Download MATH 235.9 HW QUIZ 1 (1) (a) Two nonzero real vectors u, v ∈ R 2

MATH 235.9 HW QUIZ 1
ANDREW J. HAVENS
(1) (a) Two nonzero real vectors u, v ∈ R2 are said to be colinear if and only if there is a nonzero number
λ ∈ R such that u = λv. Let a, b, c and d be real numbers, and let
u :=
a
c
,
v :=
b
d
.
Show that u and v are colinear if and only if ad − bc = 0. What does it mean geometrically for
two vectors to be colinear?
Solution:
In one direction, if u and v are colinear we have some nonzero λ ∈ R such that
u=
a
c
=λ
b
d
= λv .
Thus, a = λb and c = λd. If none of a, b, c or d are zero, then we can solve for λ in each equation,
obtaining
λ=
c
a
=
=⇒ ad = bc =⇒ ad − bc = 0 .
b
d
If any one of a = 0 or b = 0 then by the relation a = λb, λ 6= 0, both a and b must be zero, whence
ad − bc = 0, and similarly if either of c = 0 or d = 0 then both must be zero, which again shows
ad − bc = 0.
Conversely, if it is known ad − bc = 0, then ad = bc. We seek a nonzero constant λ such that
u = λv. If a, b, c, and d are each nonzero, then a/b = c/d, and one can check readily that this
ratio gives a λ 6= 0 such that u = λv. On the other hand, if any one of a, b, c, or d is zero, then
both sides of the equation ad = bc vanish. Since u and v are each nonzero vectors, at least one
component of each vector is nonzero. So, if for instance a = 0, then c 6= 0, and thus b = 0, and
one can take λ = c/d, which is well defined since d 6= 0 as well. A symmetric argument works if
c = 0 or d = 0.
The geometric meaning of colinearity is that the vectors span the same line through the origin.
In particular, the λ in u = λv is a scale factor scaling v to u. Any linear combination of colinear
vectors u and v is also a vector lying on the same line, and thus is colinear to both u and v. In
consideration of the bonus establishing the determinant ad−bc as a signed area of the parallelogram
spanned by u and v, it is obvious that colinearity of the vectors is equivalent to the vanishing of
this quantity.
1
2
ANDREW J. HAVENS
(b) Let e and f be real numbers. Under what conditions on a, b, c, d, e and f can we express
e
f
as a linear combination of the vectors u and v above?
(?)
Solution:
This is a rephrasing of the results in class regarding solutions of the pair of linear equations
(
ax + by = e
.
cx + dy = f
e
In particular, w :=
can be expressed as a linear combination of u and v if and only if
f
there are real numbers x and y such that
a
b
e
xu + yv = w ⇐⇒ x
+y
=
,
c
d
f
which can be done if and only if the system of equations (?) has at least one solution.
We know that there’s a unique solution if ad − bc 6= 0 from class, and thus in this case a unique
linear combination giving the desired vector. If not, by part (a) u and v are colinear, and so we
require the vector w to be colinear to both. In particular, the equations af − ce = 0 = bf − de
must both hold. In this case, we are guaranteed infinitely many solutions (I leave the construction
of all possible solutions as an exercise.)
Bonus: Show that ad − bc is the signed area of the parallelogram spanned by u and v, where the sign
is positive if rotating u counter-clockwise to be colinear to v sweeps into the parallelogram, and
is negative otherwise. (Most convincing arguments use some amount of trigonometry; you are
encouraged to discuss this problem with me if you are unsure of how to proceed.)
Solution:
First, let us suppose u and v are unit vectors, i.e. a2 + c2 = 1 = b2 + d2 . Geometrically, they are
vectors lying on the unit circle, and so we can express their components as trigonometric functions of
the angles they make with the x axis. Let u make an angle of α with the x axis and v make an angle of
