awkwardness of the dome model is gone in
that as a line of view slides from north to
south, no sudden changes happen. This central
projection model is uniform. So far we have
described points in projective geometry. What
about lines? What a viewer P at th
projection to this point would be along the
gray line that is parallel to the domain plane
(we say that the vanishing point is the image
of a projection from infinity). S I P For a model
that eliminates this awkwardness, cover the
projector P with a hemis
Exercises 1 Use Chis Method to find each
determinant. Topic: Chis Method 365 (a)
123456789
(b)
21400140111
10211
2 What if a1,1 is zero? 3 The Rule
of Sarrus is a mnemonic that many people
learn for the 33 determinant formula. To the
right of the matrix,
Some computer language specifications
requires that arrays be stored by column,
that is, the entire first column is stored
contiguously, then the second column, etc.
Does the code fragment given take advantage
of this, or can it be rewritten to make it fa
eigenvalue 0 and its eigenspace associated
with the eigenvalue 1 are easy to find. V0 = cfw_
0 0 c3 | c3 C V1 = cfw_ c1 c2 0
| c1, c2 C Note that V1 has dimension
two. By Lemma 3.13 if two eigenvectors ~v1
and ~v2 are associated with the same
eigenval
the same result). Subtract the second from the
first. ~0 = c1(k+1 1)~v1 + + ck(k+1
k)~vk + ck+1(k+1 k+1)~vk+1 400 Chapter
Five. Similarity The ~vk+1 term vanishes. Then
the induction hypothesis gives that c1(k+1
1) = 0, . . . , ck(k+1 k) = 0. The eigenv
0 . . . 1 . . . 0 =
0 . . . i . . . 0 has the
stated action. QED 2.5 Example To diagonalize
T = 3 2 0 1! we take T as the representation of
a transformation with respect to the standard
Section II. Similarity 391 basis RepE2,E2 (t) and
look for a basis
right. 1 1 1 1 0 0 0 1 0 0 1 0 1 0 0 0 0 1
1 0 0 0 0 1 0 1 0 0 1 0 0 0 1
1 1 1 1.2 Definition The matrices T and T
are similar if there is a nonsingular P such that
T = PTP1 . Since nonsingular matrices are
square, T and T must be square and of the
same
that central projection does not preserve
circles, that a circle may project to an ellipse.
Can a (non-circular) ellipse project to a circle?
6 Give the formula for the correspondence
between the non-equatorial part of the
antipodal modal of the projectiv
allow ~0 as an eigenvector for as long as
there are also non-~0 vectors associated with
. The key point is to disallow the trivial case
where is such that t(~v) = ~v for only the
single vector ~v = ~0. Also, note that the
eigenvalue could be 0. The issue
cfw_~0 N (t 1 ) = P0 N (t 2 ) = P1 N (t 3 ) =
P2 N (t 4 ) = P3 (later elements are the
entire space). 1.6 Example Let t : P2 P2 be
the map c0 + c1x + c2x 2 7 2c0 + c2x. As the
lemma describes, on iteration the range space
shrinks R(t 0 ) = P2 R(t) = cfw_a
the matrix representing the transformation is T
= RepE2,E2 (t) then it assumes that column
vectors are representations with respect to E2.
However T = RepB,B(t) assumes that
column vectors are representations with
respect to B, and so the column vectors t
Chapter Five. Similarity Proof Fix an eigenvalue
. Notice first that V contains the zero vector
since t(~0) = ~0, which equals ~0. So the
eigenspace is a nonempty subset of the space.
What remains is to check closure of this set
under linear combinations.
Chapter Five. Similarity Now draw the
similarity diagram R 2 wrt E2 t T R 2
wrt E2 id y id y R 2 wrt B t D R 2
wrt B and note that the matrix RepB,E2 (id) is
easy, giving this diagonalization. 3 0 0 1! = 1 1 0
1 !1 3 2 0 1! 1 1 0 1 ! In the next subsectio
matrix equivalent matrices have the same
eigenvalues? 3.40 Show that a square matrix
with real entries and an odd number of rows
has at least one real eigenvalue. 3.41
Diagonalize. 1 2 2 2 2 2 3 6 6 3.42
Suppose that P is a nonsingular n n matrix.
