Misplaced Pages

#307692

88-488: Inner products of vectors can describe finite sums (via finite-dimensional vector spaces), infinite series (via vectors in sequence spaces ), and integrals (via vectors in Hilbert spaces ). The inequality for sums was published by Augustin-Louis Cauchy  ( 1821 ). The corresponding inequality for integrals was published by Viktor Bunyakovsky  ( 1859 ) and Hermann Schwarz  ( 1888 ). Schwarz gave

176-427: A ∗ a ) ≥ φ ( a ) φ ( a ∗ ) . {\displaystyle \varphi \left(a^{*}a\right)\geq \varphi (a)\varphi \left(a^{*}\right).} This extends the fact φ ( a ∗ a ) ⋅ 1 ≥ φ ( a ) ∗ φ (

264-400: A ∗ a ) . {\displaystyle \left|\varphi \left(b^{*}a\right)\right|^{2}\leq \varphi \left(b^{*}b\right)\varphi \left(a^{*}a\right).} The next two theorems are further examples in operator algebra. Kadison–Schwarz inequality   (Named after Richard Kadison )  —  If φ {\displaystyle \varphi } is

352-412: A 1 , a 2 , … , a k , {\displaystyle a_{1},a_{2},\dots ,a_{k},} not all zero, such that where 0 {\displaystyle \mathbf {0} } denotes the zero vector. This implies that at least one of the scalars is nonzero, say a 1 ≠ 0 {\displaystyle a_{1}\neq 0} , and

440-411: A i ≠ 0 {\displaystyle a_{i}\neq 0} ), this proves that the vectors v 1 , … , v k {\displaystyle \mathbf {v} _{1},\dots ,\mathbf {v} _{k}} are linearly dependent. As a consequence, the zero vector can not possibly belong to any collection of vectors that is linearly in dependent. Now consider

528-532: A i = 0 , {\displaystyle a_{i}=0,} which means that the vectors v 1 = ( 1 , 1 ) {\displaystyle \mathbf {v} _{1}=(1,1)} and v 2 = ( − 3 , 2 ) {\displaystyle \mathbf {v} _{2}=(-3,2)} are linearly independent. In order to determine if the three vectors in R 4 , {\displaystyle \mathbb {R} ^{4},} are linearly dependent, form

616-530: A j := 0 {\displaystyle a_{j}:=0} so that consequently a j v j = 0 v j = 0 {\displaystyle a_{j}\mathbf {v} _{j}=0\mathbf {v} _{j}=\mathbf {0} } ). Simplifying a 1 v 1 + ⋯ + a k v k {\displaystyle a_{1}\mathbf {v} _{1}+\cdots +a_{k}\mathbf {v} _{k}} gives: Because not all scalars are zero (in particular,

704-881: A ) ≤ ‖ φ ( 1 ) ‖ φ ( a ∗ a ) ,  and  ‖ φ ( a ∗ b ) ‖ 2 ≤ ‖ φ ( a ∗ a ) ‖ ⋅ ‖ φ ( b ∗ b ) ‖ . {\displaystyle {\begin{aligned}\varphi (a)^{*}\varphi (a)&\leq \Vert \varphi (1)\Vert \varphi \left(a^{*}a\right),{\text{ and }}\\[5mu]\Vert \varphi \left(a^{*}b\right)\Vert ^{2}&\leq \Vert \varphi \left(a^{*}a\right)\Vert \cdot \Vert \varphi \left(b^{*}b\right)\Vert .\end{aligned}}} Finite sum Too Many Requests If you report this error to

792-448: A ) = | φ ( a ) | 2 , {\displaystyle \varphi \left(a^{*}a\right)\cdot 1\geq \varphi (a)^{*}\varphi (a)=|\varphi (a)|^{2},} when φ {\displaystyle \varphi } is a linear functional. The case when a {\displaystyle a} is self-adjoint, that is, a = a ∗ , {\displaystyle a=a^{*},}

880-454: A finite set of vectors: A finite set of vectors is linearly independent if the sequence obtained by ordering them is linearly independent. In other words, one has the following result that is often useful. A sequence of vectors is linearly independent if and only if it does not contain the same vector twice and the set of its vectors is linearly independent. An infinite set of vectors is linearly independent if every nonempty finite subset

968-458: A relation of linear dependence between u {\displaystyle \mathbf {u} } and v . {\displaystyle \mathbf {v} .} The converse was proved at the beginning of this section, so the proof is complete. ◼ {\displaystyle \blacksquare } Consider an arbitrary pair of vectors u , v {\displaystyle \mathbf {u} ,\mathbf {v} } . Define

SECTION 10

#1732780618308

1056-435: A sequence of vectors is linearly independent if and only if 0 {\displaystyle \mathbf {0} } can be represented as a linear combination of its vectors in a unique way. If a sequence of vectors contains the same vector twice, it is necessarily dependent. The linear dependency of a sequence of vectors does not depend of the order of the terms in the sequence. This allows defining linear independence for

1144-420: A subset of vectors in a vector space is linearly dependent are central to determining the dimension of a vector space. A sequence of vectors v 1 , v 2 , … , v k {\displaystyle \mathbf {v} _{1},\mathbf {v} _{2},\dots ,\mathbf {v} _{k}} from a vector space V is said to be linearly dependent , if there exist scalars

1232-401: A unital positive map, then for every normal element a {\displaystyle a} in its domain, we have φ ( a ∗ a ) ≥ φ ( a ∗ ) φ ( a ) {\displaystyle \varphi (a^{*}a)\geq \varphi \left(a^{*}\right)\varphi (a)} and φ (

1320-398: Is 0 {\displaystyle \mathbf {0} } (while the other is non-zero) then exactly one of (1) and (2) is true (with the other being false). The vectors u {\displaystyle \mathbf {u} } and v {\displaystyle \mathbf {v} } are linearly in dependent if and only if u {\displaystyle \mathbf {u} }

1408-714: Is a complex number satisfying | α | = 1 {\displaystyle |\alpha |=1} and α ⟨ u , v ⟩ = | ⟨ u , v ⟩ | {\displaystyle \alpha \langle \mathbf {u} ,\mathbf {v} \rangle =|\langle \mathbf {u} ,\mathbf {v} \rangle |} . Such an α {\displaystyle \alpha } exists since if ⟨ u , v ⟩ = 0 {\displaystyle \langle \mathbf {u} ,\mathbf {v} \rangle =0} then α {\displaystyle \alpha } can be taken to be 1. Since

