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The projection of a vector
v
∈
V
{\displaystyle v\in V}
W
∈
V
{\displaystyle W\in V}
W . In other words, we need to minimize the distance between vector v, and an arbitrary vector
w
∈
W
{\displaystyle w\in W}
‖
w
−
v
‖
2
=
‖
W
^
a
^
−
v
‖
2
{\displaystyle \|w-v\|^{2}=\|{\hat {W}}{\hat {a}}-v\|^{2}}
∂
‖
W
^
a
^
−
v
‖
2
∂
a
^
=
∂
⟨
W
^
a
^
−
v
,
W
^
a
^
−
v
⟩
∂
a
^
=
0
{\displaystyle {\frac {\partial \|{\hat {W}}{\hat {a}}-v\|^{2}}{\partial {\hat {a}}}}={\frac {\partial \langle {\hat {W}}{\hat {a}}-v,{\hat {W}}{\hat {a}}-v\rangle }{\partial {\hat {a}}}}=0}
[Projection onto space W]
a
^
=
(
W
^
T
W
^
)
−
1
W
^
T
v
{\displaystyle {\hat {a}}=({\hat {W}}^{T}{\hat {W}})^{-1}{\hat {W}}^{T}v}
For every vector
v
∈
V
{\displaystyle v\in V}
w
∈
W
{\displaystyle w\in W}
W .
w
⊥
=
x
∈
V
:
⟨
x
,
y
⟩
=
0
,
∀
y
∈
W
{\displaystyle w^{\perp }={x\in V:\langle x,y\rangle =0,\forall y\in W}}
The distance between
v
∈
V
{\displaystyle v\in V}
W is given as the minimum distance between v and an arbitrary
w
∈
W
{\displaystyle w\in W}
∂
d
(
v
,
w
)
∂
a
^
=
∂
‖
v
−
W
^
a
^
‖
∂
a
^
=
0
{\displaystyle {\frac {\partial d(v,w)}{\partial {\hat {a}}}}={\frac {\partial \|v-{\hat {W}}{\hat {a}}\|}{\partial {\hat {a}}}}=0}
Given two vector spaces V and W , what is the overlapping area between the two? We define an arbitrary vector z that is a component of both V , and W :
z
=
V
^
a
^
=
W
^
b
^
{\displaystyle z={\hat {V}}{\hat {a}}={\hat {W}}{\hat {b}}}
V
^
a
^
−
W
^
b
^
=
0
{\displaystyle {\hat {V}}{\hat {a}}-{\hat {W}}{\hat {b}}=0}
[
a
^
b
^
]
=
N
(
[
v
^
−
W
^
]
)
{\displaystyle {\begin{bmatrix}{\hat {a}}\\{\hat {b}}\end{bmatrix}}={\mathcal {N}}([{\hat {v}}-{\hat {W}}])}
Where N is the nullspace.
![{\displaystyle {\begin{bmatrix}{\hat {a}}\\{\hat {b}}\end{bmatrix}}={\mathcal {N}}([{\hat {v}}-{\hat {W}}])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b572ac2d3c18faefaaefd33f062094fd7f089b01)