Show th
t2 1 x 1 u2 y 1 1 z
= 0 = (1 u2) x + (1
t2) y + (t2u2 1) z = 0 Finding the
intersection of the two is routine. T1U1
T2U2 = t2 1 1 u2 0 (This is, of
course, a homogeneous coordinate vector of a
projective point.) The other two intersections
are similar.
complex numbers C = cfw_a + bi | a, b R and i 2
= 1. (These are often pictured on a plane
with a plotted on the horizontal axis and b on
the vertical; note that the distance of the point
from the origin is |a + bi| = a2 + b2.) In C all
quadratics factor.
point if and only if any three row vectors
representing them are linearly dependent. The
following result is more evidence of the
niceness of the geometry of the projective
plane. These two triangles are in perspective
from the point O because their corre
differ similar matrices need not have the
same eigenvectors. The next example explains.
3.6 Example These matrices are similar T = 2 0
0 0! T = 4 2 4 2 ! since T = PTP1 for this P.
P = 1 1 1 2! P 1 = 2 1 1 1 ! The matrix T has
two eigenvalues, 1 = 2 and 2
b 6= 0. 3.8 Example If S = 1 0 3! Section II.
Similarity 397 (here is not a projection map,
it is the number 3.14 . . .) then x 1 0 3 x
= (x )(x 3) so S has eigenvalues of 1 =
and 2 = 3. To find associated eigenvectors,
first plug in 1 for x 1 0 3 ! z1 z
earth right-handed and return left-handed?
[Gardner] is a nontechnical reference. [Clarke]
is a classic science fiction story about
orientation reversal. For an overview of
projective geometry see [Courant & Robbins].
The approach weve taken here, the ana
lemma, because W, X, and Y are not on the
same projective line, any homogeneous
coordinate vectors w~ 0, ~x0, and ~y0 do not
line on the same plane through the origin in R
3 and so form a spanning set for R 3 . Thus any
homogeneous coordinate vector for Z
matrix representing, with respect to some B, B,
reflection across the x-axis in R 2 . Consider
also a matrix representing, with respect to
some D, D, reflection across the y-axis. Must
they be similar? 1.17 Prove that similarity
preserves determinants and
ck ~k + ~0 + + ~0, where n is the
dimension of the domain and k is the
dimension of the range. Under this
representation the action of the map is easy to
understand because most of the matrix entries
are zero. This chapter considers the special
case where
into 21 with remainder 1 is that the
remainder is less than 4 while 4 goes 5
times, it does not go 6 times. Similarly, the
final clause of the polynomial division
statement is crucial. 1.3 Example If p(x) = 2x3
3x2 + 4x and d(x) = x 2 + 1 then q(x) = 2x
is the zero matrix. This imples that for any map
n represented by N (with respect to some B, B)
the composition n n is the zero map. This in
turn implies that for any matrix representing n
(with respect to some B, B), its square is the
zero matrix. But th
3i) (i) ! = 1 + 7i 1 1i 9 5i 3 + 3i! We shall
carry over unchanged from the previous
chapters everything that we can. For instance,
we shall call this h 1 + 0i 0 + 0i . . . 0 +
0i , . . . , 0 + 0i 0 + 0i . . . 1 + 0i
i the standard basis for C n as a vec
D The top is first. The effect of the
transformation on the starting basis B x 2 d/dx
7 2x x d/dx 7 1 1 d/dx 7 0 Section II.
Similarity 385 represented with respect to the
ending basis (also B) RepB(2x) = 0 2 0
RepB(1) = 0 0 1 RepB(0) = 0 0 0
gives the
7 6d and any higher power is the zero map.
1.2 Example This transformation of the space
M22 of 22 matrices a b c d! t 7 b a d 0!
More information on function iteration is in
the appendix. 404 Chapter Five. Similarity has
this second power a b c d! t 2 7