Assignment 4
Tahir Asif
November 24, 2021
1
Question One
Since a triangle has half the area as a square with the same base and height,
and since a parallelogram has the same area as a square with the same base
and height, a triangle has half the area of a parallelogram with the same base
and height.
~ and AC
~ form a parallelogram, then its area is
If the vectors, AB
~ AC])|
~
|det([AB
which is:
=
=
=
=
=
x 2 − x1 x3 − x1
det
y2 − y1 y3 − y1
(x2 − x1 )(y3 − y1 ) − (x3 − x1 )(y2 − y1 )
(x2 y3 − x2 y1 − x1 y3 + x1 y1 ) − (x3 y2 − x3 y1 − x1 y2 + x1 y1 )
1
1
x2 y3 − x2 y1 − x1 y3 +x
y1 − x3 y2 + x3 y1 + x1 y2 −x
y1
x1 y2 − x2 y1 − x1 y3 + x3 y1 + x2 y3 − x3 y2
The determinant given in the question can be written as:
1
(x1 y2 − y1 x2 ) − (x1 y3 − y1 x3 ) + (x2 y3 − y2 x3 )
2
1
=
[x1 y2 − x2 y1 − x1 y3 + x3 y1 + x2 y3 − x3 y2 ]
2
=
Therefore, since the expression in the square brackets is the determinant of a
parallelogram formed by the points A, B, C, then 21 of that is the area of the
triangle formed by the same points, and that is exactly the equation from
the question.
Acknowledgments:
• Aadi
2
Question Two
Answer: 1 088 649 operations
Explanation: If A is an 8 × 8 matrix, then it’s solution, ~x, will have 8
entries. Each entry needs one determinant (det Aj (~b)) plus each entry needs
the determinant of A which only needs to be calculated once. So, 9 8 × 8
determinants.
Assuming the general case, all entries of A can be any real number, so we
need to assume they’re not 0 to get an upper bound.
So, to calculate an 8 × 8 determinant using the co-factor method requires
calculating one 7 × 7 determinant 8 times (once per row/column).
Then, calculating a 7 × 7 determinant needs a 6 × 6 determinant 7 times
(once per row/column).
This pattern repeats. So, calculating an n × n determinant needs
n (n − 1 × n − 1) determinants.
Continuing this pattern down to all 2 × 2 determinants means we need:
n − 0 (n − 1 × n − 1) determinants ×
n − 1 (n − 2 × n − 2) determinants ×
n − 2 (n − 3 × n − 3) determinants and so on and so forth.
This means we need n! 2 × 2 determinants. Each 2 × 2 determinant needs 3
arithmetic operations, so an 8 × 8 matrix needs 8! × 3 arithmetic operations,
which is 120 960 operations.
But that’s not all. We need 9 8 × 8 determinants, so its 8! × 3 × 9 = 8! × 27
operations. And on top of that, each level of the co-factor expansion needs n
multiplications and n − 1 additions which is just 2n − 1 operations for each n,
and we go from n = 8 down to n = 3 (3 not 1 since the 2 × 2 matrix doesn’t
need the co-factor expansion and the 1 wont change anything because its just
times 1). So thats 5 n’s which is 2(5) − 1 = 9 extra operations, which makes
the final total:
8! × 27 + 9 = 1088649
Acknowledgments:
• Chris Singh
3
Question Three
 
 
x
t
Let ~u = y  , ~v = r ∈ V
z
s
Conditions for a vector, ~t ∈ V , are t1 , t2 , t3 ∈ R, x > 0
For V to be a vector space, it must satisfy the following properties:
1 ~u + ~v ∈ V
Since all components of ~u, ~v ∈ R and their first entries > 0, xt must give a
positive real answer, and (y 3 + r3 )1/3 and z + s + 1 must give a real answer.
This means ~u +~v ∈ V becuase the only conditions for V were that x, y, z ∈ R
and x > 0
2 ~u + ~v = ~v + ~u
xt = tx
3
3 1/3
(y + r ) = (r3 + y 3 )1/3
z+s+1=s+z+1
3 ~u + (~v + w)
~ = (~u + ~v ) + w
~
Let w
~ = (w1 , w2 , w3 )

