课程： 1.3 线性组合和斯潘

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线状组合和span
Take two arbitrary vectors, $\begin{aligned} \vec{x} and \vec{y} \end{aligned}$ then take arbitrary scalars $\begin{aligned} a and b \end{aligned}$ and write $\begin{aligned} a \vec{x} + b \vec{y} \end{aligned}$ . This is known as a linear combination.
::使用两个任意矢量, x 和 y , 然后使用任意的弧度 a 和 b , 并写入 ax by 。这被称为线性组合。

Definition A linear combination of any set of vectors is a sum of scalar multiples of each of those vectors.
::任何一套矢量的线性组合是其中每一种矢量的弧倍数总和。

For example, take $\begin{aligned} \vec{x} = [\begin{matrix} 1 \\ 1 \end{matrix}], \vec{y} = [\begin{matrix} - 1 \\ - 1 \end{matrix}] \end{aligned}$ and then $\begin{aligned} 3 [\begin{matrix} 1 \\ 1 \end{matrix}] + (- 2) [\begin{matrix} - 1 \\ - 1 \end{matrix}] = [\begin{matrix} 5 \\ 5 \end{matrix}] \end{aligned}$ so $\begin{aligned} [\begin{matrix} 5 \\ 5 \end{matrix}] \end{aligned}$ is a linear combination of $\begin{aligned} \vec{x} and \vec{y} \end{aligned}$ . For that matter, any vector of the form $\begin{aligned} [\begin{matrix} a - b \\ a - b \end{matrix}] \end{aligned}$ is a linear combination of $\begin{aligned} \vec{x} and \vec{y} \end{aligned}$ . Now, take some other vector $\begin{aligned} \vec{w} = [\begin{matrix} 3 \\ 4 \end{matrix}] \end{aligned}$ we see that $\begin{aligned} \vec{w} \end{aligned}$ is not a linear combination of   $\begin{aligned} \vec{x} and \vec{y} \end{aligned}$ or is not in the span of   $\begin{aligned} \vec{x} and \vec{y} \end{aligned}$ .
::例如,取 x[11],y[-1-1],然后取 3[11]+(-2),[-1-1]=[55],因此[55]是 x和y的线性组合。对于这个事项,形态[a-ba-b] 的任何矢量都是 x和y的线性组合。现在,取其他一些矢量 w[34],我们看到 w不是 x和y的线性组合,或者不在 x和y的跨度。

Definition The span of a set of vectors is the set of all possible linear combinations of those vectors.
::定义一组矢量的范围是这些矢量所有可能的线性组合的一组。

