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Assignment 5 Solutions
a. To prove that − −, (^) A is an inner product, we check the axioms:
(because the dot product is an inner product) = ,
A
A
u v Au Av Av Au v u
G G G G
i G G i G G
A
A A
u v w A u v Aw Au Av Aw Au Aw Av Aw u w v w
G G JG G G JG
i G G JG i G JG G JG i i G JG G JG
, A
ru v S ru Sv rSu Sv r Su Sv r u v
G G G G
i G G i G G i G G
2
(the dot product norm) 0
u u (^) A Su Su
Su
G G G G
i G
b. Let 1 2 2
: and :. u v u v u v
= ⎡^ ⎤^ =⎡^ ⎤
G G (^1) Then we see
1 1 2 2 1 2 1 2 1 2 1 2 1 2 1 2
, : cos^ sin^ cos^ sin sin cos sin cos cos sin cos sin sin cos sin cos cos sin cos sin sin
A u v u^ v u v u u v v u u v v u u v v
= ⎜⎛^ ⎡^ −^ ⎤^ ⎡^ ⎤^ ⎞ ⎛⎡^ − ⎤⎡^ ⎤
⎝ ⎣^ ⎦^ ⎣^ ⎦^ ⎠ ⎝⎣^ ⎦⎣^ ⎦
⎝ ⎣^ +^ ⎦^ ⎠ ⎝ ⎣^ + ⎦⎠
G G
i
i
( ) (^) ( )
1 2 1 2 (^2 21 1 2 )
1 2 2 2 2 1 1 2 1 2 1 1 2 2
cos sin cos (cos sin ) sin cos sin cos cos sin cos sin sin cos
u u v v u v u v u v u v u v v u u v u v
= ⎡^ ⎤ ⎡^ ⎤
⎢⎣ ⎥ ⎢⎦ ⎣^ i ⎥⎦
c. (Bonus) This inner product takes the vectors u and v
G G
maps them using the linear transformation represented by A, and then takes the dot product of the resulting vectors. Now, in part b), the matrix represents a rotation. Since the dot product measures angles and lengths and rotations change neither of these, the effect in b) makes sense.
a.
i
⎡^ − ⎤ ⎡ ⎤
i 0
b. 2 2 2
x x x x x x
( )
proj , 0 (^0 1 1 ) 1 1 1 0 1 1
v
u v u v v v
= −^ −^ ⎢^ −⎥
+ ⎢^ ⎥
G
G G
G G
G G
b. (^2 ) 1 0 1 1 1 1 1 0 0
u = u − u ⎡ ⎤ ⎡ ⎤ = ⎢ ⎥^ −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦ = ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦
JJG G JG
a. Suppose that ai is a row of A and
JG
v
G
is such that Av = 0
G G
. Consider a vi.
JG G
i
[ ]
1 1 1 2 2 2 1 2
i i i i i im m m
a v v a v v a v a a a a v v
= ⎢^ ⎥ ⎢^ ⎥^ = ⎢^ ⎥
JG G
i i "
# im
Also, let’s expand out the Av = 0
G G
11 12 1 1 2 1 2
1 2
m
i i im m n n nm
a a a v v a a a v a a a
⎢ ⎥ ⎡^ ⎤^ ⎡^ ⎤
⎢ ⎥ ⎢^ ⎥^ ⎢^ ⎥
⎢ ⎥ ⎢^ ⎥^ =⎢^ ⎥
⎢ ⎥ ⎢^ ⎥^ ⎢^ ⎥
⎢ ⎥ ⎢^ ⎥^ ⎢^ ⎥
⎢ ⎥⎣^ ⎦ ⎣^ ⎦
The calculation of the i’th coordinate in the above equation is
[ ]
1 2 i 1 i 2 im^0 m
v v a a a v
So we see that the row ai
JG
is orthogonal to v
G
. Since the row space is spanned of A and the null space is the collection of all vectors mapped to 0 by A, we see that
G
( Row(^ A )^^ )^ ⊥=Null(^ A ).
b. Using the result from a), to find the orthogonal complement of a collection of vectors, one needs only to put them into a matrix as rows and then find the null space.
1 3 (^2 3 )
2 4 3 3 2
1 2 4 3
" "" 2"" "
1 33"^ "
2
R R RR RR
R R R (^) R R
R R R R
A +↔
−
↔ −
−
⎡ −^ ⎤ ⎡ −^ −
= ⎢^ ⎥^ ⎯⎯⎯⎯→⎢
⎡ −^ −⎤
⎯⎯⎯⎯→ ⎢^ ⎥
12 33
1 3 2"" ""
R RR (^) RR − −−
⎯⎯⎯⎯→ ⎢^ ⎥
This has two free variables so the null space (and hence the orthogonal complement of W) has two vectors in its basis: 1 0 1 0 Span 0 , 1 0 0 0 1
W ⊥
⎪ ⎢^ ⎥ ⎢^ ⎥⎪
⎪ ⎢^ ⎥ ⎢^ ⎥⎪
⎪⎩ ⎢⎣^ ⎥ ⎢⎦ ⎣^ ⎥⎦⎪⎭
- First, we use the Gram-Schmidt process to construct an orthogonal set:
