The document discusses recurrence relations and matrices defined by recurrence relations. It defines recurrence relations, recurrence matrices, and their applications. It proves that arithmetic recurrence matrices have rank 2, geometric recurrence matrices have rank 1, and general homogeneous recurrence matrices have rank at most the order of the recurrence relation. It provides two examples to illustrate these results and discusses a case where the rank of an order-2 homogeneous recurrence matrix is 1.

1. Rank of Recurrence Matrices
Roman David Morales
St. Edward’s University - Austin, Texas
Senior Seminar Spring 2016
April 20, 2016

2. Definitions
• A recurrence relation defines a sequence of numbers as a
function of the preceding terms given one or more initial terms

3. Definitions
• A recurrence relation defines a sequence of numbers as a
function of the preceding terms given one or more initial terms
• An order-r homogeneous recurrence relation has the form
ak = γ1ak−1 + · · · + γr ak−r for constants γi and γr = 0

4. Definitions
• A recurrence relation defines a sequence of numbers as a
function of the preceding terms given one or more initial terms
• An order-r homogeneous recurrence relation has the form
ak = γ1ak−1 + · · · + γr ak−r for constants γi and γr = 0
• An m × n matrix whose entries read row-by-row are defined by
a recurrence relation is a recurrence matrix

5. Definitions
• A recurrence relation defines a sequence of numbers as a
function of the preceding terms given one or more initial terms
• An order-r homogeneous recurrence relation has the form
ak = γ1ak−1 + · · · + γr ak−r for constants γi and γr = 0
• An m × n matrix whose entries read row-by-row are defined by
a recurrence relation is a recurrence matrix
• Applications to dynamic response of aircraft in gusts

6. Definitions
• A recurrence relation defines a sequence of numbers as a
function of the preceding terms given one or more initial terms
• An order-r homogeneous recurrence relation has the form
ak = γ1ak−1 + · · · + γr ak−r for constants γi and γr = 0
• An m × n matrix whose entries read row-by-row are defined by
a recurrence relation is a recurrence matrix
• Applications to dynamic response of aircraft in gusts
• The rank of a matrix is the maximum number of linearly
independent columns or rows

7. Definitions
• A recurrence relation defines a sequence of numbers as a
function of the preceding terms given one or more initial terms
• An order-r homogeneous recurrence relation has the form
ak = γ1ak−1 + · · · + γr ak−r for constants γi and γr = 0
• An m × n matrix whose entries read row-by-row are defined by
a recurrence relation is a recurrence matrix
• Applications to dynamic response of aircraft in gusts
• The rank of a matrix is the maximum number of linearly
independent columns or rows
• What is the rank of a recurrence matrix?

8. Arithmetic Sequence
Theorem
Every m × n arithmetic matrix A with m, n ≥ 2 has rank 2.

9. Arithmetic Sequence
Theorem
Every m × n arithmetic matrix A with m, n ≥ 2 has rank 2.
Proof. Entries of AT (read column-by-column) are terms of an
arithmetic sequence defined by ak = a1 + (k − 1)x with a1, x = 0

10. Arithmetic Sequence
Theorem
Every m × n arithmetic matrix A with m, n ≥ 2 has rank 2.
Proof. Entries of AT (read column-by-column) are terms of an
arithmetic sequence defined by ak = a1 + (k − 1)x with a1, x = 0
(i, j)-entry of A: a(j−1)n+i = a1 + ((j − 1)n + i − 1)x

11. Arithmetic Sequence
Theorem
Every m × n arithmetic matrix A with m, n ≥ 2 has rank 2.
Proof. Entries of AT (read column-by-column) are terms of an
arithmetic sequence defined by ak = a1 + (k − 1)x with a1, x = 0
(i, j)-entry of A: a(j−1)n+i = a1 + ((j − 1)n + i − 1)x
For i = 3 . . . n, replacing Row i with
(Row i) + (i − 2)(Row 1) − (i − 1)(Row 2)
reduces entries in Row i to 0

12. Arithmetic Sequence Proof cont.
Replacing Row 2 with (Row 2) − a1+x
a1
(Row 1) reduces the jth
entry of Row 2 to

13. Arithmetic Sequence Proof cont.
Replacing Row 2 with (Row 2) − a1+x
a1
(Row 1) reduces the jth
entry of Row 2 to
−
(j − 1)nx2
a1