1496-640: Is a direct consequence of the Cauchy–Schwarz inequality, obtained by using the dot product on R n {\displaystyle \mathbb {R} ^{n}} upon substituting u i ′ = u i v i t {\displaystyle u_{i}'={\frac {u_{i}}{\sqrt {v_{i}{\vphantom {t}}}}}} and v i ′ = v i t {\displaystyle v_{i}'={\textstyle {\sqrt {v_{i}{\vphantom {t}}}}}} . This form

1584-2441: Is a generalization of this. In any inner product space , the triangle inequality is a consequence of the Cauchy–Schwarz inequality, as is now shown: ‖ u + v ‖ 2 = ⟨ u + v , u + v ⟩ = ‖ u ‖ 2 + ⟨ u , v ⟩ + ⟨ v , u ⟩ + ‖ v ‖ 2      where  ⟨ v , u ⟩ = ⟨ u , v ⟩ ¯ = ‖ u ‖ 2 + 2 Re ⁡ ⟨ u , v ⟩ + ‖ v ‖ 2 ≤ ‖ u ‖ 2 + 2 | ⟨ u , v ⟩ | + ‖ v ‖ 2 ≤ ‖ u ‖ 2 + 2 ‖ u ‖ ‖ v ‖ + ‖ v ‖ 2      using CS = ( ‖ u ‖ + ‖ v ‖ ) 2 . {\displaystyle {\begin{alignedat}{4}\|\mathbf {u} +\mathbf {v} \|^{2}&=\langle \mathbf {u} +\mathbf {v} ,\mathbf {u} +\mathbf {v} \rangle &&\\&=\|\mathbf {u} \|^{2}+\langle \mathbf {u} ,\mathbf {v} \rangle +\langle \mathbf {v} ,\mathbf {u} \rangle +\|\mathbf {v} \|^{2}~&&~{\text{ where }}\langle \mathbf {v} ,\mathbf {u} \rangle ={\overline {\langle \mathbf {u} ,\mathbf {v} \rangle }}\\&=\|\mathbf {u} \|^{2}+2\operatorname {Re} \langle \mathbf {u} ,\mathbf {v} \rangle +\|\mathbf {v} \|^{2}&&\\&\leq \|\mathbf {u} \|^{2}+2|\langle \mathbf {u} ,\mathbf {v} \rangle |+\|\mathbf {v} \|^{2}&&\\&\leq \|\mathbf {u} \|^{2}+2\|\mathbf {u} \|\|\mathbf {v} \|+\|\mathbf {v} \|^{2}~&&~{\text{ using CS}}\\&={\bigl (}\|\mathbf {u} \|+\|\mathbf {v} \|{\bigr )}^{2}.&&\end{alignedat}}} Taking square roots gives

1672-808: Is a polynomial of degree 2 {\displaystyle 2} (unless u = 0 , {\displaystyle \mathbf {u} =0,} which is a case that was checked earlier). Since the sign of p {\displaystyle p} does not change, the discriminant of this polynomial must be non-positive: Δ = 4 ( | ⟨ u , v ⟩ | 2 − ‖ u ‖ 2 ‖ v ‖ 2 ) ≤ 0. {\displaystyle \Delta =4{\bigl (}\,|\langle \mathbf {u} ,\mathbf {v} \rangle |^{2}-\Vert \mathbf {u} \Vert ^{2}\Vert \mathbf {v} \Vert ^{2}{\bigr )}\leq 0.} The conclusion follows. For

1760-395: Is a positive linear functional on a C*-algebra A , {\displaystyle A,} then for all a , b ∈ A , {\displaystyle a,b\in A,} | φ ( b ∗ a ) | 2 ≤ φ ( b ∗ b ) φ (

1848-991: Is a scalar multiple of the other. If u = c v {\displaystyle \mathbf {u} =c\mathbf {v} } where c {\displaystyle c} is some scalar then | ⟨ u , v ⟩ | = | ⟨ c v , v ⟩ | = | c ⟨ v , v ⟩ | = | c | ‖ v ‖ ‖ v ‖ = ‖ c v ‖ ‖ v ‖ = ‖ u ‖ ‖ v ‖ {\displaystyle |\langle \mathbf {u} ,\mathbf {v} \rangle |=|\langle c\mathbf {v} ,\mathbf {v} \rangle |=|c\langle \mathbf {v} ,\mathbf {v} \rangle |=|c|\|\mathbf {v} \|\|\mathbf {v} \|=\|c\mathbf {v} \|\|\mathbf {v} \|=\|\mathbf {u} \|\|\mathbf {v} \|} which shows that equality holds in

SECTION 20

#1732780618308

1936-2431: Is a vector orthogonal to the vector v {\displaystyle \mathbf {v} } (Indeed, z {\displaystyle \mathbf {z} } is the projection of u {\displaystyle \mathbf {u} } onto the plane orthogonal to v . {\displaystyle \mathbf {v} .} ) We can thus apply the Pythagorean theorem to u = ⟨ u , v ⟩ ⟨ v , v ⟩ v + z {\displaystyle \mathbf {u} ={\frac {\langle \mathbf {u} ,\mathbf {v} \rangle }{\langle \mathbf {v} ,\mathbf {v} \rangle }}\mathbf {v} +\mathbf {z} } which gives ‖ u ‖ 2 = | ⟨ u , v ⟩ ⟨ v , v ⟩ | 2 ‖ v ‖ 2 + ‖ z ‖ 2 = | ⟨ u , v ⟩ | 2 ( ‖ v ‖ 2 ) 2 ‖ v ‖ 2 + ‖ z ‖ 2 = | ⟨ u , v ⟩ | 2 ‖ v ‖ 2 + ‖ z ‖ 2 ≥ | ⟨ u , v ⟩ | 2 ‖ v ‖ 2 . {\displaystyle \|\mathbf {u} \|^{2}=\left|{\frac {\langle \mathbf {u} ,\mathbf {v} \rangle }{\langle \mathbf {v} ,\mathbf {v} \rangle }}\right|^{2}\|\mathbf {v} \|^{2}+\|\mathbf {z} \|^{2}={\frac {|\langle \mathbf {u} ,\mathbf {v} \rangle |^{2}}{(\|\mathbf {v} \|^{2})^{2}}}\,\|\mathbf {v} \|^{2}+\|\mathbf {z} \|^{2}={\frac {|\langle \mathbf {u} ,\mathbf {v} \rangle |^{2}}{\|\mathbf {v} \|^{2}}}+\|\mathbf {z} \|^{2}\geq {\frac {|\langle \mathbf {u} ,\mathbf {v} \rangle |^{2}}{\|\mathbf {v} \|^{2}}}.} The Cauchy–Schwarz inequality follows by multiplying by ‖ v ‖ 2 {\displaystyle \|\mathbf {v} \|^{2}} and then taking

