the other hand, as in the proof of Lemma 1.7
we can apply the binomial expansion to get
this. (t i) n(~v) + n 1 (t i) n1 (t j) 1
(~v) + n 2 (t i) n2 (t j) 2 (~v) + The
first term is zero because ~v N(
Thus the minimal polynomial is quadratic.
Using the method of that example to find the
minimal polynomial of a 33 matrix would
mean doing Gaussian reduction on a system
with nine equations in ten unkn
diagonal. To make those blocks into Jordan
blocks, pick each Bi to be a string basis for the
action of t i on N(t i). QED 2.14
Corollary Every square matrix is similar to the
sum of a diagonal matrix
minimal degree. Divide this polynomial by its
leading coefficient ck to get a leading 1. Hence
any map or matrix has a minimal polynomial.
Now for uniqueness. Suppose that m(x) and
m (x) both take the
x 3 then most authors explicitly write the
identity matrix f(T) = T 3I but dont write the
identity map f(t) = t 3. We shall follow this
convention. Consider again Example 1.1. The
space M22 has dimens
such that the matrix representation RepB,B(t)
is the sum of a diagonal matrix and a nilpotent
matrix. This is Jordan canonical form. IV.1
Polynomials of Maps and Matrices Recall that
the set of square
it is linked-to from page pj. The increment
depends on the importance of the linking page
pj divided by how many out-links aj are on
that page. I(pi) = X in-linking pages pj I(pj) aj
448 Chapter Five.
order the blocks, say, from longest to shortest.
2.17 Example The matrix M = 1 1 1 1 ! has an
index of nilpotency of two, as this calculation
shows. power p Mp N (Mp) 1 M = 1 1 1 1 !
cfw_ x x ! | x C
1) 3 (x 4) (c) (x 2) 2 (x 5) 2 (d) (x + 3) 2 (x
1)(x 2) 2 What is the degree of each
possibility? X 1.14 Find the minimal
polynomial of each matrix. Section IV. Jordan
Form 425 (a) 3 0 0 1 3 0 0 0 4
action of t 2 on a string basis is ~ 1 7 ~ 2
7 ~0. Thus we can represent the restriction
in the canonical form N2 = 0 0 1 0! = RepB,B(t
2) B2 = h 1 1 2 , 2 0 0 i
(other choices of basis are possible)
computation we take T to represent a
transformation t : C 2 C 2 with respect to
the standard basis (we shall do this for the rest
of the chapter). N (t 3) = cfw_ y y ! | y C N (t
3) 2 ) = C 2 428 Cha
matrix with blocks of square submatrices T1
and T2 T1 Z2 Z1 T2 ! dim(N )-many rows
dim(R)-many rows where Z1 and Z2 are blocks
of zeroes. Proof Since the two subspaces are
complementary, the concate
have the same characteristic polynomial. (b)
Show that similar matrices have the same
minimal polynomial. (c) Decide if these are
similar. 1 3 2 3 4 1 1 1 1.32 (a) Show that a
matrix is invertible if
new matrix is nilpotent; its fourth power is the
zero matrix. We could verify this with a
tedious computation or we can instead just
observe that it is nilpotent since its fourth
power is similar to N
3 0 4 1! x + 1 2 1 1! 1.9 Lemma If T is a square
matrix with characteristic polynomial c(x) then
c(T) is the zero matrix. Proof Let C be T xI, the
matrix whose determinant is the characteristic
polyno
the vectors of kind 2 and so the set C is the
set of vectors in squares.) While the vectors ~
we choose arent uniquely determined by t,
what is uniquely determined is the number of
them: it is the dim
permutations t1,(1)t2,(2) tn,(n)
sgn() 432 Chapter Five. Similarity each term
comes from a rearrangement of the column
numbers 1, . . . , n into a new order (1), . . . ,
(n). The upper right block Z2
linear map to ~0 so the fact that t n(~v) = ~0
implies that ~v = ~0. QED 2.3 Remark
Technically there is a difference between the
map t : V V and the map on the subspace t :
R(t) R(t) if the generaliz
vector, including the vector starting the
longest string, to ~0. Therefore instead
suppose that the space has a t-string basis B
where all of the strings are shorter than length
k. Because t has index
The nullities of the powers are: T 3I has
nullity two, (T 3I) 2 has nullity three, (T 3I)
3 has nullity four, and (T 3I) 4 has nullity five.
(b) The matrix S is 55 with two eigenvalues.
For the eigenv
Jordan blocks if some of the eigenvalues are
complex numbers. That is, suggest a
reasonable ordering for the complex numbers.
2.40 Let Pj(R) be the vector space over the
reals of degree j polynomials.
(Np) 1 0 0 0 0 0 1 0 0 0 0 1 1 1 1 1 0
1 0 0 0 1 0 1 1 1 cfw_ 0 0 u
v u v | u, v C 2 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 cfw_
0 y z u v | y, z, u, v C 3
zero matrix C 5 The table tell
qz is the zero map. Since m(t) sends every
vector to zero, at least one of the maps t i
sends some nonzero vectors to zero. Exactly
the same holds in the matrix case if m is
minimal for T then m(T) =
factor of m, i.e., that m() = 0. Suppose that
is an eigenvalue of T with associated
eigenvector ~v. Then T T~v = T ~v = T~v =
2~v. Similarly, T n~v = n~v. With that, we
have that for 424 Chapter Fiv
(i) qj = 0 for all j 6= i, and (ii) qi = pi. Together
these prove clause (2) because they show that
the 434 Chapter Five. Similarity degree of the
polynomial |T1 xI| is qi and the degree of
that polyn
t2 is stable after only one application the
restriction of t2 to N(t 2) is nilpotent of
index one. The restriction of t 2 to the
generalized null space acts on a string basis via
the two strings ~ 1 7