14. Arithmetic Sequence Proof cont.
Replacing Row 2 with (Row 2) − a1+x
a1
(Row 1) reduces the jth
entry of Row 2 to
−
(j − 1)nx2
a1
Thus, rank(A) = 2 because the (2, 1)-entry is 0 and all other
entries in Row 2 are nonzero since a1, x = 0

15. Geometric Sequence
Theorem
Every m × n geometric matrix G has rank 1.

16. Geometric Sequence
Theorem
Every m × n geometric matrix G has rank 1.
Proof. The entries of GT (read column-by-column) are the terms
of a geometric sequence defined by ak = apk−1

17. Geometric Sequence
Theorem
Every m × n geometric matrix G has rank 1.
Proof. The entries of GT (read column-by-column) are the terms
of a geometric sequence defined by ak = apk−1
(i, j)-entry: a(j−1)n+i = ap(j−1)n+i−1

18. Geometric Sequence
Theorem
Every m × n geometric matrix G has rank 1.
Proof. The entries of GT (read column-by-column) are the terms
of a geometric sequence defined by ak = apk−1
(i, j)-entry: a(j−1)n+i = ap(j−1)n+i−1
Row 1: a(j−1)n+1 = ap(j−1)n

19. Geometric Sequence
Theorem
Every m × n geometric matrix G has rank 1.
Proof. The entries of GT (read column-by-column) are the terms
of a geometric sequence defined by ak = apk−1
(i, j)-entry: a(j−1)n+i = ap(j−1)n+i−1
Row 1: a(j−1)n+1 = ap(j−1)n
For i = 2 . . . n, replacing Row i with (Row i) − pi−1(Row 1)
reduces entries in Row i to a(j−1)n+i−1 − pi−1ap(j−1)n = 0

20. Geometric Sequence
Theorem
Every m × n geometric matrix G has rank 1.
Proof. The entries of GT (read column-by-column) are the terms
of a geometric sequence defined by ak = apk−1
(i, j)-entry: a(j−1)n+i = ap(j−1)n+i−1
Row 1: a(j−1)n+1 = ap(j−1)n
For i = 2 . . . n, replacing Row i with (Row i) − pi−1(Row 1)
reduces entries in Row i to a(j−1)n+i−1 − pi−1ap(j−1)n = 0
Hence, rank(G) = rank(GT ) = 1

21. Upper Bound on Rank
Lee and Peterson (2014): order-r homogeneous recurrence
relations yield m × n recurrence matrices, R, of rank at most r

22. Upper Bound on Rank
Lee and Peterson (2014): order-r homogeneous recurrence
relations yield m × n recurrence matrices, R, of rank at most r
Proof. Entries of RT (read column-by-column) are defined by an
order-r homogeneous recurrence relation
ak = γ1ak−1 + · · · + γr ak−r

23. Upper Bound on Rank
Lee and Peterson (2014): order-r homogeneous recurrence
relations yield m × n recurrence matrices, R, of rank at most r
Proof. Entries of RT (read column-by-column) are defined by an
order-r homogeneous recurrence relation
ak = γ1ak−1 + · · · + γr ak−r
(i, j)-entry of RT : γ1a(j−1)n+i−1 + · · · + γr a(j−1)n+i−r

24. Upper Bound on Rank
Lee and Peterson (2014): order-r homogeneous recurrence
relations yield m × n recurrence matrices, R, of rank at most r
Proof. Entries of RT (read column-by-column) are defined by an
order-r homogeneous recurrence relation
ak = γ1ak−1 + · · · + γr ak−r
(i, j)-entry of RT : γ1a(j−1)n+i−1 + · · · + γr a(j−1)n+i−r
For i = n, ..., (r + 1), replacing Row i of RT with
(Row i) − γ1(Row (i − 1)) − · · · − γr (Row (i − r))

25. Upper Bound on Rank
Lee and Peterson (2014): order-r homogeneous recurrence
relations yield m × n recurrence matrices, R, of rank at most r
Proof. Entries of RT (read column-by-column) are defined by an
order-r homogeneous recurrence relation
ak = γ1ak−1 + · · · + γr ak−r
(i, j)-entry of RT : γ1a(j−1)n+i−1 + · · · + γr a(j−1)n+i−r
For i = n, ..., (r + 1), replacing Row i of RT with
(Row i) − γ1(Row (i − 1)) − · · · − γr (Row (i − r))
reduces entries in Row i to
a(j−1)n+i − γ1a(j−1)n+i−1 − · · · − γr a(j−1)n+i−r = 0