 

x + (y + w1 )
(x + y) + w1
[y 3 + ((r3 + w23 )1/3 )3 ]1/3  = [((y 3 + r3 )1/3 )3 + w23 ]1/3 
z + (s + w3 + 1) + 1
(z + s + 1) + w3 + 1

 

x + y + w1
x + y + w1
[y 3 + r3 + w23 ]1/3  = [y 3 + r3 + w23 ]1/3 
z + s + w3 + 2
z + s + w3 + 2
4
4 ∃~0 : ~u + ~0 = ~u
Let ~0 = (a, b, c)

  
xa
x
3
3 1/3 
~


~u + 0 = (y + b )
= y
z
z+c+1
xa = x (y 3 + b3 )1/3 = y
a = xx
y 3 +pb3 = y 3
3 − y3
a=1
b = 3 y√
b= 30
b=0
z+c+1=z
c=z−z−1
c = -1


1
∴ ~0 =  0 
−1
~ : ~u + (−~u) = ~0
5 ∃−u
Let −~u = (a, b, c)
  
xa
1
3
3 1/3 


= 0
~u + (−~u) = (y + b )
−1
z+c+1


xa = 1 (y 3 + b3 )1/3 = 0 z + c + 1 = −1
a = x1
y 3 + b3 = 0
c = −2 − z
3
3
b =p−y
b = 3 −y 3
b = −y

1/x
∴ −~u =  −y 
−2 − z
6 c~u ∈ V , where c ∈ R
xc


c~u =  c1/3 y 
cz + c − 1
5
xc > 0 for all x and c because a positive number raised to any power will
always be positive
c1/3 y will always be real as long as c and y are real, which they are
cz + c − 1 will always be real as long as c and z are real, which they are
∴ c~u ∈ V since all conditions are met
7 c (~u + ~v ) = c~u + c~v
c (~u + ~v )

xt
c (y 3 + r3 )1/3 
z+s+1


(xt)c
 c1/3 (y 3 + r3 )1/3 
c(z + s + 1) + c − 1


xc tc
 c1/3 (y 3 + r3 )1/3 
cz + cs + 2c − 1


xc tc
 c1/3 (y 3 + r3 )1/3 
cz + cs + 2c − 1

= c~u + c~v

 

xc
tc
=  c1/3 y  +  c1/3 r 
cz + c − 1
cs + c − 1


c c
xt

= 
[(c1/3 y)3 + (c1/3 r)3 ]1/3
(cz + c − 1) + (cs + c − 1) + 1


xc tc
=  [cy 3 + cr3 ]1/3 
cz + cs + 2c − 1


xc tc
=  c1/3 (y 3 + r3 )1/3 
cz + cs + 2c − 1
6
8 (c + d) ~u = c~u + d~u
(c + d) ~u = c~u + d~u
 

 

xc
xd
x
(c + d) y  =  c1/3 y  +  d1/3 y 
z
cz + c − 1
dz + d − 1




xc+d
xc+d

 = 

(c + d)1/3 y
[(c1/3 y)3 + (d1/3 y)3 ]1/3
(c + d)z + (c + d) − 1
(cz + c − 1) + (dz + d − 1) + 1




xc+d
xc+d

 =  (cy 3 + dy 3 )1/3 
(c + d)1/3 y
cz + dz + c + d − 1
cz + dz + c + d − 1




xc+d
xc+d
 =  (y 3 )1/3 (c + d)1/3 

(c + d)1/3 y
cz + dz + c + d − 1
cz + dz + c + d − 1




xc+d
xc+d
 = 


y(c + d)1/3
y(c + d)1/3
cz + dz + c + d − 1
cz + dz + c + d − 1
9 c (d~u) = (cd) ~u
c (d~u) = (cd) ~u