2024-548: Is always a non-negative real number (even if the inner product is complex-valued). By taking the square root of both sides of the above inequality, the Cauchy–Schwarz inequality can be written in its more familiar form in terms of the norm: Moreover, the two sides are equal if and only if u {\displaystyle \mathbf {u} } and v {\displaystyle \mathbf {v} } are linearly dependent . Sedrakyan's inequality , also known as Bergström 's inequality, Engel 's form, Titu 's lemma (or

2112-467: Is an n × m matrix and Λ is a column vector with m {\displaystyle m} entries, and we are again interested in A Λ = 0 . As we saw previously, this is equivalent to a list of n {\displaystyle n} equations. Consider the first m {\displaystyle m} rows of A {\displaystyle A} , the first m {\displaystyle m} equations; any solution of

2200-787: Is an index (i.e. an element of { 1 , … , k } {\displaystyle \{1,\ldots ,k\}} ) such that v i = 0 . {\displaystyle \mathbf {v} _{i}=\mathbf {0} .} Then let a i := 1 {\displaystyle a_{i}:=1} (alternatively, letting a i {\displaystyle a_{i}} be equal any other non-zero scalar will also work) and then let all other scalars be 0 {\displaystyle 0} (explicitly, this means that for any index j {\displaystyle j} other than i {\displaystyle i} (i.e. for j ≠ i {\displaystyle j\neq i} ), let

2288-606: Is especially helpful when the inequality involves fractions where the numerator is a perfect square . The real vector space R 2 {\displaystyle \mathbb {R} ^{2}} denotes the 2-dimensional plane. It is also the 2-dimensional Euclidean space where the inner product is the dot product . If u = ( u 1 , u 2 ) {\displaystyle \mathbf {u} =(u_{1},u_{2})} and v = ( v 1 , v 2 ) {\displaystyle \mathbf {v} =(v_{1},v_{2})} then

2376-493: Is henceforth assumed that v ≠ 0 . {\displaystyle \mathbf {v} \neq \mathbf {0} .} Let z := u − ⟨ u , v ⟩ ⟨ v , v ⟩ v . {\displaystyle \mathbf {z} :=\mathbf {u} -{\frac {\langle \mathbf {u} ,\mathbf {v} \rangle }{\langle \mathbf {v} ,\mathbf {v} \rangle }}\mathbf {v} .} It follows from

2464-413: Is linearly independent. Conversely, an infinite set of vectors is linearly dependent if it contains a finite subset that is linearly dependent, or equivalently, if some vector in the set is a linear combination of other vectors in the set. An indexed family of vectors is linearly independent if it does not contain the same vector twice, and if the set of its vectors is linearly independent. Otherwise,

2552-676: Is not a scalar multiple of v {\displaystyle \mathbf {v} } and v {\displaystyle \mathbf {v} } is not a scalar multiple of u {\displaystyle \mathbf {u} } . Three vectors: Consider the set of vectors v 1 = ( 1 , 1 ) , {\displaystyle \mathbf {v} _{1}=(1,1),} v 2 = ( − 3 , 2 ) , {\displaystyle \mathbf {v} _{2}=(-3,2),} and v 3 = ( 2 , 4 ) , {\displaystyle \mathbf {v} _{3}=(2,4),} then

2640-399: Is not ignored, it becomes necessary to add a third vector to the linearly independent set. In general, n linearly independent vectors are required to describe all locations in n -dimensional space. If one or more vectors from a given sequence of vectors v 1 , … , v k {\displaystyle \mathbf {v} _{1},\dots ,\mathbf {v} _{k}}

2728-688: Is only possible if c ≠ 0 {\displaystyle c\neq 0} and v ≠ 0 {\displaystyle \mathbf {v} \neq \mathbf {0} } ; in this case, it is possible to multiply both sides by 1 c {\textstyle {\frac {1}{c}}} to conclude v = 1 c u . {\textstyle \mathbf {v} ={\frac {1}{c}}\mathbf {u} .} This shows that if u ≠ 0 {\displaystyle \mathbf {u} \neq \mathbf {0} } and v ≠ 0 {\displaystyle \mathbf {v} \neq \mathbf {0} } then (1)

Cauchy–Schwarz inequality - Misplaced Pages Continue

2816-590: Is sensible, by showing that the right-hand side lies in the interval [−1, 1] and justifies the notion that (real) Hilbert spaces are simply generalizations of the Euclidean space . It can also be used to define an angle in complex inner-product spaces , by taking the absolute value or the real part of the right-hand side, as is done when extracting a metric from quantum fidelity . Let X {\displaystyle X} and Y {\displaystyle Y} be random variables . Then

2904-420: Is sometimes known as Kadison's inequality . Cauchy–Schwarz inequality   (Modified Schwarz inequality for 2-positive maps)  —  For a 2-positive map φ {\displaystyle \varphi } between C*-algebras, for all a , b {\displaystyle a,b} in its domain, φ ( a ) ∗ φ (

2992-452: Is sufficient information to describe the location, because the geographic coordinate system may be considered as a 2-dimensional vector space (ignoring altitude and the curvature of the Earth's surface). The person might add, "The place is 5 miles northeast of here." This last statement is true , but it is not necessary to find the location. In this example the "3 miles north" vector and

