Additional Topics in Linear Algebra [Lecture notes]

26 Pages·2013·0.138 MB·English

by  Joseph H. Silverman

Additional Topics
in Linear Algebra
Supplementary Material
for Math 540
Joseph H. Silverman
E-mail address: [email protected]
Mathematics Department, Box 1917 Brown University,
Providence, RI 02912 USA

Contents
Chapter 1. Characteristic Polynomials 5
1. The Characteristic Polynomial of a Linear Map 5
2. Jordan Blocks 6
3. Jordan Normal Form 9
4. The Cayley-Hamilton Theorem 10
Exercises for Chapter 1 13
Chapter 2. Linear Recursions 15
1. What is a Linear Recursion? 15
2. An Example: The Fibonacci Sequence 16
3. The Matrix Associated to a Linear Recursion 19
4. Diagonalizable Linear Recursions 21
5. Powers of Jordan Block Matrices 23
6. Closed Formulas for General Linear Recursions 24
Exercises for Chapter 2 24
3

CHAPTER 1
Characteristic Polynomials
As always, V is a finite dimensional vector space with field of
scalars F. But we do not assume that V is an inner product space.
1. The Characteristic Polynomial of a Linear Map
We begin with a fundamental definition.
Definition. Let T 2 L(V) be a linear map. The characteristic
polynomial of T is
P (z) = det(zI (cid:0)T):
T
WebeginbyshowingthatP (z)maybecomputedusinganymatrix
T
associated to T.
Proposition 1. Let fv g be a basis for V, and let
i( )
A = M T;fv g 2 Mat(n;F)
i
be the matrix for T associated to the chosen basis. Then
P (z) = det(zI (cid:0)A):
T
In particular, we can compute P (z) using the matrix for T associated
T
to any basis.
Proof. WeprovedthatifS(2 L(V))isanylinearmap, thenthede-
terminant of the matrix B = M S;fv g doesn’t depend on the choice
i
of basis. That’s what allows us to define det(S). In case you’ve forgot-
ten, the proof uses the fact that if(we choo)se some other basis fwig,
then the associated matrix C = M S;fw g satisfies C = G(cid:0)1BG for
i
some invertible matrix G, so
det(C) = det(G(cid:0)1BG) = det(G(cid:0)1)det(B)det(G)
1
= det(B)det(G) = det(B):
det(G)
We now need merely observe that
( ) ( ) ( )
M zI (cid:0)T;fv g = zM I;fv g (cid:0)M T;fv g = zI (cid:0)A;
i i i
sobydefinition,det(zI (cid:0)T) = det(zI (cid:0)A),anditdoesn’tmatterwhat
basis we use to compute A. (cid:3)
5
6 1. CHARACTERISTIC POLYNOMIALS
Example 2. Let T 2 L(R2) be given by
T(x;y) = (x+2y;3x+4y):
Then the matrix of T for the standard basis of R2 is
( )
1 2
A = :
3 4
So the characteristic polynomial of T is
(( ) ( ))
z 0 1 2
P (z) = det(zI (cid:0)A) = det (cid:0)
T 0 z 3 4
( )
z (cid:0)1 (cid:0)2
= det
(cid:0)3 z (cid:0)4
= (z (cid:0)1)(z (cid:0)4)(cid:0)6
= z2 (cid:0)5z (cid:0)2:
Proposition 3. LetT 2 L(V). Then therootsofthe characteristic
polynomial P (z) in C are exactly the eigenvalues of T.
T
Proof. We know that the eigenvalues of T are precisely the num-
bers (cid:21) 2 C for which (cid:21)I (cid:0)T is not invertible. And we also proved that
a linear map S is not invertible if and only if det(S) = 0. Hence the
eigenvalues of T are the numbers (cid:21) 2 C such that det((cid:21)I (cid:0)T) = 0,
which are exactly the roots of P (z). (cid:3)
T
2. Jordan Blocks
We know that if V has a basis of eigenvectors for a linear map
T 2 L(V), then the matrix of T for that basis is diagonal. Thus
if fv ;:::;v g is a basis for V and if
1 n
Tv = (cid:21) v for 1 (cid:20) i (cid:20) n,
i i i
then
 
