MATH 323 Lecture 19

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Norm

x→∈ℝn

‖x→‖=∑i=1nxi2, where ‖x→‖ is notation for the norm or length of x→.

‖x→‖=‖−x→‖

Distance between two points/vectors: ‖x→−y→‖

‖x→‖2=x→⋅x→=∑i=1nxi2

Theorem 5.1.1

If x→,y→∈ℝn and θ is the angle between them, then

x→⋅y→=‖x→‖‖y→‖cos⁡θ

Law of Cosines

Triangle defined by x→, y→, and x→−y→

‖y→−x→‖2=‖x→‖2+‖y→‖2−2‖x→‖‖y→‖cos⁡θ

The above can be simplified to the form

x→⋅y→=‖x→‖‖y→‖cos⁡θ


Unit Vector

u→ is a unit vector iff ‖u→‖=1

Unit vector in direction of x→ is given by

u→=x^=x→‖x→‖

Example

x→=⟨3,4⟩y→=⟨−1,7⟩u→=15⟨3,4⟩=⟨35,45⟩v→=152⟨−1,7⟩=⟨−152,752⟩cos⁡θ=u→⋅v→=…=12θ=pi4

Cauchy-Schwartz Inequality

|x→⋅y→|≤‖x‖‖y‖⟺−1≤x→⋅y→‖x→‖‖y→‖≤1

Equality iff one of the vectors is zero or one of the vectors is a multiple of another (θ=0,π)


Orthogonality

x→ and y→ are orthogonal (written x→⊥y→ if θ=π2,3π2⟺x→⋅y→=0

Scalar and Vector Projections

Given two noncolinear vectors x→ and y→

Let u→=y→‖y→‖ be a unit vector in the direction of y→ and p→=αu→ be the vector projection of x→ in the direction of y→, where α is the scalar projection of x→ in the direction of y→

‖p→‖=α=‖x→‖cos⁡θ=‖x→‖‖y→‖cos⁡θ‖y→‖=x→⋅y→‖y→‖

Exercise

Given a point P(1,4) and a line y=13x, find the point on the line closest to P.

Find vector w→ on the line: w→=⟨3,1⟩

Take vector projection of v→=P−O onto w→:

p→=v→⋅w→‖w→‖⋅w→=⟨2.1,0.7⟩


Planes in 3D Space

(See MATH 251 Lecture 5#Planes→)


Generalization of Pythagorean Theorem

Given two orthogonal vectors x→ and y→ in ℝn,

‖x→±y→‖2=(x→±y→)⋅(x→±y→)=x→⋅x→±2y→⋅x→+y→⋅y→=‖x→‖2+‖y→‖2


Orthogonal Subspaces

Two subspaces X and Y in ℝn are said to be orthogonal if x→⋅y→=0 for all x→∈X and y→∈Y.

A is m×n matrix

N(A) is subspace, x→∈ℝn, Ax→=0 iff ai1x1+ai2x2+…+ain+xn=0 for i=1,…,m.

This means that x→ is perpendicular to the ith row of A

...


Orthogonal Complement

Y⊂ℝn

Orthogonal complement given by

Y⊥={x→∈ℝn∣x→⋅y→=0∀y→∈y→}


Example

Let Y=Span(e→1) be the x-axis.

Y⊥=Span(e→2,e→3) is the yz-plane

Theorem

  1. If X⊥Y, then X∩Y={0→}
  2. If Y is a subspace of ℝn, then Y⊥ is also a subspace of ℝn

Proof

  1. Proof by contradiction. X∩Y≠{0→}⟶(x→∈X∩Y)⋅(x→∈X∩Y)=‖x‖2≠0 — CONTRADICTION!
  2. u→,v→∈Y⊥, so u→⋅y→=v→⋅y→=0∀y→∈Y. (αu→+βv→)⋅y→=0, so subspace defined by αu→+βv→∈Y⊥


Fundamental Subspaces

m×n matrix A

Ax→=b→ for some x→∈ℝn iff b→ is in col space of A.

We can write A as a linear transformation LA:ℝn→ℝn.

The column space of A is also called the range of A:

R(A)={b→∈ℝm∣b→=Ax→∃x→∈ℝn}

Range of transpose matrix

R(AT)={y→∈ℝn∣y→=ATx→∃x→∈ℝm}

Thus R(A)⊂ℝm, and R(AT)⊂ℝn, and

R(AT)⊥N(A)

Theorem 5.2.1

Fundamental Subspace Theorem

N(A)=R(AT)⊥

N(AT)=R(A)⊥