3080-1131: Is the angle between u {\displaystyle \mathbf {u} } and v {\displaystyle \mathbf {v} } . The form above is perhaps the easiest in which to understand the inequality, since the square of the cosine can be at most 1, which occurs when the vectors are in the same or opposite directions. It can also be restated in terms of the vector coordinates u 1 {\displaystyle u_{1}} , u 2 {\displaystyle u_{2}} , v 1 {\displaystyle v_{1}} , and v 2 {\displaystyle v_{2}} as ( u 1 v 1 + u 2 v 2 ) 2 ≤ ( u 1 2 + u 2 2 ) ( v 1 2 + v 2 2 ) , {\displaystyle \left(u_{1}v_{1}+u_{2}v_{2}\right)^{2}\leq \left(u_{1}^{2}+u_{2}^{2}\right)\left(v_{1}^{2}+v_{2}^{2}\right),} where equality holds if and only if

3168-735: Is the pointwise complex conjugate of g . {\displaystyle g.} In this language, the Cauchy–Schwarz inequality becomes | φ ( g ∗ f ) | 2 ≤ φ ( f ∗ f ) φ ( g ∗ g ) , {\displaystyle {\bigl |}\varphi (g^{*}f){\bigr |}^{2}\leq \varphi \left(f^{*}f\right)\varphi \left(g^{*}g\right),} which extends verbatim to positive functionals on C*-algebras: Cauchy–Schwarz inequality for positive functionals on C*-algebras  —  If φ {\displaystyle \varphi }

3256-1170: Is the field of real numbers R {\displaystyle \mathbb {R} } or complex numbers C . {\displaystyle \mathbb {C} .} Then with equality holding in the Cauchy–Schwarz Inequality if and only if u {\displaystyle \mathbf {u} } and v {\displaystyle \mathbf {v} } are linearly dependent . Moreover, if | ⟨ u , v ⟩ | = ‖ u ‖ ‖ v ‖ {\displaystyle \left|\langle \mathbf {u} ,\mathbf {v} \rangle \right|=\|\mathbf {u} \|\|\mathbf {v} \|} and v ≠ 0 {\displaystyle \mathbf {v} \neq \mathbf {0} } then u = ⟨ u , v ⟩ ‖ v ‖ 2 v . {\displaystyle \mathbf {u} ={\frac {\langle \mathbf {u} ,\mathbf {v} \rangle }{\|\mathbf {v} \|^{2}}}\mathbf {v} .} In both of

3344-396: Is the zero vector 0 {\displaystyle \mathbf {0} } then the vector v 1 , … , v k {\displaystyle \mathbf {v} _{1},\dots ,\mathbf {v} _{k}} are necessarily linearly dependent (and consequently, they are not linearly independent). To see why, suppose that i {\displaystyle i}

3432-457: Is the zero vector then u {\displaystyle \mathbf {u} } and v {\displaystyle \mathbf {v} } are necessarily linearly dependent (for example, if u = 0 {\displaystyle \mathbf {u} =\mathbf {0} } then u = c v {\displaystyle \mathbf {u} =c\mathbf {v} } where c = 0 {\displaystyle c=0} ), so

3520-714: Is true if and only if (2) is true; that is, in this particular case either both (1) and (2) are true (and the vectors are linearly dependent) or else both (1) and (2) are false (and the vectors are linearly in dependent). If u = c v {\displaystyle \mathbf {u} =c\mathbf {v} } but instead u = 0 {\displaystyle \mathbf {u} =\mathbf {0} } then at least one of c {\displaystyle c} and v {\displaystyle \mathbf {v} } must be zero. Moreover, if exactly one of u {\displaystyle \mathbf {u} } and v {\displaystyle \mathbf {v} }

3608-780: Is true in this particular case. Similarly, if v = 0 {\displaystyle \mathbf {v} =\mathbf {0} } then (2) is true because v = 0 u . {\displaystyle \mathbf {v} =0\mathbf {u} .} If u = v {\displaystyle \mathbf {u} =\mathbf {v} } (for instance, if they are both equal to the zero vector 0 {\displaystyle \mathbf {0} } ) then both (1) and (2) are true (by using c := 1 {\displaystyle c:=1} for both). If u = c v {\displaystyle \mathbf {u} =c\mathbf {v} } then u ≠ 0 {\displaystyle \mathbf {u} \neq \mathbf {0} }

Cauchy–Schwarz inequality - Misplaced Pages Continue

3696-449: Is zero. Explicitly, if v 1 {\displaystyle \mathbf {v} _{1}} is any vector then the sequence v 1 {\displaystyle \mathbf {v} _{1}} (which is a sequence of length 1 {\displaystyle 1} ) is linearly dependent if and only if v 1 = 0 {\displaystyle \mathbf {v} _{1}=\mathbf {0} } ; alternatively,

3784-771: The Cauchy–Schwarz Inequality . The case where v = c u {\displaystyle \mathbf {v} =c\mathbf {u} } for some scalar c {\displaystyle c} follows from the previous case: | ⟨ u , v ⟩ | = | ⟨ v , u ⟩ | = ‖ v ‖ ‖ u ‖ . {\displaystyle |\langle \mathbf {u} ,\mathbf {v} \rangle |=|\langle \mathbf {v} ,\mathbf {u} \rangle |=\|\mathbf {v} \|\|\mathbf {u} \|.} In particular, if at least one of u {\displaystyle \mathbf {u} } and v {\displaystyle \mathbf {v} }

3872-414: The "4 miles east" vector are linearly independent. That is to say, the north vector cannot be described in terms of the east vector, and vice versa. The third "5 miles northeast" vector is a linear combination of the other two vectors, and it makes the set of vectors linearly dependent , that is, one of the three vectors is unnecessary to define a specific location on a plane. Also note that if altitude

3960-616: The Cauchy–Schwarz inequality becomes: ⟨ u , v ⟩ 2 = ( ‖ u ‖ ‖ v ‖ cos ⁡ θ ) 2 ≤ ‖ u ‖ 2 ‖ v ‖ 2 , {\displaystyle \langle \mathbf {u} ,\mathbf {v} \rangle ^{2}={\bigl (}\|\mathbf {u} \|\|\mathbf {v} \|\cos \theta {\bigr )}^{2}\leq \|\mathbf {u} \|^{2}\|\mathbf {v} \|^{2},} where θ {\displaystyle \theta }

4048-452: The Cauchy–Schwarz inequality exist. Hölder's inequality generalizes it to L p {\displaystyle L^{p}} norms. More generally, it can be interpreted as a special case of the definition of the norm of a linear operator on a Banach space (Namely, when the space is a Hilbert space ). Further generalizations are in the context of operator theory , e.g. for operator-convex functions and operator algebras , where