β with the x axis. Then the angle between the vectors is β − α, and from the sine subtraction formula:
sin(β − α) = cos(α) sin(β) − cos(β) sin(α) = ad − bc.
Recall that the area of a parallelogram is the base times an altitude, formed by taking an orthogonal
line segment from one side to an opposite side. From a picture, one sees that the area of a parallelogram
can be expressed as the product of side lengths times the sine of the internal angle between adjacent
sides. If the sides are the unit vectors u and v, then the area is | sin(β − α)|. Thus, for unit vectors,
ad − bc is ±area, with the sign positive if the angle β − α ∈ (0, π), negative if β − α ∈ (π, 2π), and 0
if the angle β − α = 0 or π (the colinear case). Thus, for the non-colinear case, if u sweeps into the
parallelogram when rotated counterclockwise towards v, the sign is positive. Note that switching the
order of the vectors switches the sign of the determinant ad − bc, and this is consistently reflected in
the convention regarding the vectors’ orientations.
For general vectors, one scales the area of the parallelogram as well as the components, and discovers
that the scale factors for the area and the equation ad − bc are identical: e.g. if we scale u by λ, then
the area scales by λ, and so do the components:
λa
λu =
,
λc
so the determinant scales to (λa)d − b(λc) = λ(ad − bc). Thus, the determinant is the signed area,
accounting for the orientation/ordering of the two vectors.
MATH 235.9 HW QUIZ 1
3
(2) You’ve probably been made aware of the fact that two points in Euclidean space determine a line, and
given the points in Cartesian coordinates you likely can give an equation for that line. A problem in a
similar spirit from algebra and analytic geometry is to find an equation for the graph of a polynomial
given the Cartesian coordinates of some sample points.
(a) Recall that a real quadratic function has the form f (x) = ax2 + bx + c for a, b, c ∈ R, with a 6= 0,
and the graph of y = f (x) is a parabola. Suppose you are given three distinct points in the
plane R2 . Under what condition(s) can you determine a real quadratic function whose graph is a
parabola passing through these points? (Try to intuit it; no proof is required, but you should try
to explain your reasoning.)
Solution:
The condition that our parabola be realizable as a graph of a quadratic function implies that no
pair of the three points lie on the same vertical line (so the x coordinates must all be distinct).
Moreover, since it is stipulated that the quadratic function be nontrivial (the leading coefficient
a is nonzero), we cannot have the three points colinear. That a parabola cannot contain three
colinear points is intuitive, but here is a simple proof (not required in your solutions): suppose
the parabola contains three points which lie on a common line. Then this line has an equation of
the form y = dx + e for some numbers d, e ∈ R. Since the line is coincident with the parabola at
three points, we expect the difference of the quadratic and the line to have three distinct roots.
This is impossible, since the resulting difference ax2 + (b − d)x + c − e = 0 is a quadratic provided
a 6= 0, whence it has at most two distinct solutions. (This last step uses the fundamental theorem
of algebra, a theorem whose proof is nontrivial.)
(b) Assuming the necessary conditions, write down a linear system of equations to find an equation of
the form y = ax2 +bx+c for the parabola passing through the three points (x1 , y1 ), (x2 , y2 ), (x3 , y3 ) ∈
R2 . What are the variables of your linear system? What are the coefficients?
Solution:
By plugging the points in, we obtain the system

