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Projector And Projection Onto A Line
Numerical Linear Algebra
Isaac Amornortey Yowetu
NIMS-GHANA
July 10, 2020
Projection onto a Line Projection Matrix
Outline
1 Projection onto a Line
2 Projection Matrix
Projection onto a Line Projection Matrix
Graphical Example
Figure: A Graphical Example of a projection onto a line
Projection onto a Line Projection Matrix
Orthogonal Projection onto a line
Suppose we have L = span{u} is a line and v is projected onto
the line u, then ∃c ∈ R s.t:
ProjL(v) = c · u
v − ProjL(v) = v − c · u
We can obtain our orthogonal projection to be
(v − ProjL(v)) · (ProjL(v)) = (v − c · u) · (c · u) = 0 (1)
v(c · u) − (c · u)(c · u) = 0 (2)
(c · u)v = (c · u)(c · u) (3)
u · v = c(u · u) (4)
Projection onto a Line Projection Matrix
c =
u · v
u · u
Therefore
projL(v) =
u · v
u · u
u
Projection onto a Line Projection Matrix
Example
Let
v =


−2
3
−1

 and u =


−1
1
1


and let L be the line spanned by u. Compute projL(v) and
proj⊥
L (v)
Projection onto a Line Projection Matrix
Solution
projL(v) = (2)+(3)+(−1)
(1)+(1)+(1)


−1
1
1

 = 4
3


−1
1
1


proj⊥
L (v) = v − projL(v) =


−2
3
−1

 − 4
3


−1
1
1

 = 1
3


−2
5
−7


Projection onto a Line Projection Matrix
Projection Matrix
Here we would like to find a projection matrix P:
P = P2
and PT
= P
P =
uuT
uT u
projL(v) = Pv =
uuT
uT u
v
proj⊥
L (v) = v − Pv = v −
uuT
uT u
v
Projection onto a Line Projection Matrix
Example
Let
v =


−2
3
−1

 and u =


−1
1
1


and let L be the line spanned by u. Compute projL(v) and
proj⊥
L (v)
Projection onto a Line Projection Matrix
Solution
P =
uuT
uT u
u · uT
=


−1
1
1

 (−1 1 1) =


1 −1 −1
−1 1 1
−1 1 1


uT
· u = (−1 1 1)


−1
1
1

 = 3
∴ P =
1
3


1 −1 −1
−1 1 1
−1 1 1

 =


0.333 −0.333 −0.333
−0.333 0.333 0.333
−0.333 0.333 0.333


Projection onto a Line Projection Matrix
projl (v) = Pv =


0.333 −0.333 −0.333
−0.333 0.333 0.333
−0.333 0.333 0.333




−2
3
−1


=
4
3


−1
1
1


proj⊥
L (v) = v − projL(v) =


−2
3
−1

 − 4
3


−1
1
1

 = 1
3


−2
5
−7



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Projector Projection Line

  • 1. Projector And Projection Onto A Line Numerical Linear Algebra Isaac Amornortey Yowetu NIMS-GHANA July 10, 2020
  • 2. Projection onto a Line Projection Matrix Outline 1 Projection onto a Line 2 Projection Matrix
  • 3. Projection onto a Line Projection Matrix Graphical Example Figure: A Graphical Example of a projection onto a line
  • 4. Projection onto a Line Projection Matrix Orthogonal Projection onto a line Suppose we have L = span{u} is a line and v is projected onto the line u, then ∃c ∈ R s.t: ProjL(v) = c · u v − ProjL(v) = v − c · u We can obtain our orthogonal projection to be (v − ProjL(v)) · (ProjL(v)) = (v − c · u) · (c · u) = 0 (1) v(c · u) − (c · u)(c · u) = 0 (2) (c · u)v = (c · u)(c · u) (3) u · v = c(u · u) (4)
  • 5. Projection onto a Line Projection Matrix c = u · v u · u Therefore projL(v) = u · v u · u u
  • 6. Projection onto a Line Projection Matrix Example Let v =   −2 3 −1   and u =   −1 1 1   and let L be the line spanned by u. Compute projL(v) and proj⊥ L (v)
  • 7. Projection onto a Line Projection Matrix Solution projL(v) = (2)+(3)+(−1) (1)+(1)+(1)   −1 1 1   = 4 3   −1 1 1   proj⊥ L (v) = v − projL(v) =   −2 3 −1   − 4 3   −1 1 1   = 1 3   −2 5 −7  
  • 8. Projection onto a Line Projection Matrix Projection Matrix Here we would like to find a projection matrix P: P = P2 and PT = P P = uuT uT u projL(v) = Pv = uuT uT u v proj⊥ L (v) = v − Pv = v − uuT uT u v
  • 9. Projection onto a Line Projection Matrix Example Let v =   −2 3 −1   and u =   −1 1 1   and let L be the line spanned by u. Compute projL(v) and proj⊥ L (v)
  • 10. Projection onto a Line Projection Matrix Solution P = uuT uT u u · uT =   −1 1 1   (−1 1 1) =   1 −1 −1 −1 1 1 −1 1 1   uT · u = (−1 1 1)   −1 1 1   = 3 ∴ P = 1 3   1 −1 −1 −1 1 1 −1 1 1   =   0.333 −0.333 −0.333 −0.333 0.333 0.333 −0.333 0.333 0.333  
  • 11. Projection onto a Line Projection Matrix projl (v) = Pv =   0.333 −0.333 −0.333 −0.333 0.333 0.333 −0.333 0.333 0.333     −2 3 −1   = 4 3   −1 1 1   proj⊥ L (v) = v − projL(v) =   −2 3 −1   − 4 3   −1 1 1   = 1 3   −2 5 −7  