26. Upper Bound on Rank
Lee and Peterson (2014): order-r homogeneous recurrence
relations yield m × n recurrence matrices, R, of rank at most r
Proof. Entries of RT (read column-by-column) are defined by an
order-r homogeneous recurrence relation
ak = γ1ak−1 + · · · + γr ak−r
(i, j)-entry of RT : γ1a(j−1)n+i−1 + · · · + γr a(j−1)n+i−r
For i = n, ..., (r + 1), replacing Row i of RT with
(Row i) − γ1(Row (i − 1)) − · · · − γr (Row (i − r))
reduces entries in Row i to
a(j−1)n+i − γ1a(j−1)n+i−1 − · · · − γr a(j−1)n+i−r = 0
Therefore, rank (R) = rank(RT ) ≤ r

27. Example 1
Consider the order-two recurrence relation ak = 3ak−1 − 2ak−2
with initial seed a0 = 1 and a1 = 2, and corresponding recurrence
matrix
R =


1 2 4
8 16 32
64 128 256



28. Example 1
Consider the order-two recurrence relation ak = 3ak−1 − 2ak−2
with initial seed a0 = 1 and a1 = 2, and corresponding recurrence
matrix
R =


1 2 4
8 16 32
64 128 256


Each row is a multiple of the first row so we can reduce rows
beyond the first to zero.

29. Example 1
Consider the order-two recurrence relation ak = 3ak−1 − 2ak−2
with initial seed a0 = 1 and a1 = 2, and corresponding recurrence
matrix
R =


1 2 4
8 16 32
64 128 256


Each row is a multiple of the first row so we can reduce rows
beyond the first to zero.
This means the recurrence matrix will have rank 1

30. Example 1
Consider the order-two recurrence relation ak = 3ak−1 − 2ak−2
with initial seed a0 = 1 and a1 = 2, and corresponding recurrence
matrix
R =


1 2 4
8 16 32
64 128 256


Each row is a multiple of the first row so we can reduce rows
beyond the first to zero.
This means the recurrence matrix will have rank 1
This takes place because it can be rewritten to have a lower order:
ak = 2ak−1

31. Example 2
Consider the recurrence relation ak = ak−2 where a0 = 4, a1 = 0

32. Example 2
Consider the recurrence relation ak = ak−2 where a0 = 4, a1 = 0
If we construct a 3 × 3 matrix, R, from this sequence, rank(R) = 2
R =


4 0 4
0 4 0
4 0 4



33. Example 2
Consider the recurrence relation ak = ak−2 where a0 = 4, a1 = 0
If we construct a 3 × 3 matrix, R, from this sequence, rank(R) = 2
R =


4 0 4
0 4 0
4 0 4


If we construct a 4 × 4 matrix, R, from the sequence, rank(R) = 1
R =




4 0 4 0
4 0 4 0
4 0 4 0
4 0 4 0




The matrix width changes the rank of the recurrence matrix.

34. Order-2 Homogeneous Relations
• Lee and Peterson (2014): order-2 homogeneous recurrence
relations yield m × n matrices, R, of rank 1 whenever a2
a1
is an
eigenvalue of the relation with seeds a1 and a2

35. Order-2 Homogeneous Relations
• Lee and Peterson (2014): order-2 homogeneous recurrence
relations yield m × n matrices, R, of rank 1 whenever a2
a1
is an
eigenvalue of the relation with seeds a1 and a2
• The eigenvalues of an order-2 homogeneous recurrence
relation ak = γ1ak−1 + γ2ak−2 are the roots of the equation
x2 − γ1x − γ2 = 0

36. Order-2 Homogeneous Relations
• Lee and Peterson (2014): order-2 homogeneous recurrence
relations yield m × n matrices, R, of rank 1 whenever a2
a1
is an
eigenvalue of the relation with seeds a1 and a2
• The eigenvalues of an order-2 homogeneous recurrence
relation ak = γ1ak−1 + γ2ak−2 are the roots of the equation
x2 − γ1x − γ2 = 0
• We prove Lee and Peterson’s result using fundamental
solutions of the recurrence relations and find another case in
which rank drops