Going back to the example from earlier $\begin{aligned} Span {[\begin{matrix} 1 \\ 1 \end{matrix}], [\begin{matrix} - 1 \\ - 1 \end{matrix}]} = {\vec{x} \in R^{2} ∣ \vec{x} = a [\begin{matrix} 1 \\ 1 \end{matrix}] + b [\begin{matrix} - 1 \\ - 1 \end{matrix}] for some a, b \in R} \end{aligned}$ or you could simplify this by saying
::回到先前的 Span{ [11] , [-1-1-1] , [- 1- 1] x\\\\\\\\\\\\\\\\\\\\\\\\\\a[11] +b[- 1-1-1-1] , 或您可以通过说简化 </span> </p> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <p id="x-ck12-MGE1MDUzZTlhNGUxYjI0NmE0MWMxNjIyNDM5Y2Y3ZTk.-ug9"> <span style="color: #000000; font-family: Arial, Helvetica, sans-serif; font-size: 16px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: 400; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; background-color: #ffffff; text-decoration-style: initial; text-decoration-color: initial; display: inline !important; float: none;"> <span class="x-ck12-mathEditor" data-contenteditable="false" data-edithtml="" data-math-class="x-ck12-math" data-mathmethod="inline" data-tex="%5Cvec%7Bx%7D%20%5Cin%20%5Ctext%7BSpan%7D%5Cbigg%5C%7B%5Cbegin%7Bbmatrix%7D%201%20%5C%5C%201%20%5Cend%7Bbmatrix%7D%2C%20%5Cbegin%7Bbmatrix%7D%20-1%20%5C%5C%20-1%20%5Cend%7Bbmatrix%7D%5Cbigg%5C%7D%20%5Ctext%7B%20if%20and%20only%20if%20%7D%20%5Cvec%7Bx%7D%20%3D%20%5Cbegin%7Bbmatrix%7D%20c%20%5C%5C%20c%20%5Cend%7Bbmatrix%7D%20%0A"> <span class="MathJax_Preview" style="color: inherit; display: none;"> </span> <span class="MathJax_SVG" data-mathml='<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnalign="right left right left right left right left right left right left" rowspacing="3pt" columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" displaystyle="true"><mtr><mtd><mrow class="MJX-TeXAtom-ORD"><mover><mi>x</mi><mo stretchy="false">→</mo></mover></mrow><mo>∈</mo><mtext>Span</mtext><mrow class="MJX-TeXAtom-ORD"><mo maxsize="2.047em" minsize="2.047em">{</mo></mrow><mrow><mo>[</mo><mtable rowspacing="4pt" columnspacing="1em"><mtr><mtd><mn>1</mn></mtd></mtr><mtr><mtd><mn>1</mn></mtd></mtr></mtable><mo>]</mo></mrow><mo>,</mo><mrow><mo>[</mo><mtable rowspacing="4pt" columnspacing="1em"><mtr><mtd><mo>−</mo><mn>1</mn></mtd></mtr><mtr><mtd><mo>−</mo><mn>1</mn></mtd></mtr></mtable><mo>]</mo></mrow><mrow class="MJX-TeXAtom-ORD"><mo maxsize="2.047em" minsize="2.047em">}</mo></mrow><mtext> if and only if </mtext><mrow class="MJX-TeXAtom-ORD"><mover><mi>x</mi><mo stretchy="false">→</mo></mover></mrow><mo>=</mo><mrow><mo>[</mo><mtable rowspacing="4pt" columnspacing="1em"><mtr><mtd><mi>c</mi></mtd></mtr><mtr><mtd><mi>c</mi></mtd></mtr></mtable><mo>]</mo></mrow></mtd></mtr></mtable></math>' id="MathJax-Element-15-Frame" role="presentation" style="font-size: 100%; display: inline-block; position: relative;" tabindex="-1"> for some $\begin{aligned} c \in R \end{aligned}$ .
::xspan{[11],[-1-1]},如果并且只有在 x{[cc] 对于某些 cR 的情况下。 </span> </p> <hr/> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <p id="x-ck12-MDNjNDc1NmZlMGU0OTllMDM3YzIyOGM1MzU0NDMxNWE.-e8s"> <span style="color: #000000; font-family: Arial, Helvetica, sans-serif; font-size: 16px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: 400; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; background-color: #ffffff; text-decoration-style: initial; text-decoration-color: initial; display: inline !important; float: none;"> Now let's look at this span geometrically. </span> <br/> <span style="color: green; "> ::现在让我们从几何角度来看看这个 </span> </p> <div class="x-ck12-video-object" id="x-ck12-ruj"> <div class="iframe-wrapper ck12-no-annotation"> <iframe allowfullscreen="allowfullscreen" data-artifactlang="en" data-embed="PGlmcmFtZSBmcmFtZWJvcmRlcj0iMCIgc2Nyb2xsaW5nPSJubyIgdGl0bGU9IlNwYW4gUHJhY3RpY2UgSSIgc3JjPSJodHRwczovL3d3dy5nZW9nZWJyYS5vcmcvbWF0ZXJpYWwvaWZyYW1lL2lkL3huZGhwaHZ3L3dpZHRoLzEyODAvaGVpZ2h0LzY1Ny9ib3JkZXIvODg4ODg4L3Nmc2IvdHJ1ZS9zbWIvZmFsc2Uvc3RiL2ZhbHNlL3N0YmgvZmFsc2UvYWkvZmFsc2UvYXNiL2ZhbHNlL3NyaS9mYWxzZS9yYy9mYWxzZS9sZC9mYWxzZS9zZHovZmFsc2UvY3RsL2ZhbHNlIiB3aWR0aD0iMTI4MHB4IiBoZWlnaHQ9IjY1N3B4IiBzdHlsZT0iYm9yZGVyOjBweDsiPiA8L2lmcmFtZT4=" frameborder="0" height="677" id="x-ck12-MTU2NjI0ODg3NDk3Mg.." name="254841" src="https://flexbooks.ck12.org/flx/show/interactive/user%3Ajzeitlin36/https%3A//www.geogebra.org/material/iframe/id/xndhphvw/width/1280/height/657/border/888888/sfsb/true/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/false/rc/false/ld/false/sdz/false/ctl/false%3Fhash%3Dfc95dc0bfdd97c2fe3c60f09cb660555" style="max-width: 100%;" title="video" width="1300"> </iframe> </div> </div> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <p id="x-ck12-NzA4OWFkMzU0ZjZiNjEyYWFiZGY0M2YwZGYzYWI1YmU.-h0m"> First play with <span class="x-ck12-mathEditor" data-contenteditable="false" data-edithtml="" data-math-class="x-ck12-math" data-mathmethod="inline" data-tex="%5Cvec%7Bw%7D"> <span class="MathJax_Preview" style="color: inherit; display: none;"> </span> <span class="MathJax_SVG" data-mathml='<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnalign="right left right left right left right left right left right left" rowspacing="3pt" columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" displaystyle="true"><mtr><mtd><mrow class="MJX-TeXAtom-ORD"><mover><mi>w</mi><mo stretchy="false">→</mo></mover></mrow></mtd></mtr></mtable></math>' id="MathJax-Element-17-Frame" role="presentation" style="font-size: 100%; display: inline-block; position: relative;" tabindex="-1"> <svg aria-hidden="true" focusable="false" height="2.768ex" role="img" style="vertical-align: -0.809ex;" viewbox="0 -843.3 1035.3 1191.7" width="2.405ex" xmlns:xlink="http://www.w3.org/1999/xlink"> <g fill="currentColor" role="group" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"> <g role="group" transform="translate(167,0)"> <g role="group" transform="translate(-16,0)"> <g role="group" transform="translate(0,-50)"> <use x="0" xlink:href="#MJMATHI-77" y="0"> </use> <use x="676" xlink:href="#MJMAIN-20D7" y="3"> </use> </g> </g> </g> </g> </svg> <span class="MJX_Assistive_MathML" role="presentation"> <math xmlns="http://www.w3.org/1998/Math/MathML"> <mtable columnalign="right left right left right left right left right left right left" columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" displaystyle="true" rowspacing="3pt"> <mtr> <mtd> <mrow class="MJX-TeXAtom-ORD"> <mover> <mi> w </mi> <mo stretchy="false"> → </mo> </mover> </mrow> </mtd> </mtr> </mtable> </math> </span> </span> </span> and you can see that all of the vectors end up on the same line, <span class="x-ck12-mathEditor" data-contenteditable="false" data-edithtml="" data-math-class="x-ck12-math" data-mathmethod="inline" data-tex="y%3Dx"> <span class="MathJax_Preview" style="color: inherit; display: none;"> </span> <span class="MathJax_SVG" data-mathml='<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnalign="right left right left right left right left right left right left" rowspacing="3pt" columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" displaystyle="true"><mtr><mtd><mi>y</mi><mo>=</mo><mi>x</mi></mtd></mtr></mtable></math>' id="MathJax-Element-18-Frame" role="presentation" style="font-size: 100%; display: inline-block; position: relative;" tabindex="-1"> <svg aria-hidden="true" focusable="false" height="2.768ex" role="img" style="vertical-align: -0.809ex;" viewbox="0 -843.3 2722.9 1191.7" width="6.324ex" xmlns:xlink="http://www.w3.org/1999/xlink"> <g fill="currentColor" role="group" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"> <g role="group" transform="translate(167,0)"> <g role="group" transform="translate(-16,0)"> <g role="group" transform="translate(0,-47)"> <use x="0" xlink:href="#MJMATHI-79" y="0"> </use> <use x="775" xlink:href="#MJMAIN-3D" y="0"> </use> <use x="1831" xlink:href="#MJMATHI-78" y="0"> </use> </g> </g> </g> </g> </svg> <span class="MJX_Assistive_MathML" role="presentation"> <math xmlns="http://www.w3.org/1998/Math/MathML"> <mtable columnalign="right left right left right left right left right left right left" columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" displaystyle="true" rowspacing="3pt"> <mtr> <mtd> <mi> y </mi> <mo> = </mo> <mi> x </mi> </mtd> </mtr> </mtable> </math> </span> </span> </span> or the line spanned by the vector <span class="x-ck12-mathEditor" data-contenteditable="false" data-edithtml="" data-math-class="x-ck12-math" data-mathmethod="inline" data-tex="%5Cbegin%7Bbmatrix%7D1%5C%5C1%5Cend%7Bbmatrix%7D"> <span class="MathJax_Preview" style="color: inherit; display: none;"> </span> <span class="MathJax_SVG" data-mathml='<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnalign="right left right left right left right left right left right left" rowspacing="3pt" columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" displaystyle="true"><mtr><mtd><mrow><mo>[</mo><mtable rowspacing="4pt" columnspacing="1em"><mtr><mtd><mn>1</mn></mtd></mtr><mtr><mtd><mn>1</mn></mtd></mtr></mtable><mo>]</mo></mrow></mtd></mtr></mtable></math>' id="MathJax-Element-19-Frame" role="presentation" style="font-size: 100%; display: inline-block; position: relative;" tabindex="-1"> <svg aria-hidden="true" focusable="false" height="5.986ex" role="img" style="vertical-align: -2.418ex;" viewbox="0 -1536.2 2195.2 2577.4" width="5.099ex" xmlns:xlink="http://www.w3.org/1999/xlink"> <g fill="currentColor" role="group" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"> <g role="group" transform="translate(167,0)"> <g role="group" transform="translate(-16,0)"> <use xlink:href="#MJSZ3-5B"> </use> <g role="group" transform="translate(695,0)"> <g role="group" transform="translate(-16,0)"> <use x="0" xlink:href="#MJMAIN-31" y="650"> </use> <use x="0" xlink:href="#MJMAIN-31" y="-750"> </use> </g> </g> <use x="1347" xlink:href="#MJSZ3-5D" y="-1"> </use> </g> </g> </g> </svg> <span class="MJX_Assistive_MathML" role="presentation"> <math xmlns="http://www.w3.org/1998/Math/MathML"> <mtable columnalign="right left right left right left right left right left right left" columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" displaystyle="true" rowspacing="3pt"> <mtr> <mtd> <mrow> <mo> [ </mo> <mtable columnspacing="1em" rowspacing="4pt"> <mtr> <mtd> <mn> 1 </mn> </mtd> </mtr> <mtr> <mtd> <mn> 1 </mn> </mtd> </mtr> </mtable> <mo> ] </mo> </mrow> </mtd> </mtr> </mtable> </math> </span> </span> </span> . Actually if we want to express this line we do not even need the second vector, <span class="x-ck12-mathEditor" data-contenteditable="false" data-edithtml="" data-math-class="x-ck12-math" data-mathmethod="inline" data-tex="%5Cbegin%7Bbmatrix%7D-1%5C%5C-1%5Cend%7Bbmatrix%7D"> <span class="MathJax_Preview" style="color: inherit; display: none;"> </span> <span class="MathJax_SVG" data-mathml='<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnalign="right left right left right left right left right left right left" rowspacing="3pt" columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" displaystyle="true"><mtr><mtd><mrow><mo>[</mo><mtable