4136-454: The Cauchy–Schwarz inequality other than those given below. When consulting other sources, there are often two sources of confusion. First, some authors define ⟨⋅,⋅⟩ to be linear in the second argument rather than the first. Second, some proofs are only valid when the field is R {\displaystyle \mathbb {R} } and not C . {\displaystyle \mathbb {C} .} This section gives two proofs of

4224-1530: The T2 lemma), states that for real numbers u 1 , u 2 , … , u n {\displaystyle u_{1},u_{2},\dots ,u_{n}} and positive real numbers v 1 , v 2 , … , v n {\displaystyle v_{1},v_{2},\dots ,v_{n}} : ( u 1 + u 2 + ⋯ + u n ) 2 v 1 + v 2 + ⋯ + v n ≤ u 1 2 v 1 + u 2 2 v 2 + ⋯ + u n 2 v n , {\displaystyle {\frac {\left(u_{1}+u_{2}+\cdots +u_{n}\right)^{2}}{v_{1}+v_{2}+\cdots +v_{n}}}\leq {\frac {u_{1}^{2}}{v_{1}}}+{\frac {u_{2}^{2}}{v_{2}}}+\cdots +{\frac {u_{n}^{2}}{v_{n}}},} or, using summation notation, ( ∑ i = 1 n u i ) 2 / ∑ i = 1 n v i ≤ ∑ i = 1 n u i 2 v i . {\displaystyle {\biggl (}\sum _{i=1}^{n}u_{i}{\biggr )}^{2}{\bigg /}\sum _{i=1}^{n}v_{i}\,\leq \,\sum _{i=1}^{n}{\frac {u_{i}^{2}}{v_{i}}}.} It

4312-508: The Wikimedia System Administrators, please include the details below. Request from 172.68.168.132 via cp1112 cp1112, Varnish XID 940555871 Upstream caches: cp1112 int Error: 429, Too Many Requests at Thu, 28 Nov 2024 07:56:58 GMT Linear independence In the theory of vector spaces , a set of vectors is said to be linearly independent if there exists no nontrivial linear combination of

4400-481: The above computation shows that the Cauchy–Schwarz inequality holds in this case. Consequently, the Cauchy–Schwarz inequality only needs to be proven only for non-zero vectors and also only the non-trivial direction of the Equality Characterization must be shown. The special case of v = 0 {\displaystyle \mathbf {v} =\mathbf {0} } was proven above so it

4488-401: The above equation is able to be written as if k > 1 , {\displaystyle k>1,} and v 1 = 0 {\displaystyle \mathbf {v} _{1}=\mathbf {0} } if k = 1. {\displaystyle k=1.} Thus, a set of vectors is linearly dependent if and only if one of them is zero or a linear combination of

SECTION 50

#1732780618308

4576-2093: The bar notation is used for complex conjugation ), then the inequality may be restated more explicitly as follows: | ⟨ u , v ⟩ | 2 = | ∑ k = 1 n u k v ¯ k | 2 ≤ ⟨ u , u ⟩ ⟨ v , v ⟩ = ( ∑ k = 1 n u k u ¯ k ) ( ∑ k = 1 n v k v ¯ k ) = ∑ j = 1 n | u j | 2 ∑ k = 1 n | v k | 2 . {\displaystyle {\bigl |}\langle \mathbf {u} ,\mathbf {v} \rangle {\bigr |}^{2}={\Biggl |}\sum _{k=1}^{n}u_{k}{\bar {v}}_{k}{\Biggr |}^{2}\leq \langle \mathbf {u} ,\mathbf {u} \rangle \langle \mathbf {v} ,\mathbf {v} \rangle ={\biggl (}\sum _{k=1}^{n}u_{k}{\bar {u}}_{k}{\biggr )}{\biggl (}\sum _{k=1}^{n}v_{k}{\bar {v}}_{k}{\biggr )}=\sum _{j=1}^{n}|u_{j}|^{2}\sum _{k=1}^{n}|v_{k}|^{2}.} That is, | u 1 v ¯ 1 + ⋯ + u n v ¯ n | 2 ≤ ( | u 1 | 2 + ⋯ + | u n | 2 ) ( | v 1 | 2 + ⋯ + | v n | 2 ) . {\displaystyle {\bigl |}u_{1}{\bar {v}}_{1}+\cdots +u_{n}{\bar {v}}_{n}{\bigr |}^{2}\leq {\bigl (}|u_{1}|{}^{2}+\cdots +|u_{n}|{}^{2}{\bigr )}{\bigl (}|v_{1}|{}^{2}+\cdots +|v_{n}|{}^{2}{\bigr )}.} For

4664-692: The collection v 1 {\displaystyle \mathbf {v} _{1}} is linearly independent if and only if v 1 ≠ 0 . {\displaystyle \mathbf {v} _{1}\neq \mathbf {0} .} This example considers the special case where there are exactly two vector u {\displaystyle \mathbf {u} } and v {\displaystyle \mathbf {v} } from some real or complex vector space. The vectors u {\displaystyle \mathbf {u} } and v {\displaystyle \mathbf {v} } are linearly dependent if and only if at least one of

4752-812: The condition for linear dependence seeks a set of non-zero scalars, such that or Row reduce this matrix equation by subtracting the first row from the second to obtain, Continue the row reduction by (i) dividing the second row by 5, and then (ii) multiplying by 3 and adding to the first row, that is Rearranging this equation allows us to obtain which shows that non-zero a i exist such that v 3 = ( 2 , 4 ) {\displaystyle \mathbf {v} _{3}=(2,4)} can be defined in terms of v 1 = ( 1 , 1 ) {\displaystyle \mathbf {v} _{1}=(1,1)} and v 2 = ( − 3 , 2 ) . {\displaystyle \mathbf {v} _{2}=(-3,2).} Thus,

4840-403: The covariance inequality is given by: Var ⁡ ( X ) ≥ Cov ⁡ ( X , Y ) 2 Var ⁡ ( Y ) . {\displaystyle \operatorname {Var} (X)\geq {\frac {\operatorname {Cov} (X,Y)^{2}}{\operatorname {Var} (Y)}}.} After defining an inner product on the set of random variables using