37. Case 1: One repeated eigenvalue
Proof. If x2 − γ1x − γ2 = 0 has a single repeated root λ, then its
fundamental solutions are λk−1 and kλk−1 and its general solution
is ak = aλk−1 + bkλk−1

38. Case 1: One repeated eigenvalue
Proof. If x2 − γ1x − γ2 = 0 has a single repeated root λ, then its
fundamental solutions are λk−1 and kλk−1 and its general solution
is ak = aλk−1 + bkλk−1
To find a and b, we solve the system of equations
a1 = a + b, a2 = aλ + 2bλ by row reducing the augmented matrix:

39. Case 1: One repeated eigenvalue
Proof. If x2 − γ1x − γ2 = 0 has a single repeated root λ, then its
fundamental solutions are λk−1 and kλk−1 and its general solution
is ak = aλk−1 + bkλk−1
To find a and b, we solve the system of equations
a1 = a + b, a2 = aλ + 2bλ by row reducing the augmented matrix:
1 1 a1
λ 2λ a2

40. Case 1: One repeated eigenvalue
Proof. If x2 − γ1x − γ2 = 0 has a single repeated root λ, then its
fundamental solutions are λk−1 and kλk−1 and its general solution
is ak = aλk−1 + bkλk−1
To find a and b, we solve the system of equations
a1 = a + b, a2 = aλ + 2bλ by row reducing the augmented matrix:
1 1 a1
λ 2λ a2
a =
2λa1 − a2
λ
b =
a2 − λa1
λ

41. Case 1: One repeated eigenvalue
Proof. If x2 − γ1x − γ2 = 0 has a single repeated root λ, then its
fundamental solutions are λk−1 and kλk−1 and its general solution
is ak = aλk−1 + bkλk−1
To find a and b, we solve the system of equations
a1 = a + b, a2 = aλ + 2bλ by row reducing the augmented matrix:
1 1 a1
λ 2λ a2
a =
2λa1 − a2
λ
b =
a2 − λa1
λ
ak = λk−1
(2 − k)a1 + λk−2
(k − 1)a2

42. Case 1 proof cont.
(i, j)-entry of RT (read column-by-column) is:
λ(j−1)n+i−1(2 − (j − 1)n − i)a1 + λ(j−1)n+i−2((j − 1)n + i − 1)a2

43. Case 1 proof cont.
(i, j)-entry of RT (read column-by-column) is:
λ(j−1)n+i−1(2 − (j − 1)n − i)a1 + λ(j−1)n+i−2((j − 1)n + i − 1)a2
Row 1: λ(j−1)n(1 − (j − 1)n)a1 + λ(j−1)n−1((j − 1)n)a2

44. Case 1 proof cont.
(i, j)-entry of RT (read column-by-column) is:
λ(j−1)n+i−1(2 − (j − 1)n − i)a1 + λ(j−1)n+i−2((j − 1)n + i − 1)a2
Row 1: λ(j−1)n(1 − (j − 1)n)a1 + λ(j−1)n−1((j − 1)n)a2
Row 2: λ(j−1)n+1(1 − j)na1 + λ(j−1)n((j − 1)n + 1)a2

45. Case 1 proof cont.
(i, j)-entry of RT (read column-by-column) is:
λ(j−1)n+i−1(2 − (j − 1)n − i)a1 + λ(j−1)n+i−2((j − 1)n + i − 1)a2
Row 1: λ(j−1)n(1 − (j − 1)n)a1 + λ(j−1)n−1((j − 1)n)a2
Row 2: λ(j−1)n+1(1 − j)na1 + λ(j−1)n((j − 1)n + 1)a2
Replacing Row 2 with (Row 2) − a2
a1
(Row 1)
reduces entries in Row 2 to −λ(j−1)n−1(j−1)n
a1
(λa1 − a2)2

