Mathematics 3: Algebra
Workshop 8
Multiplicities and Minimal polynomials
Let V be a vector space over C. Here Jk (λ) denotes the k × k Jordan matrix with λ’s
on the diagonal, 1’s on the superdiagonal, and 0 elsewhere.
Please hand in your answer to Q3 at the Week 9 workshop.
(1) (a) Given a matrix in Jordan form, how do you read off from it the algebraic and
geometric multiplicities of its eigenvalues?
(b) Suppose that a linear map T : V → V has, with respect to a certain basis of
V , matrix A in Jordan form consisting of the following Jordan blocks:
J1 (2), J2(2), J3 (2), J3 (9), J72 (10), J73(10), J1 (12), J1(12).
Give the algebraic multiplicities ma (λ) and geometric multiplicities mg (λ) of
each of its eigenvalues λ.
(c) Write down a basis of mg (2) vectors for Ker(A − 2I).
[Recall that mg (2)=dim(Ker(A − 2I)), by definition.]
(d) Write down a basis of ma (2) vectors for Ker((A − 2I)n ), where n =dim(V ) =
what?
[Recall that ma (2)=dim(Ker((A − 2I)n )), by Lemma 10.1 of notes.]
(a) The algebraic multiplicity of a particular eigenvalue λ is the sum
of the sizes k of the Jordan blocks Jk (λ).
The geometric multiplicity of a particular eigenvalue λ is the number
of Jordan blocks Jk (λ).
(b) So ma (2) = 1 + 2 + 3 = 6, mg (2) = 3
ma (9) = 3, mg (9) = 1
ma (10) = 72 + 73 = 145, mg (10) = 2
ma (12) = 1 + 1 = 2, mg (12) = 2.
(c) The natural basis is e1 , e2 , e4 , as these are clearly independent and
lie in Ker(A−2I). This is because A−2I has columns 1, 2 and 4 equal
to 0.
(d) The natural basis for Ker((A − 2I)n ) is e1 , e2 , e3 , e4 , e5 , e6 , as (A −
2I)3 , and so any higher powers of A − 2I, has its first 6 (rows and)
columns all 0.
Also n = size of A = sum of the algebraic multiplicities = 6 + 3 +
145 + 2 = 156.
1
2
(2) (a) Write down a matrix in Jordan form having π as its only eigenvalue, with
mg (π) = 5 and ma (2) = 11.
(b) More generally, write down a matrix in Jordan form having λ as its only
eigenvalue, with mg (λ) = m, ma (λ) = M, where m ≤ M.
(c) For which values of m and M with m ≤ M is the matrix in part (b) uniquely
determined, at least up to the order of its Jordan blocks?
(a) We need 5 Jordan blocks with total size 11. So one possibility
is the matrix with 4 Jordan blocks J1 (π) and one Jordan block J7 (π).
(b) We can take m−1 Jordan blocks J1 (λ) and one Jordan block JM −m+1 (λ).
(c) If M = m the only possibility is m Jordan blocks J1 (λ).
If M = m + 1 the only possibility is m − 1 Jordan blocks J1 (λ) and
one Jordan block J2 (λ).
If M ≥ m + 2 one possibility for the matrix is m − 1 Jordan blocks
J1 (λ) and one Jordan block JM −m+1 (λ), as in (b). But another possibility
for the matrix is m−2 Jordan blocks J1 (λ), one Jordan block J2 (λ) and
one Jordan block JM −m (λ). Since M−m ≥ 2, these two matrices do not
have the same set of block sizes.
So the matrix in part (b) uniquely determined, at least up to the order
of its Jordan blocks, if and only if m ≤ M ≤ m + 1.
(3) (This question uses material from Thursday’s lecture.)
(a) Given a matrix A in Jordan form corresponding to a linear map T , explain
how you read off, in factorised form, i.e., as a product of factors (x − λ)k , both
the characteristic polynomial χT (x) of T and the minimal polynomial mT (x)
of T .
(b) For the linear map T in Q1, write down, in factorised form, both the characteristic polynomial χT (x) of T and the minimal polynomial mT (x).
(c) Given positive integers j and n with j ≤ n, write down an n × n matrix
in Jordan form having λ as its only eigenvalue, and (x − λ)j as its minimal
polynomial.
(a) If the largest Jordan block having eigenvalue λ has size k, then
(x − λ)k is the corresponding factor of mT (x).
(b) We have χT (x) = (x − 2)6 (x − 9)3 (x − 10)145 (x − 12)2 .
Then, using (a), mT (x) = (x − 2)3 (x − 9)3 (x − 10)73 (x − 12).
(c) Take the matrix in Jordan form having one Jordan block Jj (λ) and
n − j Jordan blocks J1 (λ). (So none of the latter if j = n.)