Section: 子空间、子空间测试和示例 | 4.3 子空间、子空间试验和实例

Section outline

In this lesson we will learn about subspaces which are vector spaces of vector spaces and we will study examples of them in depth.
::在这一教训中,我们将了解作为矢量空间矢量空间的亚空间,并将深入研究这些子空间的实例。

Definition of a Subspace
::子空间的定义

A subset $\begin{align*}\mathcal{S}\end{align*}$ of a vector space $\begin{align*}V\end{align*}$ is called a subspace of $\begin{align*}V\end{align*}$ if all of the following conditions hold:
::矢量空间五的子集S,如果下列所有条件都具有以下条件,则称为V的子空间:

$\begin{align*}\mathcal{S}\end{align*}$ is non-empty
::S 无空

$\begin{align*}\vec{0}\in\mathcal{S}\end{align*}$
::~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If $\begin{align*}\vec{u}\in\mathcal{S}\text{ and }\vec{v}\in\mathcal{S}\text{ and }\vec{u}+\vec{v}\in\mathcal{S}\end{align*}$
::如果"S"和"S"和"S"和"S"

From these two axioms it also follows that if $\begin{align*}\vec{x}\in\mathcal{S}\end{align*}$ and $\begin{align*}c\in\mathbb{F}\end{align*}$ then $\begin{align*}c\vec{x}\in\mathcal{S}\end{align*}$
::从这两个轴中还可以得出,如果“x”S和“c”F,c“x”S

Now to test if any set is a subspace of some given vector space, you trivially just check if all of the axioms apply (which is not so hard to do).
::现在测试任何一组是否为某些给定矢量空间的子空间, 您只需略微检查是否所有轴值都适用( 这并不难做到 ) 。

Let's look at a few examples of sets and see if they are subspaces of a given vector space. Let's further analyze their properties and try to draw general conclusions about these subspaces.
::让我们看看几个集的示例, 看看它们是否是特定矢量空间的子空间。让我们进一步分析它们的属性, 并尝试对这些子空间得出一般结论。

Example 1: The plane $\begin{align*}3(x+\frac{1}{3})+2(y-\frac{1}{2})+7(z+\frac{1}{7}) =1\end{align*}$ as a subspace of $\begin{align*}\mathbb{R}^{3}\end{align*}$
::示例1:平面3(x+13+13+2(y-12)+7(z+17)+7(z+17)=1作为R3的一个子空间

