Objection to projection is a basis for dejection!
The key to what follows is the following idea. Recall (from class and §1.2) that the
component of a vector V parallel to a (non-zero) vector u is the projection of V onto u:
u – V
prOJu(V) = u
u – u
Recall further that if you take away the component of V which is parallel to 11 away from
V, the component that is left is orthogonal to u.
1. Suppose V and u ∃≟ 0 are vectors of the same dimension. Verify that V – proju(V) is
orthogonal to u.
Hint: Use the dot product!
SOLUTION. Following the hint, we will verify that V – proju(v) is orthogonal to u by
taking the dot product of the two vectors:
u – V n – V
(V-pI‘OJu(V))-u= <V- -u=V-u- <-)u-u=V-u-u-V=O
u – u u – u
(The last bit works since the dot product is commutative.)
Since their dot product is zero, V – proju(V) is orthogonal to u. I
1 1 O 1
1 1 0 -1
Now let S ∶ −↕ , 1 , 1 , 1 be a set of four vectors 1n 4-d1mens1onal
1 1 1 1
space. We will modify this set of vectors to make it nicer in some respects.
2. Use the idea in 1 to modify the second vector in S to make it orthogonal to the first
vector in S.
1 1
SOLUTION. We will apply the formula in 1 with u ∶ and V ∶ to get a new
1 1
vector V’ which is orthogonal to u.
1 1
1 ∙ 1
1 -1 1 1
<u-V)_1_ 1 1 1
v-v prOJuV-V -u.u u- 1 -1 1 _1
1 1 ∙ 1 1
-1 -1
1 1
1 1 1 1 l
1 _ 1-1+1-1+(-1)-1+1-1 1 _ 1 −≧ 1 _ g
1 1-1+1-1+(-1)-(-1)+1-1 -1 − 1 4 -1 − g
1 1 1 1
3. Modify the third vector in S to make it orthogonal to both the first and second vectors
in S.
1 ⋛ 0
SOLUTION. We Will adapt the formula in 1 With u ∶ _11 , V’ ∶ g , and W ∶ (1)
1 ≣ 1
to get a new vector W’ which is orthogonal to both 11 and v’.
w – proju<w> – pros/(w) = w – (2)11 – v’
u – u v -v
1 0 ⋛ 0
1 ∙ 0 g ∙ 0
0 -1 1 1 g 1 l
0 _ 1 1 1 _ 5 1
1 1 1 -1 ≣↕ ≣↕ g
1 1 1 1
1 ∎ _1 ≣ ∎ g
i i
1 1
(Skipping sorne arithrnetic
0 1
9 1 _ a
1 4 -1 3 ⋚ − 0
1 1 5
We leave it ti the reader to check that W’ is indeed orthogonal to u and v’. (As well as
the original v!) I
4. Modify the fourth vector in S to make it orthogonal to all of the first three vectors in
S. [1]
1 1
SOLUTION. We will adapt the formula in 1 with u ∶ _1 , v’ ∶ g , W’ ∶ 03
1 5
and x ∶ to get a new vector x’ which is orthogonal to u, v’, and W’.
x’ = x – proju<x> – projv/ (x) – projw/ (x)
(11 – X) v’ – X , W’ – X
x – – u – 7 V – – W
11 – u v’ – v’ W’ – W’
1 1 l 1
1 ∙ 1 g ∙ 1
1 -1 1 1 g 1
1 − fl -1
1 1 1 1 i i
1 ∎ _1
i i
1 1 1 5
l: −≣ l [-1l
O3 . 1 1
2 1‘
j 2
3 . 3 3
0 0
2 Z
3 3
(Skipping more arithmetic
1 1 l −≟ 1
-1_91
1 4 -1 3 – ⊇ 0 − 0
1 1 ≷ ⇇⊰ ⊋ 0
We leave it ti the reader to check that X’ is indeed orthogonal to u, v’, and W’. (As well
as the original v and W!) I
NOTE A. It is important in the solutions to questions 3 and 4 given above to work with
the modified vectors v’ and, in 4, W’ . Working with the unmodified v and W would require
correcting for the fact that these might not be orthogonal to 11 (though W actually is), or
to each other. The difficulty is that if that they are not orthogonal, projections onto them
will overlap and interfere with each other.
5. Further modify all of your modified vectors from 274 to have length one. [1
SOLUTION. We will multiply each of u, v’, W’, and X’, in the notation of the solution
above, by the reciprocal of its length to get new vectors of length one that we’ll call a, b,
c, and 01, respectively. Note that we can reuse some numbers obtained in the solutions to
questions 2 through 4.
1 1
1 1 1 -1 1 -1
a ∶ -u ∶ -u
llull √−⊔⋅⊔⇂ √≩ 1 2 1
1 1
b 1 , 1 , 1
-V ∶ -V
-V’ √≣ g
−≟
1 , 1 , 1
-/W ∶ -W
“W \/ W’ – w’ ∖∕≣ (2) √≣ (2)
1 1 1 -1
dZ-X/Z-Z- Omittin abitof…:-
filo] ≺ g
Those who want to do more arithmetic may take the multipliers into the vectors, coordinate
by coordinate, or collect common factors of the coordinates into the multipliers. I
NOTE B. The procedure used in the solutions to 2*5 is the Gram-Schmidt process for
modifying a set of basis vectors into an orthonormal set of basis vectors. It is described in
detail in Chapter 5 of the text.
6. What might your final collection of modified vectors from 5 be good for?
SOLUTION. The vectors a, b, c, and d are unit vectors in four-dimensional space which
are mutually perpendicular. Like the vectors i, j, and k in three-dimensional space, they
can serve as the basis for a coordinate system: they provide directions for the axes, along
With a reference length of one. I
The key to what follows is the following idea. Recall (from class and §1.2) that the
component of a vector V parallel to a (non-zero) vector u is the projection of V onto u:
u – V
pI‘OJu(V) ∶ u
u – u
Recall further that if you take away the component of V which is parallel to 11 away from
V, the component that is left is orthogonal to u.
1. Suppose V and u ∃≟ 0 are vectors of the same dimension. Verify that V – proju(V) is
orthogonal to u.
Hint: Use the dot product!
1 1 O 1
1 1 0 -1
Now let S ∶ −↕ , 1 , 1 , 1 be a set of four vectors 1n 4-d1mens1onal
1 1 1 1
space. We will modify this set of vectors to make it nicer in some respects.
2. Use the idea in 1 to modify the second vector in S to make it orthogonal to the first
vector in S.
3. Modify the third vector in S to make it orthogonal to both the first and second vectors
in S.
4. Modify the fourth vector in S to make it orthogonal to all of the first three vectors in
S. [1
5. Further modify all of your modified vectors from 274 to have length one. [1
6. What might your final collection of modified vectors from 5 be good for?
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