rowspacing="4pt" columnspacing="1em"><mtr><mtd><mo>−</mo><mn>1</mn></mtd></mtr><mtr><mtd><mo>−</mo><mn>1</mn></mtd></mtr></mtable><mo>]</mo></mrow></mtd></mtr></mtable></math>' id="MathJax-Element-20-Frame" role="presentation" style="font-size: 100%; display: inline-block; position: relative;" tabindex="-1"> <svg aria-hidden="true" focusable="false" height="5.986ex" role="img" style="vertical-align: -2.418ex;" viewbox="0 -1536.2 2973.7 2577.4" width="6.907ex" xmlns:xlink="http://www.w3.org/1999/xlink"> <g fill="currentColor" role="group" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"> <g role="group" transform="translate(167,0)"> <g role="group" transform="translate(-16,0)"> <use xlink:href="#MJSZ3-5B"> </use> <g role="group" transform="translate(695,0)"> <g role="group" transform="translate(-16,0)"> <g role="group" transform="translate(0,650)"> <use x="0" xlink:href="#MJMAIN-2212" y="0"> </use> <use x="778" xlink:href="#MJMAIN-31" y="0"> </use> </g> <g role="group" transform="translate(0,-750)"> <use x="0" xlink:href="#MJMAIN-2212" y="0"> </use> <use x="778" xlink:href="#MJMAIN-31" y="0"> </use> </g> </g> </g> <use x="2126" xlink:href="#MJSZ3-5D" y="-1"> </use> </g> </g> </g> </svg> <span class="MJX_Assistive_MathML" role="presentation"> <math xmlns="http://www.w3.org/1998/Math/MathML"> <mtable columnalign="right left right left right left right left right left right left" columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" displaystyle="true" rowspacing="3pt"> <mtr> <mtd> <mrow> <mo> [ </mo> <mtable columnspacing="1em" rowspacing="4pt"> <mtr> <mtd> <mo> − </mo> <mn> 1 </mn> </mtd> </mtr> <mtr> <mtd> <mo> − </mo> <mn> 1 </mn> </mtd> </mtr> </mtable> <mo> ] </mo> </mrow> </mtd> </mtr> </mtable> </math> </span> </span> </span> and by this we call <span class="x-ck12-mathEditor" data-contenteditable="false" data-edithtml="" data-math-class="x-ck12-math" data-mathmethod="inline" data-tex="%5Cbigg%5C%7B%5Cbegin%7Bbmatrix%7D%201%20%5C%5C%201%20%5Cend%7Bbmatrix%7D%2C%20%5Cbegin%7Bbmatrix%7D%20-1%20%5C%5C%20-1%20%5Cend%7Bbmatrix%7D%5Cbigg%5C%7D"> <span class="MathJax_Preview" style="color: inherit; display: none;"> </span> <span class="MathJax_SVG" data-mathml='<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnalign="right left right left right left right left right left right left" rowspacing="3pt" columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" displaystyle="true"><mtr><mtd><mrow class="MJX-TeXAtom-ORD"><mo maxsize="2.047em" minsize="2.047em">{</mo></mrow><mrow><mo>[</mo><mtable rowspacing="4pt" columnspacing="1em"><mtr><mtd><mn>1</mn></mtd></mtr><mtr><mtd><mn>1</mn></mtd></mtr></mtable><mo>]</mo></mrow><mo>,</mo><mrow><mo>[</mo><mtable rowspacing="4pt" columnspacing="1em"><mtr><mtd><mo>−</mo><mn>1</mn></mtd></mtr><mtr><mtd><mo>−</mo><mn>1</mn></mtd></mtr></mtable><mo>]</mo></mrow><mrow class="MJX-TeXAtom-ORD"><mo maxsize="2.047em" minsize="2.047em">}</mo></mrow></mtd></mtr></mtable></math>' id="MathJax-Element-21-Frame" role="presentation" style="font-size: 100%; display: inline-block; position: relative;" tabindex="-1"> a linearly dependent set.
::第一场与 w 播放, 您可以看到所有的矢量最终都在同一行上, y=x 或矢量所跨越的线条 [11] 。实际上, 如果我们想要表达这条线的话, 我们甚至不需要第二个矢量, [-1-1] , 这样我们称之为 {[ 11] , [- 1- 1)] 线性依附集。 </span> </p> <hr/> <p id="x-ck12-ZDQxZDhjZDk4ZjAwYjIwNGU5ODAwOTk4ZWNmODQyN2U.-dyv"> </p> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <p id="x-ck12-OGE3OTM3MTE5MmNkNDJkNzFlMTcwYWU4OWNjZDE2NTE.-52c"> <span class="x-ck12-underline"> Definition: </span> Formally speaking, a linearly dependent set is a set of vectors <span class="x-ck12-mathEditor" data-contenteditable="false" data-edithtml="" data-math-class="x-ck12-math" data-mathmethod="inline" data-tex="%5C%7B%5Cvec%7Bv_%7B1%7D%7D%2C%20%5Cvec%7Bv_%7B2%7D%7D%2C%20%5Cvec%7Bv_%7B3%7D%7D%2C%20%5Ccdot%5Ccdot%5Ccdot%2C%20%5Cvec%7Bv_%7Bn%7D%7D%5C%7D"> <span class="MathJax_Preview" style="color: inherit; display: none;"> </span> <span class="MathJax_SVG" data-mathml='<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnalign="right left right left right left right left right left right left" rowspacing="3pt" columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" displaystyle="true"><mtr><mtd><mo fence="false" stretchy="false">{</mo><mrow class="MJX-TeXAtom-ORD"><mover><msub><mi>v</mi><mrow class="MJX-TeXAtom-ORD"><mn>1</mn></mrow></msub><mo stretchy="false">→</mo></mover></mrow><mo>,</mo><mrow class="MJX-TeXAtom-ORD"><mover><msub><mi>v</mi><mrow class="MJX-TeXAtom-ORD"><mn>2</mn></mrow></msub><mo stretchy="false">→</mo></mover></mrow><mo>,</mo><mrow class="MJX-TeXAtom-ORD"><mover><msub><mi>v</mi><mrow class="MJX-TeXAtom-ORD"><mn>3</mn></mrow></msub><mo stretchy="false">→</mo></mover></mrow><mo>,</mo><mo>⋅</mo><mo>⋅</mo><mo>⋅</mo><mo>,</mo><mrow class="MJX-TeXAtom-ORD"><mover><msub><mi>v</mi><mrow class="MJX-TeXAtom-ORD"><mi>n</mi></mrow></msub><mo stretchy="false">→</mo></mover></mrow><mo fence="false" stretchy="false">}</mo></mtd></mtr></mtable></math>' id="MathJax-Element-22-Frame" role="presentation" style="font-size: 100%; display: inline-block; position: relative;" tabindex="-1"> such that there exists a linear dependence relation, a relation such that there exists scalars $\begin{aligned} a_{1}, a_{2}, a_{3}, \cdot \cdot \cdot, a_{n} \end{aligned}$ such that $\begin{aligned} a_{1} \vec{v_{1}} + a_{2} \vec{v_{2}} + a_{3} \vec{v_{3}} + \cdot \cdot \cdot + a_{n} \vec{v_{n}} = \vec{0} . \end{aligned}$
::定义:从形式上说,线性依赖的一组是一组矢量 {v1},v2},v3},,vn} ,这样就存在线性依赖关系,这种关系就存在 calars a1,a2,a3,a3,,a1v1a2}a2v2}a3v3anvn0}。