46. Case 1 proof cont.
(i, j)-entry of RT (read column-by-column) is:
λ(j−1)n+i−1(2 − (j − 1)n − i)a1 + λ(j−1)n+i−2((j − 1)n + i − 1)a2
Row 1: λ(j−1)n(1 − (j − 1)n)a1 + λ(j−1)n−1((j − 1)n)a2
Row 2: λ(j−1)n+1(1 − j)na1 + λ(j−1)n((j − 1)n + 1)a2
Replacing Row 2 with (Row 2) − a2
a1
(Row 1)
reduces entries in Row 2 to −λ(j−1)n−1(j−1)n
a1
(λa1 − a2)2
This is 0 iff λ = a2
a1
, so rank(R) = rank (RT ) = 1 iff λ = a2
a1

47. Case 2: Two distinct eigenvalues
If x2 − γ1x − γ2 = 0 has two distinct roots, λ1 and λ2, then its
fundamental solutions are λk−1
1 and λk−1
2 , and its general solution
is ak = aλk−1
1 + bλk−1
2

48. Case 2: Two distinct eigenvalues
If x2 − γ1x − γ2 = 0 has two distinct roots, λ1 and λ2, then its
fundamental solutions are λk−1
1 and λk−1
2 , and its general solution
is ak = aλk−1
1 + bλk−1
2
To find a and b, we solve the system of equations a1 = a + b,
a2 = aλ1 + bλ2 by row reducing the augmented matrix:

49. Case 2: Two distinct eigenvalues
If x2 − γ1x − γ2 = 0 has two distinct roots, λ1 and λ2, then its
fundamental solutions are λk−1
1 and λk−1
2 , and its general solution
is ak = aλk−1
1 + bλk−1
2
To find a and b, we solve the system of equations a1 = a + b,
a2 = aλ1 + bλ2 by row reducing the augmented matrix:
1 1 a1
λ1 λ2 a2

50. Case 2: Two distinct eigenvalues
If x2 − γ1x − γ2 = 0 has two distinct roots, λ1 and λ2, then its
fundamental solutions are λk−1
1 and λk−1
2 , and its general solution
is ak = aλk−1
1 + bλk−1
2
To find a and b, we solve the system of equations a1 = a + b,
a2 = aλ1 + bλ2 by row reducing the augmented matrix:
1 1 a1
λ1 λ2 a2
a =
λ2a1 − a2
λ2 − λ1
b =
a2 − λ1a1
λ2 − λ1

51. Case 2: Two distinct eigenvalues
If x2 − γ1x − γ2 = 0 has two distinct roots, λ1 and λ2, then its
fundamental solutions are λk−1
1 and λk−1
2 , and its general solution
is ak = aλk−1
1 + bλk−1
2
To find a and b, we solve the system of equations a1 = a + b,
a2 = aλ1 + bλ2 by row reducing the augmented matrix:
1 1 a1
λ1 λ2 a2
a =
λ2a1 − a2
λ2 − λ1
b =
a2 − λ1a1
λ2 − λ1
ak =
λ2λk−1
1 − λ1λk−1
2
λ2 − λ1
a1 +
λk−1
2 − λk−1
1
λ2 − λ1
a2

52. Case 2 proof cont.
(i, j)-entry of RT (read column-by-column) is:
λ2λ
(j−1)n+i−1
1 − λ1λ
(j−1)n+i−1
2
λ2 − λ1
a1 +
λ
(j−1)n+i−1
2 − λ
(j−1)n+i−1
1
λ2 − λ1
a2

53. Case 2 proof cont.
(i, j)-entry of RT (read column-by-column) is:
λ2λ
(j−1)n+i−1
1 − λ1λ
(j−1)n+i−1
2
λ2 − λ1
a1 +
λ
(j−1)n+i−1
2 − λ
(j−1)n+i−1
1
λ2 − λ1
a2
Row 1:
λ2λ
(j−1)n
1 −λ1λ
(j−1)n
2
λ2−λ1
a1 +
λ
(j−1)n
2 −λ
(j−1)n
1
λ2−λ1
a2

54. Case 2 proof cont.
(i, j)-entry of RT (read column-by-column) is:
λ2λ
(j−1)n+i−1
1 − λ1λ
(j−1)n+i−1
2
λ2 − λ1
a1 +
λ
(j−1)n+i−1
2 − λ
(j−1)n+i−1
1
λ2 − λ1
a2
Row 1:
λ2λ
(j−1)n
1 −λ1λ
(j−1)n
2
λ2−λ1
a1 +
λ
(j−1)n
2 −λ
(j−1)n
1
λ2−λ1
a2
Row 2:
λ2λ
(j−1)n+1
1 −λ1λ
(j−1)n+1
2
λ2−λ1
a1 +
λ
(j−1)n+1
2 −λ
(j−1)n+1
1
λ2−λ1
a2