 2
2

x1
ax1 + bx1 + c = y1
2

x22
←→
ax2 + bx2 + c = y2

 2
x23
ax3 + bx3 + c = y3
x1
x2
x3

1 y1
1 y2  .
1 y3
The coefficients are determined by the given points, and a, b, and c are the variables of the linear
system.
(c) Find an equation of the form y = ax2 + bx + c for the parabola containing the points (1, −1),
(2, 1), and (−2, 5).
Solution:
The associated system is




1 1 1 −1
a + b + c = −1
4a + 2b + c = 1 ←→ 4 2 1 1  .


4 −2 1 5
4a − 2b + c = 5
Applying Gauss-Jordan gives the solution a = 1, b = −1, and c = −1, so the equation is y =
x2 − x − 1.
(d) Describe how you would determine an equation of the form y = ax3 + bx2 + cx + d for a cubic
curve given some sample points. How many points are needed? Can you guess what conditions
on the sample points are necessary for the resulting system to yield a unique solution?
Solution:
Since there are four coefficients to find, we need four equations, and thus four distinct points.
Moreover, since we want the cubic curve to be a graph, we require that no two points lie on the
same vertical line (equivalently, all of the x values of the coordinates are distinct). There are
4
ANDREW J. HAVENS
two additional algebraic conditions: the four points cannot lie on a common line, or on the graph
of a quadratic function. The proof of this is in the same spirit of the proof that three colinear
points do not lie on the graph of a nontrivial quadratic. These conditions ensure that the resulting
linear system is consistent (i.e. that it may be solved, without producing contradictory equations).
Given the necessary conditions on our sample points the linear system necessary to determine the
coefficients has the augmented matrix

 3
x1 x21 x1 1 y1
x32 x22 x2 1 y2 

 3
x3 x23 x3 1 y3  .
x34 x24 x4 1 y4
Note that the coefficient matrix is eerily reminiscent of the Vandermonde matrix in problem
3. If we chose to reorder our variables, then we would permute the columns, and realize that
Vandermonde matrices are precisely the matrices which arise from the above kinds of polynomial
interpolation problems.
Bonus:
Suppose the points (x1 , y1 ), (x2 , y2 ), and (x3 , y3 ) satisfy the conditions you found in part (a).
A theorem of geometry states that there is then also a unique circle which contains these three
points. Write down (with justification) a linear system in some choice of variables from which
one can determine an equation for the circle (in particular, you should be able to determine its
center and radius.)
For extra points, write down explicit formulae for the center and radius in terms of (x1 , y1 ), (x2 , y2 ),
and (x3 , y3 ). You may even consider writing a program: it should accept three points as inputs,
warn the user if the points fail the criteria from part (a), and otherwise let the user choose to see
the equation of a circle or a parabola through these points, perhaps with the corresponding graph.
Solution:
We can in fact recover a circle so long as the three points are not all colinear. Indeed, when
three points are not a colinear set, they are the vertices of a triangle, and there is a unique
circle circumscribing this triangle, called the circumcircle. Thus, the task at hand is to know the
circumcircle of a given triangle.
The equation for a circle in R2 with center (b, c) and radius r is
(x − b)2 + (y − c)2 = r2 .
Expanding this, one has
x2 − 2bx + b2 + y 2 − 2cy + c2 = r2 .
Given b and c, we can recover r from the constant term in this equation, which after rearrangement
we find to be the quantity r2 − b2 − c2 . Let a := r2 − b2 − c2 . Then, given points (x1 , y1 ), (x2 , y2 ),
and (x3 , y3 ) which are not colinear, we obtain three equations



2
2
2
2

1 2x1 2y1 x21 + y12
a + 2bx1 + 2cy1 = x1 + y1
a + 2bx22 + 2cy22 = x22 + y22 ←→ 1 2x2 2y2 x22 + y22  .