Now we can say that this essentially means that the vectors can be expressed as linear combinations of the other vectors, because if there exists a linear dependence relation then every $\begin{aligned} \vec{v_{i}} = - \frac{1}{a_{i}} (a_{1} \vec{v_{1}} + \cdot \cdot \cdot + \vec{v_{i - 1}} + \vec{v_{i + 1}} + \cdot \cdot \cdot + a_{n} \vec{v_{n}}) \end{aligned}$ by basic algebraic manipulation.
::现在我们可以说,这基本上意味着矢量可以作为其他矢量的线性组合表示,因为如果存在线性依赖关系,那么通过基本的代数操控,每个 v1i(a1v1_i-1vi+1anvn)就会出现。

When a set is linearly dependent there are redundant vectors in the set that do not help you span the space you are trying to span. Take an example in the 2-dimensional space, $\begin{aligned} R^{2} \end{aligned}$ take $\begin{aligned} {[\begin{matrix} 7 \\ 7 \end{matrix}], [\begin{matrix} - 2 \\ - 2 \end{matrix}]} \end{aligned}$ .
::当一组线性依赖时, 集中有多余的矢量无法帮助您跨越您试图跨越的空间。举二维空间的例子, R2 以 {[ 77] , [- 2 - 2]} 。

The span of this set is just $\begin{aligned} Span {[\begin{matrix} 7 \\ 7 \end{matrix}], [\begin{matrix} - 2 \\ - 2 \end{matrix}]} = {\vec{x} ∣ \vec{x} = [\begin{matrix} 7 a - 2 b \\ 7 a - 2 b \end{matrix}], a, b \in R} \end{aligned}$ because $\begin{aligned} 7 a - 2 b \end{aligned}$ can hit every real number for some $\begin{aligned} a and b \end{aligned}$ . Hence the span of the vector $\begin{aligned} [\begin{matrix} 1 \\ 1 \end{matrix}] \end{aligned}$ and in addition this set is linearly dependent because $\begin{aligned} \frac{2}{7} [\begin{matrix} 7 \\ 7 \end{matrix}] + [\begin{matrix} - 2 \\ - 2 \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}] = \vec{0} \end{aligned}$ and the vector $\begin{aligned} [\begin{matrix} - 2 \\ - 2 \end{matrix}] \end{aligned}$ is redundant.
::本集的宽度仅为Span{[77],[-2-2]xx}[7a-2b7a-2b7a-2b],a,bR},因为7a-2b可以击中某些a和b.的每个实际数字。因此,矢量的宽度[11],此外,由于27[77]+[-2-2-2]=[00]=0和矢量[-2-2]是多余的,因此本集的宽度线性依赖。