4928-1914: The covariance inequality using the Cauchy–Schwarz inequality, let μ = E ⁡ ( X ) {\displaystyle \mu =\operatorname {E} (X)} and ν = E ⁡ ( Y ) , {\displaystyle \nu =\operatorname {E} (Y),} then | Cov ⁡ ( X , Y ) | 2 = | E ⁡ ( ( X − μ ) ( Y − ν ) ) | 2 = | ⟨ X − μ , Y − ν ⟩ | 2 ≤ ⟨ X − μ , X − μ ⟩ ⟨ Y − ν , Y − ν ⟩ = E ⁡ ( ( X − μ ) 2 ) E ⁡ ( ( Y − ν ) 2 ) = Var ⁡ ( X ) Var ⁡ ( Y ) , {\displaystyle {\begin{aligned}{\bigl |}\operatorname {Cov} (X,Y){\bigr |}^{2}&={\bigl |}\operatorname {E} ((X-\mu )(Y-\nu )){\bigr |}^{2}\\&={\bigl |}\langle X-\mu ,Y-\nu \rangle {\bigr |}^{2}\\&\leq \langle X-\mu ,X-\mu \rangle \langle Y-\nu ,Y-\nu \rangle \\&=\operatorname {E} \left((X-\mu )^{2}\right)\operatorname {E} \left((Y-\nu )^{2}\right)\\&=\operatorname {Var} (X)\operatorname {Var} (Y),\end{aligned}}} where Var {\displaystyle \operatorname {Var} } denotes variance and Cov {\displaystyle \operatorname {Cov} } denotes covariance . There are many different proofs of

5016-537: The determinant of A {\displaystyle A} , which is Since the determinant is non-zero, the vectors ( 1 , 1 ) {\displaystyle (1,1)} and ( − 3 , 2 ) {\displaystyle (-3,2)} are linearly independent. Otherwise, suppose we have m {\displaystyle m} vectors of n {\displaystyle n} coordinates, with m < n . {\displaystyle m<n.} Then A

5104-430: The difference of the right and the left hand side is 1 2 ∑ i = 1 n ∑ j = 1 n ( u i v j − u j v i ) 2 ≥ 0 {\displaystyle {\tfrac {1}{2}}\sum _{i=1}^{n}\sum _{j=1}^{n}(u_{i}v_{j}-u_{j}v_{i})^{2}\geq 0} or by considering

5192-1109: The domain and/or range are replaced by a C*-algebra or W*-algebra . An inner product can be used to define a positive linear functional . For example, given a Hilbert space L 2 ( m ) , m {\displaystyle L^{2}(m),m} being a finite measure, the standard inner product gives rise to a positive functional φ {\displaystyle \varphi } by φ ( g ) = ⟨ g , 1 ⟩ . {\displaystyle \varphi (g)=\langle g,1\rangle .} Conversely, every positive linear functional φ {\displaystyle \varphi } on L 2 ( m ) {\displaystyle L^{2}(m)} can be used to define an inner product ⟨ f , g ⟩ φ := φ ( g ∗ f ) , {\displaystyle \langle f,g\rangle _{\varphi }:=\varphi \left(g^{*}f\right),} where g ∗ {\displaystyle g^{*}}

5280-1142: The equality case, notice that Δ = 0 {\displaystyle \Delta =0} happens if and only if p ( t ) = ( t ‖ u ‖ + ‖ v ‖ ) 2 . {\displaystyle p(t)={\bigl (}t\Vert \mathbf {u} \Vert +\Vert \mathbf {v} \Vert {\bigr )}^{2}.} If t 0 = − ‖ v ‖ / ‖ u ‖ , {\displaystyle t_{0}=-\Vert \mathbf {v} \Vert /\Vert \mathbf {u} \Vert ,} then p ( t 0 ) = ⟨ t 0 α u + v , t 0 α u + v ⟩ = 0 , {\displaystyle p(t_{0})=\langle t_{0}\alpha \mathbf {u} +\mathbf {v} ,t_{0}\alpha \mathbf {u} +\mathbf {v} \rangle =0,} and hence v = − t 0 α u . {\displaystyle \mathbf {v} =-t_{0}\alpha \mathbf {u} .} Various generalizations of

5368-581: The expectation of their product, ⟨ X , Y ⟩ := E ⁡ ( X Y ) , {\displaystyle \langle X,Y\rangle :=\operatorname {E} (XY),} the Cauchy–Schwarz inequality becomes | E ⁡ ( X Y ) | 2 ≤ E ⁡ ( X 2 ) E ⁡ ( Y 2 ) . {\displaystyle {\bigl |}\operatorname {E} (XY){\bigr |}^{2}\leq \operatorname {E} (X^{2})\operatorname {E} (Y^{2}).} To prove

SECTION 60

#1732780618308

5456-476: The fact that n {\displaystyle n} vectors in R n {\displaystyle \mathbb {R} ^{n}} are linearly independent if and only if the determinant of the matrix formed by taking the vectors as its columns is non-zero. In this case, the matrix formed by the vectors is We may write a linear combination of the columns as We are interested in whether A Λ = 0 for some nonzero vector Λ. This depends on

5544-417: The family is said to be linearly dependent . A set of vectors which is linearly independent and spans some vector space, forms a basis for that vector space. For example, the vector space of all polynomials in x over the reals has the (infinite) subset {1, x , x , ...} as a basis. A person describing the location of a certain place might say, "It is 3 miles north and 4 miles east of here." This

5632-712: The following quadratic polynomial in x {\displaystyle x} ( u 1 x + v 1 ) 2 + ⋯ + ( u n x + v n ) 2 = ( ∑ i u i 2 ) x 2 + 2 ( ∑ i u i v i ) x + ∑ i v i 2 . {\displaystyle (u_{1}x+v_{1})^{2}+\cdots +(u_{n}x+v_{n})^{2}={\biggl (}\sum _{i}u_{i}^{2}{\biggr )}x^{2}+2{\biggl (}\sum _{i}u_{i}v_{i}{\biggr )}x+\sum _{i}v_{i}^{2}.} Since

5720-490: The following is true: If u = 0 {\displaystyle \mathbf {u} =\mathbf {0} } then by setting c := 0 {\displaystyle c:=0} we have c v = 0 v = 0 = u {\displaystyle c\mathbf {v} =0\mathbf {v} =\mathbf {0} =\mathbf {u} } (this equality holds no matter what the value of v {\displaystyle \mathbf {v} } is), which shows that (1)

