension. Also commutativity and associativity are trivial, because $\begin{align*}\vec{0} + \vec{0} = \vec{0} + \vec{0} = \vec{0}, \vec{0} + (\vec{0} + \vec{0}) = (\vec{0} + \vec{0}) + \vec{0}\end{align*}$ . Also, $\begin{align*}\vec{0}\end{align*}$ is both the identity element and it's own inverse, because $\begin{align*}\vec{0} + \vec{0} = \vec{0}\end{align*}$ so it's the inverse and the identity. Also, all of the scalar multiplication axioms trivially follow because this is a vector space over any field $\begin{align*}\mathbb{F}\end{align*}$ . Zero multiplied by any element of a field is zero so you will always get the zero vector.
::这是一个非常次要的矢量空间。显然, 添加性封闭存在, 因为在任何维度 00000000000000000+( 00000) =( 000000) 。另外, 0 既是身份元素,又是其自身的反向元素, 因为%00000000000000000.0。另外, 所有的弧度乘法乘法xxions都很少跟进, 因为这是任何字段的矢量空间 F. 零乘以字段中的任何元素为零, 所以您总能得到零矢量。

More specifically, $\begin{align*}c\vec{0} = \vec{0}, c(\vec{0} + \vec{0}) = \vec{0}, (c+d)\vec{0} = c\vec{0} + d\vec{0} = \vec{0} + \vec{0} = \vec{0}, c(d\vec{0}) = d(c\vec{0}) = \vec{0}\end{align*}$ .
::更具体地说,c00,c0,c0,(c+d)0=c0+d0,000,c(d0)=d(c0)0。

Example 4: $\begin{align*}M_{n\times m}(\mathbb{F})\end{align*}$
::实例4:Mnxm(F)

$\begin{align*}M_{n\times m}(\mathbb{F})\end{align*}$ is the set of all $\begin{align*}n\times m\end{align*}$ matrices with coefficients in some field $\begin{align*}\mathbb{F}\end{align*}$ , usually $\begin{align*}\mathbb{R}\text{ or sometimes }\mathbb{Q}\end{align*}$ . So here each n by m matrix is really treated as a vector where addition and scalar multiplication are defined as they would be when adding or scaling matrices. Proving that this is a vector space is similar to the proof that $\begin{align*}\mathbb{R}^{n}\end{align*}$ is a vector space. Because we have already shown that any field $\begin{align*}\mathbb{F}\end{align*}$ is a vector space over itself, and because the entries are a field we can use the fact that all of these operations are defined component-wise so the proofs are relatively easy.
::Mnxm( F) 是所有 nxm 矩阵的集集, 其中含有某些 F 字段中的系数, 通常是 R 或有时 Q 。因此, 这里每 n x m 矩阵都真正被视为一个矢量, 在添加或缩放矩阵时, 添加和缩放乘法的定义与它们的定义相同。证明这是一个矢量空间与 Rn 是矢量空间的证明相似。因为我们已经显示任何字段 F 是它本身的矢量空间, 并且由于条目是一个字段, 我们可以使用以下事实, 即所有这些操作都是定义成份的, 因此证明相对容易。

Addition and scalar multiplication are obviously closed, because they are closed in $\begin{align*}\mathbb{F}\end{align*}$ so it is closed for each component so it is closed for each matrix of n by m dimensions. Similarly, resorting to our proofs from example one commutativity, associativity, identities and additive inverses are all axioms of a field so if this is satisfied for each entry of a matrix then it is satisfied for every matrix.
::添加和计算乘法显然被关闭,因为它们在F中是封闭的,因此每个部件都是封闭的,因此每个 n 维度的矩阵都是封闭的。同样,利用我们从示例中得出的一个共振、关联性、身份和添加反差的证明都是字段的暗喻,因此,如果对每个矩阵的条目都满意,那么每个矩阵就满足了。

Similarly, because addition is closed, scalar multiplication must be closed because it is just repeated addition (and multiplication is closed in $\begin{align*}\mathbb{F}\end{align*}$ ). From this all of the scalar axioms are trivially satisfied because they are satisfied in $\begin{align*}\mathbb{F}\end{align*}$ where we just generalize this to tuples of tuples of $\begin{align*}\mathbb{F}\end{align*}$ where the same principles apply.
::同样,由于加法是封闭的,因此,由于它只是重复加法(而倍增在F中是封闭的),因此必须关闭增法的倍增。从这一点上,所有calar 轴轴都是微不足道的满足,因为它们在F中是满意的,我们只是将它推广到适用相同原则的F型象牙的象牙中。

Example 5: $\begin{align*}\mathbb{P}_{n}(x)\end{align*}$
::实例5:pn(x)

$\begin{align*}\mathbb{P}_{n}(x)\end{align*}$ is defined to be the set of polynomials in terms of $\begin{align*}x\end{align*}$ with degree equal to $\begin{align*}n\end{align*}$ , of course with coefficients in $\begin{align*}\mathbb{F}\end{align*}$ where $\begin{align*}\mathbb{F}\end{align*}$ is usually just $\begin{align*}\mathbb{R}\end{align*}$ .
::Pn(x)的定义是按x表示的一组多数值,其程度等于n,当然与F的系数相同,F通常只有R。

Now we will prove that $\begin{align*}\mathbb{P}_{n}(x)\end{align*}$ is a vector space over some field $\begin{align*}\mathbb{F}\end{align*}$ .
::现在我们将证明 Pn(x) 是某些 F 字段的矢量空间。

The first axiom of a vector space holds, because addition is closed in $\begin{align*}\mathbb{F}\end{align*}$ so some polynomial of degree n plus some other polynomial of degree n is also a polynomial of degree n, because degree only changes when you multiply or divide polynomials. However, even if you subtract polynomials such that the coefficient of $\begin{align*}x^{n}\end{align*}$ is 0 the polynomial is still an element of of $\begin{align*}\mathbb{P}_{n}(x)\end{align*}$ because by definition $\begin{align*}0\in\mathbb{F}\end{align*}$ so all of the coefficients are still in $\begin{align*}\mathbb{F}\end{align*}$ .
::矢量空间的第一个等同值, 因为在 F 中加法是封闭的, 所以某种多数值 n 加上其他一些多数值 n 也是一种多数值 n, 因为度只在您乘或除多数值时才会变化。但是, 即使您减去了多数值 xn 系数为 0 的多数值, 多数值仍然是 Pn( x) 的一个元素, 因为根据定义 0F , 所有系数都仍然在 F 中。

Next, commutativity, associativity, zero and inverses are all trivial because it all holds in $\begin{align*}\mathbb{F}\end{align*}$ and if it holds for the coefficients then it holds for
::其次,交流、联合、零和逆向都是微不足道的,因为它都保留F,如果它保留系数,那么它也保留

Now, scalar multiplication is also closed, because it is closed in $\begin{align*}\mathbb{F}\end{align*}$ so the polynomial stays in the same vector space and the coefficients are still in $\begin{align*}\mathbb{F}\end{align*}$ .
::现在,天平乘法也关闭了, 因为它在 F 中关闭, 所以多向量在同一个矢量空间中停留, 而系数仍然在 F 中。

Now, left as an exercise for the reader, proving the last of the axioms that this is a vector space is relatively trivial and follow from our proof of any field $\begin{align*}\mathbb{F}\end{align*}$ being a vector space.
::现在,作为给读者的练习, 证明最后的轴是这是矢量空间相对来说是微不足道的从我们证明F是矢量空间的证据来看