55. Case 2 proof cont.
(i, j)-entry of RT (read column-by-column) is:
λ2λ
(j−1)n+i−1
1 − λ1λ
(j−1)n+i−1
2
λ2 − λ1
a1 +
λ
(j−1)n+i−1
2 − λ
(j−1)n+i−1
1
λ2 − λ1
a2
Row 1:
λ2λ
(j−1)n
1 −λ1λ
(j−1)n
2
λ2−λ1
a1 +
λ
(j−1)n
2 −λ
(j−1)n
1
λ2−λ1
a2
Row 2:
λ2λ
(j−1)n+1
1 −λ1λ
(j−1)n+1
2
λ2−λ1
a1 +
λ
(j−1)n+1
2 −λ
(j−1)n+1
1
λ2−λ1
a2
Replacing Row 2 with (Row 2) − a2
a1
(Row 1)
reduces entries in Row 2 to
λ
(j−1)n
1 −λ
(j−1)n
2
(λ2−λ1)a1
(a2 − λ1a1)(a2 − λ2a1)

56. Case 2 proof cont.
(i, j)-entry of RT (read column-by-column) is:
λ2λ
(j−1)n+i−1
1 − λ1λ
(j−1)n+i−1
2
λ2 − λ1
a1 +
λ
(j−1)n+i−1
2 − λ
(j−1)n+i−1
1
λ2 − λ1
a2
Row 1:
λ2λ
(j−1)n
1 −λ1λ
(j−1)n
2
λ2−λ1
a1 +
λ
(j−1)n
2 −λ
(j−1)n
1
λ2−λ1
a2
Row 2:
λ2λ
(j−1)n+1
1 −λ1λ
(j−1)n+1
2
λ2−λ1
a1 +
λ
(j−1)n+1
2 −λ
(j−1)n+1
1
λ2−λ1
a2
Replacing Row 2 with (Row 2) − a2
a1
(Row 1)
reduces entries in Row 2 to
λ
(j−1)n
1 −λ
(j−1)n
2
(λ2−λ1)a1
(a2 − λ1a1)(a2 − λ2a1)
This is 0 iff λ1 = a2
a1
, λ2 = a2
a1
, or λn
1 = λn
2

57. Case 2 proof cont.
(i, j)-entry of RT (read column-by-column) is:
λ2λ
(j−1)n+i−1
1 − λ1λ
(j−1)n+i−1
2
λ2 − λ1
a1 +
λ
(j−1)n+i−1
2 − λ
(j−1)n+i−1
1
λ2 − λ1
a2
Row 1:
λ2λ
(j−1)n
1 −λ1λ
(j−1)n
2
λ2−λ1
a1 +
λ
(j−1)n
2 −λ
(j−1)n
1
λ2−λ1
a2
Row 2:
λ2λ
(j−1)n+1
1 −λ1λ
(j−1)n+1
2
λ2−λ1
a1 +
λ
(j−1)n+1
2 −λ
(j−1)n+1
1
λ2−λ1
a2
Replacing Row 2 with (Row 2) − a2
a1
(Row 1)
reduces entries in Row 2 to
λ
(j−1)n
1 −λ
(j−1)n
2
(λ2−λ1)a1
(a2 − λ1a1)(a2 − λ2a1)
This is 0 iff λ1 = a2
a1
, λ2 = a2
a1
, or λn
1 = λn
2
Thus, rank(RT ) = 1 iff λ1 = a2
a1
, λ2 = a2
a1
, or
λ1
λ2
n
= 1

58. Generalization of Lee & Peterson’s Theorem
Bozlee (2015): If R is an m × n matrix with m, n ≥ 2 whose
entries (read row-by-row) are given by an order-2 homogeneous
recurrence relation ak = γ1ak−1 + γ2ak−2, then
rank(R) =



0 if a0 = a1 = 0

59. Generalization of Lee & Peterson’s Theorem
Bozlee (2015): If R is an m × n matrix with m, n ≥ 2 whose
entries (read row-by-row) are given by an order-2 homogeneous
recurrence relation ak = γ1ak−1 + γ2ak−2, then
rank(R) =