5808-401: The following theorem: Cauchy–Schwarz inequality  —  Let u {\displaystyle \mathbf {u} } and v {\displaystyle \mathbf {v} } be arbitrary vectors in an inner product space over the scalar field F , {\displaystyle \mathbb {F} ,} where F {\displaystyle \mathbb {F} }

5896-654: The full list of equations must also be true of the reduced list. In fact, if ⟨ i 1 ,..., i m ⟩ is any list of m {\displaystyle m} rows, then the equation must be true for those rows. Furthermore, the reverse is true. That is, we can test whether the m {\displaystyle m} vectors are linearly dependent by testing whether for all possible lists of m {\displaystyle m} rows. (In case m = n {\displaystyle m=n} , this requires only one determinant, as above. If m > n {\displaystyle m>n} , then it

5984-459: The function p : R → R {\displaystyle p:\mathbb {R} \to \mathbb {R} } defined by p ( t ) = ⟨ t α u + v , t α u + v ⟩ {\displaystyle p(t)=\langle t\alpha \mathbf {u} +\mathbf {v} ,t\alpha \mathbf {u} +\mathbf {v} \rangle } , where α {\displaystyle \alpha }

6072-1856: The inner product is positive-definite, p ( t ) {\displaystyle p(t)} only takes non-negative real values. On the other hand, p ( t ) {\displaystyle p(t)} can be expanded using the bilinearity of the inner product: p ( t ) = ⟨ t α u , t α u ⟩ + ⟨ t α u , v ⟩ + ⟨ v , t α u ⟩ + ⟨ v , v ⟩ = t α t α ¯ ⟨ u , u ⟩ + t α ⟨ u , v ⟩ + t α ¯ ⟨ v , u ⟩ + ⟨ v , v ⟩ = ‖ u ‖ 2 t 2 + 2 | ⟨ u , v ⟩ | t + ‖ v ‖ 2 {\displaystyle {\begin{aligned}p(t)&=\langle t\alpha \mathbf {u} ,t\alpha \mathbf {u} \rangle +\langle t\alpha \mathbf {u} ,\mathbf {v} \rangle +\langle \mathbf {v} ,t\alpha \mathbf {u} \rangle +\langle \mathbf {v} ,\mathbf {v} \rangle \\&=t\alpha t{\overline {\alpha }}\langle \mathbf {u} ,\mathbf {u} \rangle +t\alpha \langle \mathbf {u} ,\mathbf {v} \rangle +t{\overline {\alpha }}\langle \mathbf {v} ,\mathbf {u} \rangle +\langle \mathbf {v} ,\mathbf {v} \rangle \\&=\lVert \mathbf {u} \rVert ^{2}t^{2}+2|\langle \mathbf {u} ,\mathbf {v} \rangle |t+\lVert \mathbf {v} \rVert ^{2}\end{aligned}}} Thus, p {\displaystyle p}

6160-596: The inner product itself. The Cauchy–Schwarz inequality allows one to extend the notion of "angle between two vectors" to any real inner-product space by defining: cos ⁡ θ u v = ⟨ u , v ⟩ ‖ u ‖ ‖ v ‖ . {\displaystyle \cos \theta _{\mathbf {u} \mathbf {v} }={\frac {\langle \mathbf {u} ,\mathbf {v} \rangle }{\|\mathbf {u} \|\|\mathbf {v} \|}}.} The Cauchy–Schwarz inequality proves that this definition

6248-510: The inner product on the vector space C n {\displaystyle \mathbb {C} ^{n}} is the canonical complex inner product (defined by ⟨ u , v ⟩ := u 1 v 1 ¯ + ⋯ + u n v n ¯ , {\displaystyle \langle \mathbf {u} ,\mathbf {v} \rangle :=u_{1}{\overline {v_{1}}}+\cdots +u_{n}{\overline {v_{n}}},} where

6336-748: The inner product space of square-integrable complex-valued functions , the following inequality holds. | ∫ R n f ( x ) g ( x ) ¯ d x | 2 ≤ ∫ R n | f ( x ) | 2 d x ∫ R n | g ( x ) | 2 d x . {\displaystyle \left|\int _{\mathbb {R} ^{n}}f(x){\overline {g(x)}}\,dx\right|^{2}\leq \int _{\mathbb {R} ^{n}}{\bigl |}f(x){\bigr |}^{2}\,dx\int _{\mathbb {R} ^{n}}{\bigl |}g(x){\bigr |}^{2}\,dx.} The Hölder inequality

6424-1403: The latter polynomial is nonnegative, it has at most one real root, hence its discriminant is less than or equal to zero. That is, ( ∑ i u i v i ) 2 − ( ∑ i u i 2 ) ( ∑ i v i 2 ) ≤ 0. {\displaystyle {\biggl (}\sum _{i}u_{i}v_{i}{\biggr )}^{2}-{\biggl (}\sum _{i}{u_{i}^{2}}{\biggr )}{\biggl (}\sum _{i}{v_{i}^{2}}{\biggr )}\leq 0.} If u , v ∈ C n {\displaystyle \mathbf {u} ,\mathbf {v} \in \mathbb {C} ^{n}} with u = ( u 1 , … , u n ) {\displaystyle \mathbf {u} =(u_{1},\ldots ,u_{n})} and v = ( v 1 , … , v n ) {\displaystyle \mathbf {v} =(v_{1},\ldots ,v_{n})} (where u 1 , … , u n ∈ C {\displaystyle u_{1},\ldots ,u_{n}\in \mathbb {C} } and v 1 , … , v n ∈ C {\displaystyle v_{1},\ldots ,v_{n}\in \mathbb {C} } ) and if

6512-1035: The linearity of the inner product in its first argument that: ⟨ z , v ⟩ = ⟨ u − ⟨ u , v ⟩ ⟨ v , v ⟩ v , v ⟩ = ⟨ u , v ⟩ − ⟨ u , v ⟩ ⟨ v , v ⟩ ⟨ v , v ⟩ = 0. {\displaystyle \langle \mathbf {z} ,\mathbf {v} \rangle =\left\langle \mathbf {u} -{\frac {\langle \mathbf {u} ,\mathbf {v} \rangle }{\langle \mathbf {v} ,\mathbf {v} \rangle }}\mathbf {v} ,\mathbf {v} \right\rangle =\langle \mathbf {u} ,\mathbf {v} \rangle -{\frac {\langle \mathbf {u} ,\mathbf {v} \rangle }{\langle \mathbf {v} ,\mathbf {v} \rangle }}\langle \mathbf {v} ,\mathbf {v} \rangle =0.} Therefore, z {\displaystyle \mathbf {z} }

