这部分是专门定义和证明柯西-Schwarz不等式的向量形式。据/p>
适用于所有载体据span class="katex">
X据/span>和据span class="katex">
y据/span>实内积空间,据/p>
|据/span>⟨据/span>X据/span>那据/span>y据/span>⟩据/span>|据/span>2据/span>≤.据/span>⟨据/span>X据/span>那据/span>X据/span>⟩据/span>⋅据/span>⟨据/span>y据/span>那据/span>y据/span>⟩据/span>那据/span>
在哪里据span class="katex">
⟨据/span>⋅据/span>那据/span>⋅据/span>⟩据/span>是内积。据/p>
同等,据span class="katex">
|据/span>⟨据/span>X据/span>那据/span>y据/span>⟩据/span>|据/span>≤.据/span>∥据/span>X据/span>∥据/span>⋅据/span>∥据/span>y据/span>∥据/span>.据/span>
平等保持当且仅当所述2个向量是线性相关的。据/p>
对于一个复杂的向量空间,我们有据/p>
|据/span>|据/span>|据/span>|据/span>|据/span>一世据/span>=据/span>1据/span>σ.据/span>N据/span>X据/span>一世据/span>y据/span>ˉ据/span>一世据/span>|据/span>|据/span>|据/span>|据/span>|据/span>2据/span>≤.据/span>j据/span>=据/span>1据/span>σ.据/span>N据/span>|据/span>X据/span>j据/span>|据/span>2据/span>K.据/span>=据/span>1据/span>σ.据/span>N据/span>|据/span>y据/span>K.据/span>|据/span>2据/span>.据/span>
等式成立当且仅当据span class="katex">
X据/span>和据span class="katex">
y据/span>是线性相关的,即,一个是另一个的标量倍数(其包括的情况下,当一个或两个是零)。据/p>
现在,我们将证明由柯西 - 施瓦茨的载体形式的上述要求。据/p>
如果据span class="katex">
V.据/span>=据/span>0.据/span>那据/span>很显然,我们有平等的,在这种情况下据span class="katex">
你据/span>和据span class="katex">
V.据/span>也是平凡的线性相关。今后,我们假设据span class="katex">
V.据/span>是非零。我们还假设据span class="katex">
⟨据/span>你据/span>那据/span>V.据/span>⟩据/span>据/span>=据/span>0.据/span>否则不等式显然成立。据/p>
让据/p>
Z.据/span>=据/span>你据/span>-据/span>⟨据/span>V.据/span>那据/span>V.据/span>⟩据/span>⟨据/span>你据/span>那据/span>V.据/span>⟩据/span>V.据/span>.据/span>
从向量的理论,我们知道据span class="katex">
Z.据/span>是的投影据span class="katex">
你据/span>到正交的飞机上据span class="katex">
V.据/span>.我们将证明这是真的,通过展示据span class="katex">
Z.据/span>正交于这个向量吗据span class="katex">
V.据/span>,或等价说据span class="katex">
⟨据/span>Z.据/span>那据/span>V.据/span>⟩据/span>=据/span>0.据/span>.通过在它的第一个参数的内积的线性,一个具有据/p>
⟨据/span>Z.据/span>那据/span>V.据/span>⟩据/span>=据/span>⟨据/span>你据/span>-据/span>⟨据/span>V.据/span>那据/span>V.据/span>⟩据/span>⟨据/span>你据/span>那据/span>V.据/span>⟩据/span>V.据/span>那据/span>V.据/span>⟩据/span>=据/span>⟨据/span>你据/span>那据/span>V.据/span>⟩据/span>-据/span>⟨据/span>V.据/span>那据/span>V.据/span>⟩据/span>⟨据/span>你据/span>那据/span>V.据/span>⟩据/span>⟨据/span>V.据/span>那据/span>V.据/span>⟩据/span>=据/span>0.据/span>.据/span>
由于这些向量是正交的,我们可以运用勾股定理据/p>
你据/span>=据/span>⟨据/span>V.据/span>那据/span>V.据/span>⟩据/span>⟨据/span>你据/span>那据/span>V.据/span>⟩据/span>V.据/span>+据/span>Z.据/span>那据/span>
这使据/p>
∥据/span>你据/span>∥据/span>2据/span>=据/span>|据/span>|据/span>|据/span>|据/span>⟨据/span>V.据/span>那据/span>V.据/span>⟩据/span>⟨据/span>你据/span>那据/span>V.据/span>⟩据/span>|据/span>|据/span>|据/span>|据/span>2据/span>∥据/span>V.据/span>∥据/span>2据/span>+据/span>∥据/span>Z.据/span>∥据/span>2据/span>=据/span>∥据/span>V.据/span>∥据/span>2据/span>|据/span>⟨据/span>你据/span>那据/span>V.据/span>⟩据/span>|据/span>2据/span>+据/span>∥据/span>Z.据/span>∥据/span>2据/span>≥据/span>∥据/span>V.据/span>∥据/span>2据/span>|据/span>⟨据/span>你据/span>那据/span>V.据/span>⟩据/span>|据/span>2据/span>.据/span>
