Linear Algebra, 2020a

442 Chapter Five. Similarity
polynomial then it has a nonzero leading coefficient. Dividing through by that
leading coefficient would make it a polynomial that takes the map or matrix to
zero, has leading coefficient 1, and is of smaller degree than m and ˆm (because
in the subtraction the leading 1’s cancel). That would contradict the minimality
of the degree of m and ˆm. Thusm(x)− ˆm(x) is the zero polynomial and the
two are equal.
QED
1.6 Example One way to compute the minimal polynomial for the matrix of
Example 1.3 is to find the powers of T up to n 2 = 4.
(
T 2 −1/2 − √ ) ( ) ( √ )
3/2
= √ T 3 −1 0
=
T 4 −1/2 3/2
=
3/2 −1/2 0 −1
− √ 3/2 −1/2
Put c 4 T 4 + c 3 T 3 + c 2 T 2 + c 1 T + c 0 I equal to the zero matrix.
Apply Gauss’ Method.
−(1/2)c 4 − c 3 − (1/2)c 2 + (1/2)c 1 + c 0 = 0
( √ 3/2)c 4 −( √ 3/2)c 2 −( √ 3/2)c 1 = 0
−( √ 3/2)c 4 +( √ 3/2)c 2 +( √ 3/2)c 1 = 0
−(1/2)c 4 − c 3 − (1/2)c 2 + (1/2)c 1 + c 0 = 0
−(1/2)c 4 − c 3 −(1/2)c 2 +(1/2)c 1 + c 0 = 0
−( √ 3)c 3 − √ 3c 2 + √ 3c 0 = 0
With an eye toward making the degree of the polynomial as small as possible,
note that setting c 4 , c 3 , and c 2 to zero forces c 1 and c 0 to also come out as zero
so the equations won’t allow a degree one minimal polynomial. Instead, set c 4
and c 3 to zero. The system
−(1/2)c 2 +(1/2)c 1 + c 0 = 0
− √ 3c 2 + √ 3c 0 = 0
has the solution set c 1 =−c 0 and c 2 = c 0 . Taking the leading coefficient to
be c 2 = 1 gives the minimal polynomial x 2 − x + 1.
That computation is ungainly. We shall develop an alternative.
1.7 Lemma Suppose that the polynomial f(x) =c n x n + ···+ c 1 x + c 0 factors
as k(x − λ 1 ) q1 ···(x − λ z ) q z
.Ift is a linear transformation then these two are
equal maps.
c n t n + ···+ c 1 t + c 0 = k · (t − λ 1 ) q 1
◦···◦(t − λ z ) q z
Consequently, if T is a square matrix then f(T) and k · (T − λ 1 I) q1 ···(T − λ z I) q z
are equal matrices.