6600-549: The matrix equation, Row reduce this equation to obtain, Rearrange to solve for v 3 and obtain, This equation is easily solved to define non-zero a i , where a 3 {\displaystyle a_{3}} can be chosen arbitrarily. Thus, the vectors v 1 , v 2 , {\displaystyle \mathbf {v} _{1},\mathbf {v} _{2},} and v 3 {\displaystyle \mathbf {v} _{3}} are linearly dependent. An alternative method relies on

6688-448: The modern proof of the integral version. The Cauchy–Schwarz inequality states that for all vectors u {\displaystyle \mathbf {u} } and v {\displaystyle \mathbf {v} } of an inner product space where ⟨ ⋅ , ⋅ ⟩ {\displaystyle \langle \cdot ,\cdot \rangle } is the inner product . Examples of inner products include

6776-559: The others. A sequence of vectors v 1 , v 2 , … , v n {\displaystyle \mathbf {v} _{1},\mathbf {v} _{2},\dots ,\mathbf {v} _{n}} is said to be linearly independent if it is not linearly dependent, that is, if the equation can only be satisfied by a i = 0 {\displaystyle a_{i}=0} for i = 1 , … , n . {\displaystyle i=1,\dots ,n.} This implies that no vector in

6864-712: The proof of the Equality Characterization given above; that is, it proves that if u {\displaystyle \mathbf {u} } and v {\displaystyle \mathbf {v} } are linearly dependent then | ⟨ u , v ⟩ | = ‖ u ‖ ‖ v ‖ . {\displaystyle {\bigl |}\langle \mathbf {u} ,\mathbf {v} \rangle {\bigr |}=\|\mathbf {u} \|\|\mathbf {v} \|.} By definition, u {\displaystyle \mathbf {u} } and v {\displaystyle \mathbf {v} } are linearly dependent if and only if one

6952-399: The proofs given below, the proof in the trivial case where at least one of the vectors is zero (or equivalently, in the case where ‖ u ‖ ‖ v ‖ = 0 {\displaystyle \|\mathbf {u} \|\|\mathbf {v} \|=0} ) is the same. It is presented immediately below only once to reduce repetition. It also includes the easy part of

7040-707: The real and complex dot product ; see the examples in inner product . Every inner product gives rise to a Euclidean ℓ 2 {\displaystyle \ell _{2}} norm , called the canonical or induced norm , where the norm of a vector u {\displaystyle \mathbf {u} } is denoted and defined by ‖ u ‖ := ⟨ u , u ⟩ , {\displaystyle \|\mathbf {u} \|:={\sqrt {\langle \mathbf {u} ,\mathbf {u} \rangle }},} where ⟨ u , u ⟩ {\displaystyle \langle \mathbf {u} ,\mathbf {u} \rangle }

7128-437: The sequence can be represented as a linear combination of the remaining vectors in the sequence. In other words, a sequence of vectors is linearly independent if the only representation of 0 {\displaystyle \mathbf {0} } as a linear combination of its vectors is the trivial representation in which all the scalars a i {\textstyle a_{i}} are zero. Even more concisely,

7216-433: The special case where the sequence of v 1 , … , v k {\displaystyle \mathbf {v} _{1},\dots ,\mathbf {v} _{k}} has length 1 {\displaystyle 1} (i.e. the case where k = 1 {\displaystyle k=1} ). A collection of vectors that consists of exactly one vector is linearly dependent if and only if that vector

7304-475: The square root. Moreover, if the relation ≥ {\displaystyle \geq } in the above expression is actually an equality, then ‖ z ‖ 2 = 0 {\displaystyle \|\mathbf {z} \|^{2}=0} and hence z = 0 ; {\displaystyle \mathbf {z} =\mathbf {0} ;} the definition of z {\displaystyle \mathbf {z} } then establishes

7392-717: The standard inner product, which is the dot product , the Cauchy–Schwarz inequality becomes: ( ∑ i = 1 n u i v i ) 2 ≤ ( ∑ i = 1 n u i 2 ) ( ∑ i = 1 n v i 2 ) . {\displaystyle {\biggl (}\sum _{i=1}^{n}u_{i}v_{i}{\biggr )}^{2}\leq {\biggl (}\sum _{i=1}^{n}u_{i}^{2}{\biggr )}{\biggl (}\sum _{i=1}^{n}v_{i}^{2}{\biggr )}.} The Cauchy–Schwarz inequality can be proved using only elementary algebra in this case by observing that

7480-449: The three vectors are linearly dependent. Two vectors: Now consider the linear dependence of the two vectors v 1 = ( 1 , 1 ) {\displaystyle \mathbf {v} _{1}=(1,1)} and v 2 = ( − 3 , 2 ) , {\displaystyle \mathbf {v} _{2}=(-3,2),} and check, or The same row reduction presented above yields, This shows that

7568-414: The triangle inequality: ‖ u + v ‖ ≤ ‖ u ‖ + ‖ v ‖ . {\displaystyle \|\mathbf {u} +\mathbf {v} \|\leq \|\mathbf {u} \|+\|\mathbf {v} \|.} The Cauchy–Schwarz inequality is used to prove that the inner product is a continuous function with respect to the topology induced by

7656-453: The vector ( u 1 , u 2 ) {\displaystyle \left(u_{1},u_{2}\right)} is in the same or opposite direction as the vector ( v 1 , v 2 ) {\displaystyle \left(v_{1},v_{2}\right)} , or if one of them is the zero vector. In Euclidean space R n {\displaystyle \mathbb {R} ^{n}} with

7744-403: The vectors that equals the zero vector. If such a linear combination exists, then the vectors are said to be linearly dependent . These concepts are central to the definition of dimension . A vector space can be of finite dimension or infinite dimension depending on the maximum number of linearly independent vectors. The definition of linear dependence and the ability to determine whether

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