(cid:21) 0 0 (cid:1)(cid:1)(cid:1) 0
1
0 (cid:21) 0 (cid:1)(cid:1)(cid:1) 0 
( )  2 
M T;fvig = 0 0 (cid:21)3 (cid:1)(cid:1)(cid:1) 0 :
 ... ... ... 
0 0 0 (cid:1)(cid:1)(cid:1) (cid:21)
n
Unfortunately, not every linear transformation has a basis of eigenvec-
tors. Our goal in the next two sections is to find bases that are almost
as good.
2. JORDAN BLOCKS 7
Example 4. Consider the linear map T 2 L(F2) given by the
formula
T(x ;x ) = (x +x ;x ):
1 2 1 2 2
The matrix of T for the standard basis of F2 is (1 1), so the character-
0 1
istic polynomial of T is
( )
z (cid:0)1 (cid:0)1
det(zI (cid:0)T) = det = (z (cid:0)1)2;
0 z (cid:0)1
so Proposition 3 says that the only eigenvalue of T is (cid:21) = 1. But one
easily checks that the only vectors satisfying Tv = v are multiples
of (1;0), so F2 does not have a basis consisting of eigenvectors for T.
Generalizing Example 4, we are led to look at matrices of the fol-
lowing form.
Definition. The Jordan matrix (or Jordan block) of dimension m
and eigenvalue (cid:21) is the m-by-m matrix
 
(cid:21) 1 0 0 (cid:1)(cid:1)(cid:1) 0
0 (cid:21) 1 0 (cid:1)(cid:1)(cid:1) 0
 
0 0 (cid:21) 1 (cid:1)(cid:1)(cid:1) 0
 
Jm((cid:21)) =  ... ... ... ...
 
0 0 0 (cid:1)(cid:1)(cid:1) (cid:21) 1
0 0 0 0 (cid:1)(cid:1)(cid:1) (cid:21)
Thus J ((cid:21)) is an m-by-m matrix with (cid:21)’s on its main diagonal and
m
with 1’s just above the main diagonal, and all other entries are 0.
Proposition 5. Let J ((cid:21)) be a Jordan matrix. The only eigen-
m
value of J ((cid:21)) is (cid:21), and the only eigenvectors of J ((cid:21)) are multiples
m m
of the standard basis vector e .
1
Proof. Let J = J ((cid:21)). The matrix J is upper triangular, so its
m
characteristic polynomial is
P (z) = det(zI (cid:0)J) = (z (cid:0)(cid:21))m:
J
Proposition 3 tells us that (cid:21) is the only eigenvalue of J. Next suppose
that Jv = (cid:21)v, so
(J (cid:0)(cid:21)I)v = 0:
8 1. CHARACTERISTIC POLYNOMIALS
But looking at the matrix J (cid:0)(cid:21)I, we see that
    
0 1 0 0 (cid:1)(cid:1)(cid:1) 0 x x
1 2
0 0 1 0 (cid:1)(cid:1)(cid:1) 0 x  x 
  2   3
0 0 0 1 (cid:1)(cid:1)(cid:1) 0 x  x 
(J (cid:0)(cid:21)I)v =  ... ... ... ... ...3  =  ...4:
    
0 0 0 (cid:1)(cid:1)(cid:1) 0 1 x x
m(cid:0)1 m
0 0 0 0 (cid:1)(cid:1)(cid:1) 0 x 0
m
Hence (J (cid:0)(cid:21)I)v = 0 if and only if
x = x = (cid:1)(cid:1)(cid:1) = x = 0;
2 3 m
which is just another way of saying that v is a multiple of e . (cid:3)
1
We note that the application of J = J ((cid:21)) to the standard basis
m
vectors of Fm exhibits a sort of shift effect,
Je = (cid:21)e
1 1
Je = e + (cid:21)e
2 1 2
Je = e + (cid:21)e
3 2 3
... ...
Je = e + (cid:21)e
m m(cid:0)1 m
It is often convenient to write a Jordan block as
 
0 1 0 0 (cid:1)(cid:1)(cid:1) 0
0 0 1 0 (cid:1)(cid:1)(cid:1) 0
 
0 0 0 1 (cid:1)(cid:1)(cid:1) 0
 
J = (cid:21)I +N with N =  ... ... ... ...: (1)
 