Test: First off, it is nonempty and the zero vector is contained in this plane as plugging in the point $\begin{align*}(0,0,0)\end{align*}$ we get $\begin{align*}1-1+1=0\end{align*}$ . Now, taking two points in the plane and adding them it stays in the plane. To prove this we simplify the equation of the plane to $\begin{align*}3x+2y+7z=0\end{align*}$ . So if $\begin{align*}(x_{1},y_{1},z_{1})\end{align*}$ is a point on this plane and $\begin{align*}(x_{2},y_{2},z_{2})\end{align*}$ is another point on the plane, then the point $\begin{align*}(x_{1}+x_{2},y_{1}+y_{2},z_{1}+z_{2})\end{align*}$ is also a point on the plane. To see this we plug it into the equation of the plane and get $\begin{align*}3(x_{1}+x_{2}) + 2(y_{1}+y_{2}) + 7(z_{1}+z_{2})\\ 3x_{1}+3x_{2}+2y_{1}+2y_{2}+7z_{1}+7z_{2}\\ (3x_{1}+2y_{1}+7z_{1}) + (3x_{2} + 2y_{2} + 7z_{2})\end{align*}$ and because both points $\begin{align*}(x_{1},y_{1},z_{1})\text{ and }(x_{2},y_{2},z_{2})\end{align*}$ satisfy $\begin{align*}3x_{1}+2y_{1}+7z_{1} = 0\text{ and }3x_{2}+2y_{2}+7z_{2}=0,\text{ so } 3(x_{1}+x_{2}) + 2(y_{1}+y_{2}) + 7(z_{1}+z_{2}) = 0\end{align*}$ . Hence, the sum of the points is also on the plane and this satisfies closure under addition. Now, lastly we look for scalar multiplication.
::测试 : 首先, 它不是空, 并且此平面中含有零矢量, 作为在点( 0, 0, 0) 中的插头, 我们得到 1 - 1+1=0 。现在, 在平面中取两个点, 并在平面中添加它们。要证明这一点, 我们简化平面的方程为 3x+2y+7z=0. 。所以如果 (x1, y1, z1) 将平面的方程式简化为 3x+2y+7z2+2y2+7z1+7z2( 3x1+2y1+2y1+7z1+7z1+3x+2y2+2y1+2y2x+7z1+3x+3x2+2+2y2+2y1+2+2z1+3x+2+3x2+2+3x7z1+3x1+3x2) +3xlxlusxlxxxxxxxlmlxxxxxxxxxxlxxxxxxxxxxxxxxxxxxlxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxlllllllllllllxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Take $\begin{align*}(x_{0},y_{0},z_{0})\end{align*}$ and any scalar $\begin{align*}c\end{align*}$ on the plane we get $\begin{align*}3x_{0}+2y_{0}+7z_{0}=0\end{align*}$ we get that $\begin{align*}(cx_{0},cy_{0},cz_{0})\end{align*}$ and plugging it in we get $\begin{align*}3cx_{0} + 2cy_{0} + 7cz_{0}\\ c(3x_{0}+2y_{0}+7z_{0}) = c\cdot{0} = 0\end{align*}$ hence any scalar multiple of a point in the plane is also in the plane, hence this plane is a subspace.
::采取 (x0, y0, z0) 和任何在飞机上的 calar c 我们得到 3x0+2y0+7z0=0 我们得到的(cx0, cyro0, cz0) 并插入 3cx0+2cy0+7cz0c( 3x0+2y0+2y0+7z0) = cx0=0 因此飞机上的任何一个点的 calar 倍数也是在飞机上, 因此这架飞机是一个子空间。

Example 2: 2x2 Matrices with real entries.
::实例2: 2x2 带真实条目的矩阵。

First, show that it is nonempty and the zero vector is in the set. Then, try and finish the rest of the proof on your own.
::首先, 显示它不是空的, 零矢量在集中。然后, 尝试自己完成其余的证据。

Example 3: Polynomials
::实例3:多面体

The set $\begin{align*}\mathbb{P}_{n}(x)\end{align*}$ is the set of all polynomials of degree $\begin{align*}n\end{align*}$ with the variable $\begin{align*}x\end{align*}$ over some field $\begin{align*}\mathbb{F}\end{align*}$ . Now, we know that the zero vector is going to be in $\begin{align*}\mathbb{P}_{n}\end{align*}$ , because

$\begin{align*} 0 = 0x^{n} + 0x^{n-1} + 0x^{n-2} + \cdot\cdot\cdot + 0x^{2} + 0x + 0 \in \mathbb{P}_{n}(x) \end{align*}$
::设定 Pn( x) 是所有多多边度 n 的集, 以及某些字段 F 上的变量 x 。现在我们知道零矢量将在 Pn 中, 因为 0= 0xn+0xn - 1+0xn - 1+0xn - 2\\\\ 0x2+0x+0x+0}Pn( x)

Next, because summing two polynomials keeps their degree constant, so $\begin{align*}f(x) \in \mathbb{P}_{n} \text{ and } g(x) \in \mathbb{P}_{n} \text{ sums to } f(x) + g(x)\end{align*}$ and from that
::下一位, 因为将两个多圆性数相制成两个不同度常量, 所以 f( x) Pn 和 g( x) Pn 和 f( x) +g( x) , 然后从 f( x) +g( x)