乘以后据span class="katex">
|据/span>|据/span>V.据/span>|据/span>|据/span>2据/span>,我们得到柯西 - 施瓦茨不等式。据/p>
如果在上述表达式中的关系实际上是一个相等,则据span class="katex">
|据/span>|据/span>Z.据/span>|据/span>|据/span>2据/span>=据/span>0.据/span>因此据span class="katex">
Z.据/span>=据/span>0.据/span>.从定义据span class="katex">
Z.据/span>作为投影矢量,这意味着没有部分据span class="katex">
你据/span>垂直于据span class="katex">
V.据/span>,因此这两个向量是平行的。这建立定理。据span class="katex">
□据/span>
以代理方式证明这一点的另一种方法是下面给出的。据/p>
考虑两个序列据span class="katex">
一种据/span>=据/span>{据/span>一种据/span>一世据/span>}据/span>和据span class="katex">
B.据/span>=据/span>{据/span>B.据/span>j据/span>}据/span>和一个矩阵,对于每个元件,其构成的规则是据span class="katex">
m据/span>一世据/span>j据/span>=据/span>一种据/span>一世据/span>2据/span>.据/span>B.据/span>j据/span>2据/span>.对于具有同等数量方面的序列,我们将有据/p>
|据/span>|据/span>|据/span>|据/span>|据/span>|据/span>|据/span>|据/span>|据/span>一种据/span>1据/span>2据/span>.据/span>B.据/span>1据/span>2据/span>一种据/span>2据/span>2据/span>.据/span>B.据/span>1据/span>2据/span>⋮据/span>一种据/span>N据/span>2据/span>.据/span>B.据/span>1据/span>2据/span>一种据/span>1据/span>2据/span>.据/span>B.据/span>2据/span>2据/span>一种据/span>2据/span>2据/span>.据/span>B.据/span>2据/span>2据/span>⋮据/span>一种据/span>N据/span>2据/span>.据/span>B.据/span>2据/span>2据/span>......据/span>......据/span>⋱据/span>......据/span>一种据/span>1据/span>2据/span>.据/span>B.据/span>N据/span>2据/span>一种据/span>2据/span>2据/span>.据/span>B.据/span>N据/span>2据/span>⋮据/span>一种据/span>N据/span>2据/span>.据/span>B.据/span>N据/span>2据/span>|据/span>|据/span>|据/span>|据/span>|据/span>|据/span>|据/span>|据/span>|据/span>.据/span>
如果我们尝试将矩阵上的所有术语汇总到分配属性,则将其作为据/p>
S.据/span>m据/span>=据/span>(据/span>一世据/span>=据/span>1据/span>σ.据/span>N据/span>一种据/span>一世据/span>2据/span>)据/span>(据/span>j据/span>=据/span>1据/span>σ.据/span>N据/span>B.据/span>j据/span>2据/span>)据/span>.据/span>
现在考虑两个相关的序列据span class="katex">
一种据/span>'据/span>=据/span>{据/span>一种据/span>一世据/span>B.据/span>一世据/span>}据/span>和据span class="katex">
B.据/span>'据/span>=据/span>{据/span>一种据/span>j据/span>B.据/span>j据/span>}据/span>和矩阵,其构成规则的每个元素是据span class="katex">
m据/span>一世据/span>j据/span>'据/span>=据/span>一种据/span>一世据/span>B.据/span>一世据/span>一种据/span>j据/span>B.据/span>j据/span>, 如下据/p>
|据/span>|据/span>|据/span>|据/span>|据/span>|据/span>|据/span>|据/span>|据/span>一种据/span>1据/span>2据/span>B.据/span>1据/span>2据/span>一种据/span>2据/span>B.据/span>2据/span>一种据/span>1据/span>B.据/span>1据/span>⋮据/span>一种据/span>1据/span>B.据/span>1据/span>一种据/span>N据/span>B.据/span>N据/span>一种据/span>1据/span>B.据/span>1据/span>一种据/span>2据/span>B.据/span>2据/span>一种据/span>2据/span>2据/span>B.据/span>2据/span>2据/span>⋮据/span>一种据/span>2据/span>B.据/span>2据/span>一种据/span>N据/span>B.据/span>N据/span>......据/span>......据/span>⋱据/span>......据/span>一种据/span>1据/span>B.据/span>1据/span>一种据/span>N据/span>B.据/span>N据/span>一种据/span>2据/span>B.据/span>2据/span>一种据/span>N据/span>B.据/span>N据/span>⋮据/span>一种据/span>N据/span>2据/span>B.据/span>N据/span>2据/span>|据/span>|据/span>|据/span>|据/span>|据/span>|据/span>|据/span>|据/span>|据/span>.据/span>
