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Orthogonality for Subspaces and Bases
Recall that x ⊥ y if and only if x · y = 0, so it’s just a dot product based definition
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Definition: The sets P ∈ Rn and Q ∈ Rn are orthogonal if
p · q for all p ∈ P and all q ∈ Q,
so all of P is orthogonal to all of Q (vice-versa, etc)
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The subspaces U and W of V are U ⊥ W if all
u ∈ U and all w ∈ W have u · w = 0
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The spans of




−1
0
1
those sets will also be orthogonal to each other
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Property: The orthogonal complement of a subspace is also a subspace
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• 0 ∈ V has 0 · u for all u ∈ U , so it’s in U ⊥
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(x + y) · u = x · u + y · u = 0 + 0 = 0

for any u ∈ U
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• Take x ∈ U ⊥ and a ∈ R
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all three conditions hold, it’s a subspace
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Property: The intersection between U and U ⊥ , U

∩

U ⊥ is just the zero vector
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Example: In R3 , here we have two subspaces:
 

 x

U =  y  | x − 2y − 3z = 0


z
1

  

1


V = a 1  , a ∈ R ,


2

What are their complements?
For  , it’ll the set of all vectors in R3 orthogonal to the plane itself, so the subspace
be
U
1 

Span  −2 
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In R3 , the orthogonal complement of a line is a plane, the complement of a plane is a line
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If you want an orthogonal complement of the span of two vectors,
you’d want the span of a set of vectors orthogonal to those two
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I
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Property: If U = Span {u1 , u2 ,
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Corollary: If U = Span {u1 , u2 ,
...

then
U ⊥ = {x ∈ Rn such that Ax = 0}
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

...


1



−1
Example: If U = Span 
 −2


3

Row(A)⊥ = Null(A), and (using a bunch of transposes)



−1
−1
 3  2


 0  1
−2
−2


1 

  0 
⊥


  −3  then find a basis for U
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Doing so makes the
orthogonal complement equal to the null space of the matrix
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2

How To Solve for All At Once,

OPTIONAL

The annoyance of the previous section is that our usual reduction of A yields a basis of
the row space, column space (taking the original vectors in the pivot columns) and, with
some effort, the null space
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We’re missing the complement of the
column space
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Take your matrix A and put it into an augmented matrix of the form [A|I] with I the
identity matrix with the same height
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Row reduce the A side until in RREF
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3
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(b) The basis of Col(A) will be the original A columns corresponding to the pivot
columns
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(d) The basis of Null(AT ), of Col(A)⊥ , will be the row vectors on the right corresponding to zero rows on the left
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1 −1/2 0 0 0 1
0 0 0
1 −1 1
There are several ways to analyze this algorithm
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The justification for the right hand side is that we are actually solving it for an arbitrary
right hand side, with each column on the RHS representing a different coefficient of the b
vector
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1
−1/2
0
0
0
1
b3
Once row reduced on the left, each zero row represents a row that COULD cause a contradiction
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So,
the column space of A is the orthogonal complement to that one vector, so


⊥

1 
1 


Col(A) = Span  −1 
=⇒ Col(A)⊥ = Span  −1 




1
1
using the (U ⊥ )⊥ = U equality
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  
 1
Example: The set  0  

0

is orthogonal if every pair of different vectors in the set is
     
  
0
0 
0
0 
 1
  0  is orthogonal but  0   1   1  is not
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Proof: This is surprisingly easy
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, vn }
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What do we do? We inner product v1 on each side for
a1 v1 · v1 + a2 v2 · v1 + · · · + an vn · v1 = 0 · v1
which works out to be
a 1 v1 · v1 = 0
which makes a1 = 0 since v1 ̸= 0 which makes v1 · v1 ̸= 0
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A set of n non-zero orthogonal vectors is automatically a basis of Rn , of course
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Note too that the basic unit vectors of Rn form an orthogonal basis
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Recall that when x ∈ Span of the
basis {v1 , v2 ,
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Having the v form an orthogonal basis makes this even easier
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We start with the above equation and use v1 (as before) for
x · v1 = a1 v1 · v1 + a2 v2 · v1 + · · · + an vn · v1
= a1 v1 · v1
x·v
so a1 = v1 ·v11
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We can get
this for ANY of the vectors and coefficients:
x · vi
i ∈ 1
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ai =
vi · vi

which works out to
x = Projv1 (x) + Projv2 (x) + · · · + Projvn (x)
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1
0
2
1
4

They’re orthogonal, and so linearly independent, so they span R2
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0
−1
2
1
The helpful thing here is that we can actually expand this into projections, projections
on to subspaces (U ) and orthogonal complements (U ⊥ )
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with x1 · x2 = 0, and so on
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, vm } then
ProjV (x) = Projv1 (x) + Projv2 (x) + · · · + Projvm (x)
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Recall the section about projections
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We use the same
principle
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Recall that V is spanned by the v vectors
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So:
(
)
Projv1 (x) + Projv2 (x) + · · · + Projvm (x) − x · vi
= Projvi (x) · vi − x · vi
(
)
vi · x
vi · vi − x · vi
=
vi · vi
vi · x
vi · vi − x · vi
=
vi · vi
= x · vi − x · vi = 0
and done
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5: 2
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b), 6
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Section 4
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df),2, 9

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