To contrast this, a linearly independent set on the other hand is a set such that there exists no linear dependence relation.
::与此形成对照的是,另一面的线性独立套装是一套没有线性依赖关系的套装。

Looking at 2 dimensional space specifically, any set of two linearly independent vectors spans all of two dimensional space:
::具体地看两维空间, 任何一组两个线性独立的矢量都横跨所有两个维空间:

In this example you have two arbitrary vectors in space and then a third vector which is the sum of scalar multiples of the first two vectors and when dragging these two vectors and changing the scalars you see that this vector can hit anywhere in space.
::在此示例中, 您在空间中有两个任意矢量, 然后是第三个矢量, 即前两个矢量的弧倍数之和, 当拖曳这两个矢量并改变天体时, 您可以看到此矢量可以在空间中任何地方撞击。

On the other hand we have here, two linearly dependent vectors as the value depends on each other. The sum of any scalar multiple is just another scalar multiple of that vector illustrated in this interactive:
::另一方面,我们这里有两种线性依赖矢量,因为值是互相依赖的。任何卡路里倍数的总和只是该互动中显示的该矢量的另一个卡路里倍数:

I think now that it is fair to conclude that in $\begin{aligned} R^{2} \end{aligned}$ a set of vectors is linearly independent if and only if the vectors are not scalar multiples of each other and also there are only two vectors in the set.
::我认为,现在可以公正地得出结论,在R2中,如果并且只有在矢量不是相互的卡路里倍数,而且集中只有两个矢量时,一套矢量是线性独立的。