总结所有的条款据/p>
S.据/span>m据/span>'据/span>=据/span>(据/span>K.据/span>=据/span>1据/span>σ.据/span>N据/span>一种据/span>K.据/span>B.据/span>K.据/span>)据/span>2据/span>=据/span>一世据/span>=据/span>1据/span>σ.据/span>N据/span>一种据/span>一世据/span>B.据/span>一世据/span>j据/span>=据/span>1据/span>σ.据/span>N据/span>一种据/span>j据/span>B.据/span>j据/span>.据/span>
用一个和减去另一个和抵消了对角线项和收益据/p>
S.据/span>m据/span>-据/span>S.据/span>m据/span>'据/span>=据/span>1据/span>≤.据/span>一世据/span>据据/span>j据/span>σ.据/span>j据/span>据据/span>一世据/span>≤.据/span>N据/span>j据/span>=据/span>1据/span>σ.据/span>N据/span>一种据/span>一世据/span>一种据/span>一世据/span>B.据/span>j据/span>B.据/span>j据/span>-据/span>一种据/span>一世据/span>B.据/span>一世据/span>一种据/span>j据/span>B.据/span>j据/span>.据/span>
因为每对一对据span class="katex">
一世据/span>据据/span>j据/span>,存在两个术语据span class="katex">
一种据/span>一世据/span>一种据/span>一世据/span>B.据/span>j据/span>B.据/span>j据/span>-据/span>一种据/span>一世据/span>B.据/span>一世据/span>一种据/span>j据/span>B.据/span>j据/span>和据span class="katex">
一种据/span>j据/span>一种据/span>j据/span>B.据/span>一世据/span>B.据/span>一世据/span>-据/span>一种据/span>j据/span>B.据/span>j据/span>一种据/span>一世据/span>B.据/span>一世据/span>,上述差异可以重新编写为据/p>
S.据/span>m据/span>-据/span>S.据/span>m据/span>'据/span>=据/span>一世据/span>=据/span>1据/span>σ.据/span>j据/span>-据/span>1据/span>j据/span>=据/span>1据/span>σ.据/span>N据/span>一种据/span>一世据/span>一种据/span>一世据/span>B.据/span>j据/span>B.据/span>j据/span>-据/span>一种据/span>一世据/span>B.据/span>一世据/span>一种据/span>j据/span>B.据/span>j据/span>+据/span>一种据/span>j据/span>一种据/span>j据/span>B.据/span>一世据/span>B.据/span>一世据/span>-据/span>一种据/span>j据/span>B.据/span>j据/span>一种据/span>一世据/span>B.据/span>一世据/span>=据/span>一世据/span>=据/span>1据/span>σ.据/span>j据/span>-据/span>1据/span>j据/span>=据/span>1据/span>σ.据/span>N据/span>(据/span>一种据/span>一世据/span>B.据/span>j据/span>-据/span>一种据/span>j据/span>B.据/span>一世据/span>)据/span>2据/span>.据/span>
因为正方形是非负的,因此据span class="katex">
S.据/span>m据/span>≥据/span>S.据/span>m据/span>'据/span>,因此据/p>
(据/span>一世据/span>=据/span>1据/span>σ.据/span>N据/span>一种据/span>一世据/span>2据/span>)据/span>(据/span>j据/span>=据/span>1据/span>σ.据/span>N据/span>B.据/span>j据/span>2据/span>)据/span>≥据/span>(据/span>一世据/span>=据/span>1据/span>σ.据/span>N据/span>一种据/span>一世据/span>B.据/span>一世据/span>)据/span>2据/span>.据/span>□据/span>
下面的例子是由柯西 - 施瓦茨的矢量形式的动机。据/p>
什么是最大的功能据span class="katex">
一种据/span>罪据/span>θ据/span>+据/span>B.据/span>COS.据/span>θ据/span>还是据/span>
由Cauchy-Schwarz,据/p>
(据/span>一种据/span>罪据/span>θ据/span>+据/span>B.据/span>COS.据/span>θ据/span>)据/span>2据/span>≤.据/span>(据/span>一种据/span>2据/span>+据/span>B.据/span>2据/span>)据/span>(据/span>罪据/span>2据/span>θ据/span>+据/span>COS.据/span>2据/span>θ据/span>)据/span>=据/span>一种据/span>2据/span>+据/span>B.据/span>2据/span>.据/span>
因此,最大值为据span class="katex">
一种据/span>2据/span>+据/span>B.据/span>2据/span>
,这是实现的据span class="katex">
棕褐色据/span>θ据/span>=据/span>B.据/span>一种据/span>.据span class="katex">
□据/span>