1 2x3 2y3 x23 + y32
a + 2bx23 + 2cy32 = x23 + y32
There’s another, more geometric (or perhaps just more esoteric), method to arrive at a set of
linear equations from which one can recover the circumcircle from the triangle. It is a theorem
of Euclid (the proof of which is a good exercise) that the circumcenter is the intersection of
the perpendicular bisectors of the sides of the triangle. Of course, one only needs two of the
perpendicular bisectors to find the point of intersection of the three bisectors. The radius is then
easily recovered using the distance formula from the circumcenter to any one of the given points.
This has the advantage of producing a pair of linear equations, but at the cost of computing
the equations of two perpendicular bisectors. If one is armed with a compass and straightedge,
intersecting perpendicular bisectors is an efficient way to go, but armed with a computer and
Gauss Jordan, the above augmented matrix does things about as efficiently as possible for this
problem.
MATH 235.9 HW QUIZ 1
5
In the interest of reviewing some basic facts about lines in R2 , I discuss the Euclidean method of
finding the center. Observe that at most one side of the triangle made by our points is horizontal,
so we may assume, possibly after a relabeling of the points, that neither of the sides meeting at
the vertex (x1 , y1 ) are horizontal. Thus, these sides are segments along lines with slopes
m2 =
y2 − y1
,
x2 − x1
y3 − y1
.
x3 − x1
If we add the assumption that the points have distinct x coordinates, then we may assume none
of the sides are vertical so that both of these slopes are defined. This assumption isn’t necessary
if we later rewrite our equations in standard form. We make this assumption temporarily as it
eases our considerations of slopes. Recall that lines with nonzero, defined slopes m and m0 are
orthogonal if mm0 = −1. Thus the lines we desire have slopes
m3 =
m02 = −
x2 − x1
,
y2 − y1
x3 − x1
.
y3 − y1
Further, recall that one obtains the midpointof two points by averaging their coordinates. Thus,
1 yi +y1
the midpoints of these sides are xi +x
for i = 1, 2. We can then write down the equations
2 ,
2
of the perpendicular bisectors as
x2 − x1
x2 + x1
y2 + y1
y−
=−
x−
,
2
y2 − y1
2
x3 − x1
x3 + x1
y3 + y1
=−
x−
.
y−
2
y3 − y1
2
m03 = −
These equations are only defined if the slopes are, but the standard forms are defined even if we
admit a horizontal edge. Thus, one might write the system as
(
2(x2 − x1 )x + 2(y2 − y1 )y = x22 − x21 + y22 − y12
.
2(x3 − x1 )x + 2(y3 − y1 )y = x23 − x21 + y32 − y12
These are precisely the result of subtracting the first row of the augmented matrix from each of
the second and third rows! We may apply the formula from our initial investigation of solutions
to two linear equations, and obtain the coordinates of the center as
(y3 − y1 )(x22 − x21 + y22 − y12 ) − (y2 − y1 )(x23 − x21 + y32 − y12 )
2(x2 − x1 )(y3 − y1 ) − 2(x3 − x1 )(y2 − y1 )
y3 − y2 + (x2 + x1 )m02 − (x3 + x1 )m03
=
,
2m02 − 2m03
b=
(x2 − x1 )(x23 − x21 + y32 − y12 ) − (x3 − x1 )(x22 − x21 + y22 − y12 )
2(x2 − x1 )(y3 − y1 ) − 2(x3 − x1 )(y2 − y1 )
x3 − x2 + (y3 + y1 )m3 − (y2 + y1 )m2
=
.
2m3 − 2m2
c=
Thus, the radius can be obtained as
"
2
(y3 − y1 )(x22 − x21 + y22 − y12 ) − (y2 − y1 )(x23 − x21 + y32 − y12 )
r=
x1 −
2(x2 − x1 )(y3 − y1 ) − 2(x3 − x1 )(y2 − y1 )
(x2 − x1 )(x23 − x21 + y32 − y12 ) − (x3 − x1 )(x22 − x21 + y22 − y12 )
+ y1 −
2(x2 − x1 )(y3 − y1 ) − 2(x3 − x1 )(y2 − y1 )
2 #1/2
.
6
ANDREW J. HAVENS
The reduced row echelon form of the augmented matrix, for your viewing pleasure:


y32 x2 y1 +y32 (−y2 )x1 +y3 x22 x1 −y3 x2 x21 −y3 x2 y12 +y3 y22 x1 −x22 x3 y1 +x2 x23 y1 −y22 x3 y1 −y2 x23 x1 +y2 x3 x21 +y2 x3 y12
1 0 0
−y3 x2 +y3 x1 +x2 y1 +y2 x3 −y2 x1 −x3 y1


y32 y2 −y32 y1 −y3 x22 −y3 y22 +y3 x21 +y3 y12 +x22 y1 +y22 y1 +y2 x23 −y2 x21 −y2 y12 −x23 y1
,
0 1 0
−


2(y3 x2 −y3 x1 −x2 y1 −y2 x3 +y2 x1 +x3 y1 )
2
2
2
2
2
2
2
2
(2x2 −2x1 )(y3 +x3 −x1 −y1 )−(2x3 −2x1 )(x2 +y2 −x1 −y1 )
0 0 1
4y3 x2 −4y3 x1 −4x2 y1 −4y2 x3 +4y2 x1 +4x3 y1
(3) The matrix

1 a a2 a3
 1 b b 2 b3 

V := 
 1 c c2 c3 
1 d d2 d3
is an example of a Vandermonde matrix. Let vi,j denote the (i, j)th-entry of V.
(a) Assuming a, b, c and d are distinct real numbers, how many pivots does the matrix V have? If
a = c and b = d, how many pivots does V possess?