First of all, if a set has more than two vectors and the first two vectors are not scalar multiples then the third vector has to be a linear combination of the first two. Consider $\begin{aligned} {[\begin{matrix} x_{1} \\ x_{2} \end{matrix}], [\begin{matrix} x_{3} \\ x_{4} \end{matrix}], [\begin{matrix} x_{5} \\ x_{6} \end{matrix}]} \end{aligned}$ now we want to show that $\begin{aligned} [\begin{matrix} x_{5} \\ x_{6} \end{matrix}] \in Span {[\begin{matrix} x_{1} \\ x_{2} \end{matrix}], [\begin{matrix} x_{3} \\ x_{4} \end{matrix}]} \end{aligned}$ . This is only true if we can find some $\begin{aligned} a, b such that a \cdot [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] + b \cdot [\begin{matrix} x_{3} \\ x_{4} \end{matrix}] = [\begin{matrix} x_{5} \\ x_{6} \end{matrix}] \end{aligned}$ meaning we have to find solutions to these scalars by solving the system of equations
::首先,如果一组的矢量超过两个矢量,而前两个矢量不是弧度乘数,则第三个矢量必须是前两个矢量的线性组合。考虑 {[x1x2],[x3x4],[x5x6],[x5x6]},我们现在要显示 [x5x6],{{{[x1x2],[x3x4]}]。只有当我们找到 a,b,这样 a\[x1x2],+b}[x3x4]=[x5x6],我们才能通过解析方程式系统找到这些星量的解决方案。

$\begin{aligned} a x_{1} + b x_{3} = x_{5} \\ a x_{2} + b x_{4} = x_{6} \end{aligned}$ .
::ax1+bx3=x5ax2+bx4=x6。

$\begin{aligned} a x_{1} + b x_{3} = x_{5} \\ a x_{2} + b x_{4} = x_{6} \end{aligned}$
::ax1+bx3=x5ax2+bx4=x6 =x6 =x5ax2+bx4=x6

$\begin{aligned} - \frac{x_{2}}{x_{1}} (a x_{1} + b x_{3} = x_{5}) \to - a x_{2} - b \frac{x_{2} x_{3}}{x_{1}} = - \frac{x_{2} x_{5}}{x_{1}} \end{aligned}$
::−x2x1( 轴1+bx3=x5) ax2- bx2x3x1 x2x5x1

$\begin{aligned} a x_{2} + b x_{4} = x_{6} \end{aligned}$
::ax2+bx4=x6 轴

Adding the two equations together,
::将两个方程加在一起,

$\begin{aligned} b (x_{4} - \frac{x_{3} x_{2}}{x_{1}}) = x_{6} - \frac{x_{2} x_{5}}{x_{1}} \\ b (\frac{x_{4} x_{1} - x_{3} x_{2}}{x_{1}}) = \frac{x_{6} x_{1} - x_{2} x_{5}}{x_{1}} \\ b = (\frac{x_{6} x_{1} - x_{2} x_{5}}{x_{4} x_{1} - x_{3} x_{2}}) \end{aligned}$
::b(x4-x3x2x1)=x6-x2x5x1b(x4x1-x3x2x1)=x6x1-x2x5x1b=(x6x1-x2x5x5x4x4x1-x3x2)

Now plugging back in and solving for $\begin{aligned} a \end{aligned}$ , we get
::现在再插插进去,解决一个,我们得到

$\begin{aligned} a x_{2} + b x_{4} = x_{6} \\ a x_{2} + x_{4} (\frac{x_{6} x_{1} - x_{2} x_{5}}{x_{4} x_{1} - x_{3} x_{2}}) = x_{6} \\ a x_{2} = x_{6} - (\frac{x_{1} x_{4} x_{6} - x_{2} x_{4} x_{5}}{x_{1} x_{4} - x_{2} x_{3}}) \\ a = \frac{1}{x_{2}} (\frac{x_{2} x_{4} x_{5} - x_{2} x_{3} x_{6}}{x_{4} x_{1} - x_{2} x_{3}}) = \frac{x_{4} x_{5} - x_{3} x_{6}}{x_{4} x_{1} - x_{2} x_{3}} \end{aligned}$
::ax2+bx4=x6Ax2+x6Ax2+x4(x6x1-x2x5x4x1-x3x2)=x6Ax2=x6(x1x4x6-x2x2x4x5x1x4x4x4-x2x3)a=1x2(x2x4x5-x2x3x3x6x6x4x4x4x1-x2x3)=x4x5x3x6x6x4x1x2x3)a=1x2(x2x2x4x4x4x5-x4x4x5x4x4x4x4x4x4x4x4x4-x4x3)a=1x2(x2x2x2x2x5-x3x6x6x6x6x4x4x4x4x4x4x4x4x4x1-x3)=x4x5x5x6x6x3