$\begin{align*}f(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + \cdot\cdot\cdot + a_{1}x + a_{0} \\ g(x) = b_{n}x^{n} + b_{n-1}x^{n-1} + \cdot\cdot\cdot + b_{1}x + b_{0} \\\end{align*}$
:x) = anxn+an- 1xn- 1a1x+a0g(x) = bnxn+bn- 1xn- 1xn- 1b1x+b0)

$\begin{align*} f(x) + g(x) = (a_{n}+b_{n})x^{n} + (a_{n-1}+b_{n-1})x^{n-1} + \cdot\cdot\cdot + (a_{1}+b_{1})x + a_{0} + b_{0} \in \mathbb{P}_{n}(x)\end{align*}$
:x)+g(x) = (an+bn)xn+(an- 1+bn- 1)xn-1(a1+b1)x+a0+b0}Pn(x)

So now we know that non-emptiness holds and similarly closure under addition holds and also the zero vector is in the set. Lastly, we must prove that this set is closed under scalar multiplication. To prove this we see given $\begin{align*}c\in\mathbb{F}\end{align*}$ and $\begin{align*}f(x) \in \mathbb{P}_{n}(x)\end{align*}$ we see that $\begin{align*}\\ f(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + \cdot\cdot\cdot + a_{1}x + a_{0}\end{align*}$
::现在我们知道无空状态和在附加下同样关闭存在, 并且设定的矢量为零。最后, 我们必须证明这个设置是在缩放乘法下关闭的。为了证明这一点, 我们可以看到给定的 cF 和 f( x) Pn( x) 我们可以看到 f( x) = anxn+an - 1xn - 1\\\ 1\\\ a1x+a0

and $\begin{align*}cf(x) = ca_{n}x^{n} + ca_{n-1}x^{n-1} + \cdot\cdot\cdot + ca_{1}x + ca_{0} \in \mathbb{P}_{n}(x)\end{align*}$ so closure under scalar multiplication holds.
::和(x) =canxn+can- 1xn- 1ca1x+ca0_Pn(x) ,因此在计算乘法下关闭。

Non-example: Unit Circle as a subspace of the reals in two dimensions
::非示例: 单位圆圆作为两个维度真实物的子空间

The unit circle is the set of all points $\begin{align*}(x,y)\end{align*}$ satisfies $\begin{align*}x^{2} + y^{2} = 1\end{align*}$ . Now, we know that this set is nonempty, because it contains the points $\begin{align*}\{(0,1), (1,0),(-1,0),(0,-1)\}\end{align*}$ ; however, there is no zero vector.
::单位圆是所有点的集合( x,y) 满足 x2+y2=1. 现在,我们知道这组是非空的, 因为它包含 {( 0, 1,0,0,0,0,0- 1) 的点; 但是, 没有零矢量。

Also, we can see that given two points on the unit circle it is not necessarily true that the sum of these two points on the unit circle. Here, we define addition has adding the sum of the coordinates.
::此外,我们可以看到,考虑到单位圆的两点,单位圆的这两点之和不一定是肯定的。在这里,我们定义的加法增加了坐标之和。

So, if $\begin{align*}P_{1} = (x_{1},y_{1})\end{align*}$ and $\begin{align*}P_{2} = (x_{2},y_{2})\end{align*}$ satisfy the equation
$\begin{align*}x_{1}^{2} + y_{1}^{2} = 1 \\ x_{2}^{2} + y_{2}^{2} = 1\end{align*}$
::所以,如果 P1=(x1,y1) 和 P2=(x2,y2) 符合公式 x21+y21=1x22+y22=1