0 0 0 (cid:1)(cid:1)(cid:1) 0 1
0 0 0 0 (cid:1)(cid:1)(cid:1) 0
In other words, J is the sum of a multiple of the identity matrix and
a matrix N that has 1’s just above the diagonal, and 0’s everywhere
else. When we compute the powers of the matrix N, we find something
surprising.
Proposition 6. Let N be the m-by-m matrix described in (1).
Then
Nm = 0:
Proof. The effect of N on the list e ;e ;:::;e of standard basis
1 2 m
vectors for Fm is to shift the list to the right, deleting e and putting
m
a zero vector in front. In other words,
Ne = 0; Ne = e ; Ne = e ; Ne = e ; ::: Ne = e :
1 2 1 3 2 4 3 m m(cid:0)1
3. JORDAN NORMAL FORM 9
What happens if we apply N again? The list shifts again, yielding
N2e = 0; N2e = 0; N2e = e ; N2e = e ; ::: N2e = e :
1 2 3 1 4 2 m m(cid:0)2
Applying N again, the list shifts yet one step further. So if we apply N
a total of m times, all of the e are shifted out of the list, and we’re
i
just left with zero vectors,
Nme = 0; Nme = 0; Nme = 0; Nme = 0; ::: Nme = 0:
1 2 3 4 m
ThisshowsthatNm sendsallofthebasisvectorse ;:::;e to0, soNm
1 m
is the zero matrix. (cid:3)
Thereisanameformatrices(orlinearmaps)thathavetheproperty
described in Proposition 6.
Definition. A linear map T 2 L(V) is said to be nilpotent if there
is some integer j (cid:21) 1 such that Tj = 0. Similarly, a square matrix A
is said to be nilpotent if there is some integer j (cid:21) 1 such that Aj = 0.
3. Jordan Normal Form
We can now define matrices that are “almost, but not necessarily
entirely, diagonal.”
Definition. A matrix is in Jordan Normal Form if it looks like
 
J 0 0 (cid:1)(cid:1)(cid:1) 0
1
0 J 0 (cid:1)(cid:1)(cid:1) 0
 2 
A = 0 0 J3 (cid:1)(cid:1)(cid:1) 0;
 ... ... ... 
0 0 0 (cid:1)(cid:1)(cid:1) J
r
where each J is a Jordan block. Thus J is an m -by-m Jordan block
i i i i
with eigenvalue (cid:21) ,
i
 
(cid:21) 1 0 0 (cid:1)(cid:1)(cid:1) 0
i
0 (cid:21) 1 0 (cid:1)(cid:1)(cid:1) 0
 i 
0 0 (cid:21) 1 (cid:1)(cid:1)(cid:1) 0
 i 
Ji =  ... ... ... ... :
 
0 0 0 (cid:1)(cid:1)(cid:1) (cid:21) 1
i
0 0 0 0 (cid:1)(cid:1)(cid:1) (cid:21)
i
10 1. CHARACTERISTIC POLYNOMIALS
Example 7. The 6-by-6 matrix
 
3 1 0 0 0 0
0 3 0 0 0 0
 
0 0 5 0 0 0
A =  ;
0 0 0 5 1 0
 
0 0 0 0 5 1
0 0 0 0 0 5
consists of three Jordan blocks. The first is a 2-by-2 Jordan block with
eigenvalue 3, the second is a 1-by-1 Jordan block with eigenvalue 5,
and the third is a 3-by-3 Jordan block with eigenvalue 5. Thus
   
( )
J1 0 0 3 1 ( ) 5 1 0
A = 0 J 0 with J = ; J = 5 ; J = 0 5 1 :
2 1 0 3 2 3
0 0 J 0 0 5
3
Note that different blocks need not have different eigenvalues.
As we have seen, not every linear transformation T can be diago-
nalized. But we can come close in the sense that we can always find a
basisthatputsT intoJordannormalform, asdescribedinthefollowing
important result.
Theorem 8. (Jordan Normal Form Theorem) Let V be a vector
space over C, and let T 2 L(V) be a linea(r transfo)rmation. Then there
is a basis fv g for V so that the matrix M T;fv g is in Jordan normal
i i
form.
Proof. We won’thave time in class to prove this theorem, but you
can find a proof in the book if you’re interested. You may also someday
see it proven in a more advanced mathematics course as a special case
of a general theorem on finitely generated modules over principal ideal
domains. (cid:3)
4. The Cayley-Hamilton Theorem
We are now ready to state and prove a useful theorem about linear
maps. The proof uses the Jordan Normal Form Theorem (Theorem 8).
Theorem 9. (Cayley-Hamilton Theorem) Let T 2 L(V) and let
P (z) be its characteristic polynomial. Then P (T) = 0.
T T
Example 10. Before proving the Cayley-Hamilton Theorem, we
illustrate it using the linear map T in Example 2. That map had