xcd
c  d1/3 y  =  (cd)1/3 y 
cdz + cd − 1
dz + d − 1




dc
x
xcd

 =  (cd)1/3 y 
c1/3 d1/3 y
c(dz + d − 1) + c − 1
cdz + cd − 1




dc
x
xcd
 (cd)1/3 y  =  (cd)1/3 y 
cdz + cd − 1
cdz + cd − 1

x
d
7
10 1~u = ~u
 
x
1~u = 1 y 
z


x1
=  11/3 y 
1z + 1 − 1
 
x

y
=
z
Acknowledgments:
• Chris Singh
8
Question Four
A)
Let P = V ∩ W
For P to be a subspace of Rn , it must satisfy the following conditions:
1 ~0 ∈ P
If V and W are subspaces of Rn , then they each need to satisfy this condition,
so each has ~0 in it. Therefore, P must also have ~0 in it.
2 ~v + ~u ∈ P , where ~v , ~u ∈ P
For ~v and ~u to be in P , they must both be in W and V . Now since W and
V are subspaces of Rn , they must also be closed under addition. So ~v + ~u
must be in both W and V , and therefore in P
3 c~v ∈ P
For c~v to be in P , it must both be in W and V , which are closed under scaler
multiplication therefore c~v must be in W and V , and therefore P
B)
For some vector, ~v , to be in V ∪ W , it must be in either V or W . So the
properties mentioned in part A need to be true for both V and W , which
they are since they are subspaces of Rn . Therefore, V ∪ W is a subspace of
Rn .
Alternate explanation in case the first doesn’t make sense:
For V ∪W to be a subspace of Rn , V and W must be subspaces of Rn because
that would mean all vectors in V and W satisfy all properties from part A,
therefore their union does too. V and W are subspaces of Rn , and therefore
so is V ∪ W .
Acknowledgments:
• Chris Singh
9
Question Five
For V to be a subspace of R4 , it must satisfy the following conditions:
1 ~0 ∈ V
For ~0 ∈ V , it must be orthogonal to all columns of A, which means:
~0 · ~a = 0 which it does since any vector dotted with the ~0 vector of the
same size is 0. This is true since the zero vector and vector addition and
multiplication are defined normally since it’s in R4 .
2 ~v + ~u ∈ V , where ~v , ~u ∈ V
For ~v + ~u to be in V , it must be orthogonal to A, so it must satisfy this
equation for all column vectors of A:
a1 (v1 + u1 ) + a2 (v2 + u2 ) + a3 (v3 + u3 ) + a4 (v4 + u4 ) = 0
where (a1 , a2 , a3 , a4 ) is a column vector of A. This then becomes:
a1 v1 + a1 u1 + a2 v2 + a2 u2 + a3 v3 + a3 u3 + a4 v4 + a4 u4 = 0
a1 v1 + a2 v2 + a3 v3 + a4 v4 + a1 u1 + a2 u2 + a3 u3 + a4 u4 = 0
The first 4 terms of this equation add to 0, since ~a · ~v = 0 since ~v ∈ V . The
last 4 terms also add to 0 for the same reason but with u instead of v. This
means the entire equation adds to 0, which confirms the property.
3 c~v ∈ V
If ~v is orthogonal to all column vectors of A, then it satisfies this: ~v · ~a = 0,
where a is a column vector of A. So, c~v would need to satisfy (c~v ) · ~a = 0. It
does because (c~v ) · ~a = 0 can be written as:
(cv1 )a1 + (cv2 )a2 + (cv3 )a3 + (cv4 )a4 = 0
which, after factoring out the c, becomes,
c(v1 a1 + v2 a2 + v3 a3 + v4 a4 ) = 0
10
The ~v · ~a dot product in the brackets equals 0, so this is just c(0) = 0, so
(c~v ) · ~a = 0, and therefore c~v ∈ V since it is orthogonal to all column vectors
of A
To find a basis for V , we need to find a set of vectors that are linearly
independent of each other and span all of R4 , and additionally, they all need
to be orthogonal to the columns of A.
x1 a~1 + x2 a~2 + x3 a~3 + x4 a~4 = 0
A~x = 0
~x ∈ null(A)
To find the null space:

22 24
42 44

62 64
82 84
26
46
66
86


28
1
0
48
RREF
 ====⇒ 
0
68
88
0

0 −1 −2
1 2
3

0 0
0
0 0
0
Using the null space algorithm, you get the vectors:
 
 
1
2
−2
−3

 
v~1 = 
 1  v~2 =  0 
0
1
Therefore, v~1 and v~2 are basis vectors for V .
Acknowledgments:
• Chris Singh
• Aadi
• Tutorial
11