Now, $\begin{align*}P_{1} \oplus P_{2} = (x_{1} + x_{2}, y_{1} + y_{2})\end{align*}$ and $\begin{align*}(x_{1}+x_{2})^{2} + (y_{1} + y_{2})^{2} \\ = x_{1}^{2} + 2x_{1}x_{2} + x_{2}^{2} + y_{1}^{2} + 2y_{1}y_{2} + y_{2}^{2} \\ = (x_{1}^{2} + y_{1}^{2}) + (x_{2}^{2} + y_{2}^{2}) + 2x_{1}x_{2} + 2y_{1}y_{2} \\ = 2 + 2x_{1}x_{2} + 2y_{1}y_{2}\end{align*}$
::P1P2=(x1+x2,y1+y2)和(x1+x2)2+(y1+y2)2+(y1+y2)2=x21+2x1x2+x22+y21+2y1y2+y22=(x21+y2+2y2+y22=(x21+y2+y2+y2+y22)+(x22+y22+y22)+2x1+2x2+y1+y2=2+2x1x2x2+2x2+2y1y2

and this does not always equal 1, so it does not satisfy the equation. Similarly because of this it cannot be true that this set is closed under scalar multiplication. Now, none of these axioms hold true so we are done and this is not a subspace.
::这并不总是等于 1, 所以它不能满足等式。同样, 由于这个原因, 这组无法在天平乘法下关闭。现在, 这些正数中没有一个是真实的, 因此我们完成了, 这不是一个子空间。

*Note it only must be true that one of these does not hold for this to not be a subspace, but it helps to check all three.
::* 仅须注意的是,其中之一并不认为它不是一个子空间,而是有助于检查所有三个空间。

Non-example: The plane of the form: $\begin{align*}2x+5y+7z=1\end{align*}$ as a subspace of the reals in two dimensions
::非示例:窗体的平面: 2x+5y+7z=1 作为两个维的实物的子空间

First off, the zero vector is not in this set because after we plug in the coordinate $\begin{align*}(0,0,0)\end{align*}$ we get $\begin{align*}0 = 1\end{align*}$ . However, this set is nonempty for example, the point $\begin{align*}\big(1,1,-\frac{6}{7}\big)\end{align*}$ is on the plane.
::首先,零矢量不在此设置中,因为在插入坐标(0,0,0)后,我们得到 0=1。但是,这个设置是非空的,例如,点(1,1,-67)在平面上。

In this example also, this set is not closed under scalar multiplication and addition of points. Take, for example, the intercepts. The x,y and z intercepts are
::在这个例子中,这一组不是在标度乘法和加点的情况下封闭的。例如,拦截。 X,y 和 z 拦截是

$\begin{align*}P_{1} = \big(0,0,\frac{1}{7}\big) \\ P_{2} = \big(0,\frac{1}{5},0\big) \\ P_{3} = \big(\frac{1}{2},0,0\big)\end{align*}$ Summing some of the points we can get
::P1=(0,0,0,17P2=(0,1,15,0)P3=(12,0)

$\begin{align*}P_{1} \oplus P_{2} = \big(0,\frac{1}{5}, \frac{1}{7} \big) \\ P_{1} \oplus P_{3} = \big(\frac{1}{2}, 0, \frac{1}{7}\big) \\ P_{2} \oplus P_{3} = \big(\frac{1}{2},\frac{1}{5},0 \big)\end{align*}$
::P1P2=( 0, 15, 17, P1P3=( 12, 10, 17, P2P3=( 12, 15, 0)

Now, plugging each of these into the equation of the plane, we get $\begin{align*}P_{1} \oplus P_{2} \to 2\big(0\big) + 5\big(\frac{1}{5}\big) + 7\big(\frac{1}{7}\big) = 2 \neq 1 \\ P_{1} \oplus P_{3} \to 2\big(\frac{1}{2}\big) + 5\big(0\big) + 7\big(\frac{1}{7}\big) = 2 \neq 1 \\ P_{2} \oplus P_{3} \to 2\big(\frac{1}{2}\big) + 5\big(\frac{1}{5}\big) + 7\big(0\big) = 2 \neq 1\end{align*}$
::现在,将其中每一个插入平面方程, 我们得到 P1P22( 0)+5( 15)+7( 17)=21P1}P32( 12)+5( 7)+7( 17)=21P2( 2( 12)+5( 12)+5( 15)+7( 0)=21

Now, this is always going to be true that the sum of two points in this set is never in the set. Take $\begin{align*}P_{1} = (x_{1},y_{1},z_{1})\end{align*}$ to wit $\begin{align*}2x_{1} + 5y_{1} + 7z_{1} = 1\end{align*}$ and similarly $\begin{align*}P_{2} = (x_{2},y_{2},z_{2})\end{align*}$ and $\begin{align*}2x_{2} + 5y_{2} + 7z_{2} = 1\end{align*}$ .
::现在, 这始终会是真实的, 此集中的两点之和从未出现在集中。请使用 P1 = (x1, y1, z1) 来理解 2x1+5y1+7z1=1 和类似的 P2 = (x2, y2, z2) 和 2x2+5y2+7z2=1 。

Now, taking $\begin{align*}P_{1} \oplus P_{2}\end{align*}$ we get $\begin{align*}(x_{1}+x_{2},y_{1}+y_{2},z_{1}+z_{2})\end{align*}$ , but plugging this into the equation of this plane we get

$\begin{align*}\\ 2(x_{1}+x_{2}) + 5(y_{1}+y_{2}) + 7(z_{1} + z_{2}) \end{align*}$

$\begin{align*}\\ = (2x_{1} + 5y_{1} + 7z_{1}) + (2x_{2} + 5y_{2} + 7z_{2}) \end{align*}$

$\begin{align*}\\ = 1 + 1 \end{align*}$

$\begin{align*}\\ = 2\end{align*}$
::现在,使用 P1P2 我们得到 (x1+x2, y1+x2, y1+y2, z1+z2) , 但是将它插入到此平面的方程中, 我们得到 2( x1+x2)+5(y1+y2)+5(y1+y2)+7(z1+z2) =( 2x1+5y1+7z1+7z1)+(2x2+5y2+7z2) =1+1=2

Now, obviously because summing two points is not closed this set is obviously not closed under scalar multiplication. However, let's just take verify this ourselves:
::现在,很明显,因为两个点的缩放没有关闭, 这个组合显然没有在缩放乘法下关闭。然而,让我们自己来验证一下:

Let $\begin{align*}P_{0} = (x_{0},y_{0},z_{0})\end{align*}$ such that $\begin{align*}2x_{0} + 5y_{0} + 7z_{0} = 1\end{align*}$ , however, take $\begin{align*}P_{1} = c \cdot P_{0} = (cx_{0}, cy_{0}, cz_{0})\end{align*}$ and plugging this into the original equation we get

$\begin{align*}\\ 2cx_{0} + 5cy_{0} + 7cz_{0} = c(2x_{0} + 5y_{0} + 7z_{0}) = c \cdot (1) = c \\\end{align*}$
::Let P0= (x0, y0, z0) 等 2x0+50+5y0+7z0=1, 然而, 采用 P1=cQP0=( cx0, cyro0, cz0) 并将其插入原始方程式中, 我们得到 2x0+5cy0+7cz0=c( 2x0+5y0+7z0+7z0) = c=c=l=c

And this only equals one if $\begin{align*}c=1 \to P_{1} = P_{0}\end{align*}$ , so being closed under scalar multiplication is false.
::如果 c=1QP1=P0, 则此值只等于 1, 所以在计算乘法下关闭是虚假的。

Why do we need the zero vector?
::为什么我们需要零矢量?