Now we see that there exists solutions if and only if $\begin{aligned} x_{4} x_{1} - x_{3} x_{2} \neq 0 \end{aligned}$ which would only be true if the vectors $\begin{aligned} [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] and [\begin{matrix} x_{3} \\ x_{4} \end{matrix}] \end{aligned}$ because then   $\begin{aligned} [\begin{matrix} c x_{1} \\ c x_{2} \end{matrix}] = [\begin{matrix} x_{3} \\ x_{4} \end{matrix}] and then x_{4} x_{1} - x_{3} x_{2} = c x_{3} x_{4} - c x_{4} x_{3} = 0 \end{aligned}$ . However, because the vectors are not scalar multiples we are okay and there always exist a solution.
::现在我们看到,如果而且只有在 x4x1 - x3x2=0 的情况下,才会有解决办法,而只有矢量[x1x2] 和 [x3x4] 才是真实的,因为当时[cx1cx2] = [x3x4] = [x3x4] ,然后 x4x1 -x3x2= cx3x4 - cx4x4=cx4x3x3=0。然而,由于矢量不是卡路里倍数,所以我们没有问题,而且总是有解决办法。

Now let's move up to 3 dimensional space.
::现在,让我们向上移动到三维空间。

Again, the same definitions hold in all n-dimensions, but here things get a bit harder.
::同样的,同样的定义在所有n-diensions都持有, 但在这里,事情变得有点困难。

One vector spans a line:
::一个矢量横跨一条线 :

Now let's look at two linearly independent vectors spanning a plane:
::现在让我们看看两个线性独立的矢量横跨平面:

You see here that if you have two linearly independent vectors then all linear combinations must stay in the plane going through those two vectors and the origin. No vector can go outside of that plane because multiplying a vector by a scalar just keeps the new vector going in the same direction and then adding two vectors in the same plane stays in the same plane. This is because all planes can be expressed in the form $\begin{aligned} z = a x + b y \end{aligned}$ and if a vector of the form $\begin{aligned} [\begin{matrix} x^{'} \\ y^{'} \\ z^{'} \end{matrix}] \end{aligned}$ is in the plane then $\begin{aligned} a x^{'} + b y^{'} = z^{'} \end{aligned}$ . So if we have $\begin{aligned} [\begin{matrix} x_{0} \\ y_{0} \\ z_{0} \end{matrix}] in the plane then a x_{0} + b y_{0} = z_{0} \end{aligned}$ and if   $\begin{aligned} [\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix}] in the plane then a x_{1} + b y_{1} = z_{1} \end{aligned}$ . Now we need to show that $\begin{aligned} [\begin{matrix} x_{0} + x_{1} \\ y_{0} + y_{1} \\ z_{0} + z_{1} \end{matrix}] \end{aligned}$ is in the plane. We simply show this by adding the equations of the first two vectors to get $\begin{aligned} a x_{0} + a x_{1} + b y_{0} + b y_{1} = z_{0} + z_{1} \to a (x_{0} + x_{1}) + b (x_{0} + x_{1}) = z_{0} + z_{1} \end{aligned}$ which proves that that sum is in the plane.
::这里您可以看到, 如果您有两个线上独立的矢量, 那么所有线性组合都必须保留在通过这两个矢量和来源的平面中。没有任何矢量可以留在通过这两个矢量和源头的平面中。因为将一个矢量乘以一个星标, 使新的矢量保持同一方向, 然后在同一平面中添加两个矢量, 在同一平面中停留在同一平面中。这是因为所有方都可以以 z=ax+Ax+Ax+Ax+Ax+Ax1+Z1+Z1 。这是因为所有方能以 z=z0+Z0+Ax0+X0+Z0+Z0+Z1, 并且如果[x1+1z1] 使新的矢量保持同一方向, 然后在同一平面上添加两个方口号, 我们只需添加第一个两个矢量的方程式, 以获得 ax0+Ax1+Ax1+x1+x1+x1+x1+Z1。

Now let's show that 3 linearly independent vectors spans 3-space.
::现在让我们展示一下 3个线性独立的矢量横跨3个空格

When you play with this linear combination you see that it does not just go through the planes spanned by two of the three vectors there and that it goes through all of three space. We can prove this algebraically later when we have more tools.
::当你玩这个线性组合时,你会看到它不仅穿过由三向量中两个向量所跨越的平面,而且穿过全部三个空间。当我们有更多的工具时,我们可以稍后证明这个代数。

Food for thought: Does a set of n linearly independent vectors spans all of $\begin{aligned} R^{n} \end{aligned}$ and what is the relationship between the number of vectors and dimension of the vectors implying whether or not a set is linearly independent or linearly dependent?
::启发思考:一组 n 线性独立矢量是否覆盖了所有Rnand 矢量数量与矢量尺寸之间的关系,意味着一组矢量是否线性独立或线性依赖?

章节大纲

常规

线状组合和span