Solution:
If a, b, c and d are distinct, there are four pivots (see part b for a sequence of row operations proving
that each diagonal entry is a pivot position). If a = c and b = d, there are only two pivots, as the
matrix then contains two pairs of twin rows, which means two rows can each be made into the
zero row.
(b) A square matrix is said to be upper triangular if all nonzero entries occur on or above the main
diagonal (the main diagonal is the entries whose row and column numbers match, i.e. vi,i for
i = 1, 2, 3, 4).
Find a sequence of row operations to convert V into an upper triangular matrix. (Hint: use a
modified Gauss-Jordan algorithm; you should not change entries on the diagonal to 1 as you do
when putting a matrix into REF or RREF–doing so makes V a strictly upper triangular matrix,
but you will need to track the values on the diagonal if you intend to do the bonus.)
Solution:
See the solution for the bonus; I give a row reduction which simultaneously solves this and achieves
the property listed in the bonus.
Bonus: Find a series of row operations on V making V upper triangular such that the product of the
entries on the main diagonal of the final matrix is, up to a sign (depending on the row operations
you chose), the product of the differences of the distinct numbers a, b, c and d: i.e. if Ṽ is the final
upper triangular matrix, with entries ṽi,j , i, j = 1, 2, 3, 4, then
4
Y
ṽi,i = ±(a − b)(a − c)(a − d)(b − c)(b − d)(c − d) .
i=1
Solution:
Before proceeding, note that I have simplified my calculations using two factorizations: the difference of perfect squares factorization, and the difference of perfect cube factorization. For example,
b2 − a2 = (b − a)(b + a), and c3 − a3 = (c − a)(c2 + ac + a2 ). The sequence of row operations is


R2 − R1 7→ R2 ,
(i) R3 − R1 7→ R3 ,


R4 − R1 7→ R4 ,
(
R3 − c−a
b−a R2 7→ R3 ,
(ii)
R4 − d−a
b−a R2 7→ R4 ,
(d − a)(d − b)
(iii) R4 −
R3 7→ R4 .
(c − a)(c − b)
MATH 235.9 HW QUIZ 1
7
The pivotal moments (pun intended) in the sequence of augmented matrices are shown below:




1
a
a2
a3
1 a a2 a3
 1 b b2 b3  (i)  0 b − a b2 − a2 b3 − a3 




 1 c c2 c3  7−→  0 c − a c2 − a2 c3 − a3 
0 d − a d2 − a2 d3 − a3
1 d d2 d3


1
a
a2
a3

(ii)  0 b − a
b2 − a2
b3 − a 3

7−→ 
 0
0
(c − a)(c − b) (c − a)(c − b)(a + b + c) 
0
0
(d − a)(d − b) (d − a)(d − b)(a + b + d)


1
a
a2
a3

(iii)  0 b − a
b2 − a 2
b3 − a 3

7−→ 
 0
0
(c − a)(c − b) (c − a)(c − b)(a + b + c) 
0
0
0
(d − a)(d − b)(d − c)
The final matrix is upper-triangular, and the product of diagonal entries is
4
Y
ṽi,i = (b − a)(c − a)(c − b)(d − a)(d − b)(d − c)
i=1
= (−1)6 (a − b)(a − c)(a − d)(b − c)(b − d)(c − d)
= (a − b)(a − c)(a − d)(b − c)(b − d)(c − d) ,
as desired. Note that the product vanishes if and only if a, b, c and d are not all distinct, in which
case the matrix has less than four pivots.