The zero vector is imperative to have in a subspace. First off, without the zero vector we would have a significant number of problems, i.e., the empty set could be considered a subspace (which we don't want), addition and scalar multiplication would not be closed in the vector space.
::零矢量必须在子空间中存在。首先, 如果没有零矢量, 我们就会遇到很多问题, 也就是说, 空的集可以被视为子空间( 我们不希望这样) , 添加和计算乘法不会在矢量空间中关闭。

The trivial example of this is just that $\begin{align*}\mathbb{F}\end{align*}$ , over the vector space and then multiplying a vector inside the subspace by zero we get the zero vector. Hence because if we do not have the zero vector in there we do not have closure under scalar multiplication.
::次要的例子就是F,在矢量空间上空,然后在子空间内将矢量乘以零,我们就得到零矢量。因此,如果我们在那里没有零矢量,我们就不会在斜度乘法下关闭。

Also, if we have some vector $\begin{align*}\vec{v} \in \mathcal{S}\end{align*}$ were $\begin{align*}\mathcal{S}\end{align*}$ is a subspace of some vector space $\begin{align*}V\end{align*}$ then we know that $\begin{align*}-\vec{v} \in \mathcal{S}\end{align*}$ so then $\begin{align*}\vec{v} + (-\vec{v})\end{align*}$ should be in $\begin{align*}\mathcal{S}\end{align*}$ , but the sum of a vector and its negative is the zero vector which is not in the subspace.
::而且,如果我们有某种矢量“vS”是S是某些矢量空间V的一个子空间,那么我们就知道“vS”应该是S,但矢量和负的和是零矢量,不在子空间。

Hence, the zero vector being in the subspace is very much a necessity.
::因此,在子空间中的零矢量是非常必要的。

Spans and Subspaces
::Spaans 和子空间

Theorem—Given any $\begin{align*}k\text{ }(\neq 0)\end{align*}$ vectors in $\begin{align*}\mathbb{R}^{n}\end{align*}$ , denoted by $\begin{align*}\{\vec{v_{1}}, \vec{v_{2}}, \cdot\cdot\cdot, \vec{v_{k}}\}\end{align*}$ , the span of those vectors is a subspace of $\begin{align*}\mathbb{R}^{n}\end{align*}$

Proof: Now, first let's denote the span of this by our subset notation $\begin{align*}\mathcal{S}\end{align*}$ . Next, we know that the set is nonempty as we are assuming this set of vectors is nonempty and it is also obvious that this set contains the zero vector.

This is because the span is the set of all linear combinations of these vectors and because $\begin{align*}0\in\mathbb{F}\end{align*}$ for all fields $\begin{align*}\mathbb{F}\end{align*}$ we can just make the coefficients of each vector equal to 0 in the linear combination.

Next, we can show that the span is closed under addition. Take two elements of $\begin{align*}\mathcal{S}\end{align*}$ , denoted by $\begin{align*}\vec{x_{a}}\text{ and }\vec{x_{b}}\end{align*}$ equivalent to

$\begin{align*}\\ \vec{x_{a}} = a_{1}\vec{v_{1}} + a_{2}\vec{v_{2}} + \cdot\cdot\cdot + a_{k}\vec{v_{k}} \\ \vec{x_{b}} = b_{1}\vec{v_{1}} + b_{2}\vec{v_{2}} + \cdot\cdot\cdot + b_{k}\vec{v_{k}} \\\end{align*}$

$\begin{align*}\\ \vec{x_{a}} + \vec{x_{b}} = (a_{1}+b_{1})v_{1} + (a_{2}+b_{2})\vec{v_{2}} + \cdot\cdot\cdot (a_{k}+b_{k})\vec{v_{k}} = \vec{x_{a+b}} \in \mathcal{S}\end{align*}$ So now, it is obvious that this set is closed under addition. Following an analogous argument for scalar multiplication we can see that given

$\begin{align*}\vec{x_{c}} = c_{1}\vec{v_{1}} + c_{2}\vec{v_{2}} + \cdot\cdot\cdot + c_{k}\vec{v_{k}} \in \mathcal{S}\end{align*}$ and some constant $\begin{align*}c\in\mathbb{F}\end{align*}$ we get that

$\begin{align*}c\vec{x_{c}} = cc_{1}\vec{v_{1}} + cc_{2}\vec{v_{2}} + \cdot\cdot\cdot + cc_{k}\vec{v_{k}} \in \mathcal{S}\end{align*}$

so we are done!