0 if a0 = a1 = 0
1 if a2
1 − γ1a1a0 − γ2a2
0 = 0

60. Generalization of Lee & Peterson’s Theorem
Bozlee (2015): If R is an m × n matrix with m, n ≥ 2 whose
entries (read row-by-row) are given by an order-2 homogeneous
recurrence relation ak = γ1ak−1 + γ2ak−2, then
rank(R) =



0 if a0 = a1 = 0
1 if a2
1 − γ1a1a0 − γ2a2
0 = 0
1 if γ2
1 + 4γ2 = 0 and
γ1+
√
γ2
1 +4γ2
γ1−
√
γ2
1 +4γ2
n
= 1

61. Generalization of Lee & Peterson’s Theorem
Bozlee (2015): If R is an m × n matrix with m, n ≥ 2 whose
entries (read row-by-row) are given by an order-2 homogeneous
recurrence relation ak = γ1ak−1 + γ2ak−2, then
rank(R) =



0 if a0 = a1 = 0
1 if a2
1 − γ1a1a0 − γ2a2
0 = 0
1 if γ2
1 + 4γ2 = 0 and
γ1+
√
γ2
1 +4γ2
γ1−
√
γ2
1 +4γ2
n
= 1
2 else

62. Generalization of Lee & Peterson’s Theorem
Bozlee (2015): If R is an m × n matrix with m, n ≥ 2 whose
entries (read row-by-row) are given by an order-2 homogeneous
recurrence relation ak = γ1ak−1 + γ2ak−2, then
rank(R) =



0 if a0 = a1 = 0
1 if a2
1 − γ1a1a0 − γ2a2
0 = 0
1 if γ2
1 + 4γ2 = 0 and
γ1+
√
γ2
1 +4γ2
γ1−
√
γ2
1 +4γ2
n
= 1
2 else
So rank drops when recurrence relation has lower order or when
the nth powers of eigenvalues coincide

63. Example 1 revisited
Consider the recurrence relation ak = 3ak−1 − 2ak−2 with initial
seeds a0 = 1 and a1 = 2

64. Example 1 revisited
Consider the recurrence relation ak = 3ak−1 − 2ak−2 with initial
seeds a0 = 1 and a1 = 2
Its corresponding characteristic polynomial is: x2 − 3x + 2

65. Example 1 revisited
Consider the recurrence relation ak = 3ak−1 − 2ak−2 with initial
seeds a0 = 1 and a1 = 2
Its corresponding characteristic polynomial is: x2 − 3x + 2
Solving the equation for x2 − 3x + 2 = 0 gives us the roots:
x = 2, 1

66. Example 1 revisited
Consider the recurrence relation ak = 3ak−1 − 2ak−2 with initial
seeds a0 = 1 and a1 = 2
Its corresponding characteristic polynomial is: x2 − 3x + 2
Solving the equation for x2 − 3x + 2 = 0 gives us the roots:
x = 2, 1
Notice the ration of the relation first two seeds a1
a0
= 2
1 = 2, which
is an eigenvalue of the recurrence relation.

67. Example 1 revisited
Consider the recurrence relation ak = 3ak−1 − 2ak−2 with initial
seeds a0 = 1 and a1 = 2
Its corresponding characteristic polynomial is: x2 − 3x + 2
Solving the equation for x2 − 3x + 2 = 0 gives us the roots:
x = 2, 1
Notice the ration of the relation first two seeds a1
a0
= 2
1 = 2, which
is an eigenvalue of the recurrence relation.
Hence, the rank will drop to 1 as shown in the example

68. Example 2 revisited
Consider the recurrence relation ak = ak−2 with initial seeds
a0 = 4 and a1 = 0

69. Example 2 revisited
Consider the recurrence relation ak = ak−2 with initial seeds
a0 = 4 and a1 = 0
Its corresponding characteristic polynomial is: x2 − 1

70. Example 2 revisited
Consider the recurrence relation ak = ak−2 with initial seeds
a0 = 4 and a1 = 0
Its corresponding characteristic polynomial is: x2 − 1
Solving the equation for x2 − 1 = 0 gives us the roots: x = −1, 1

71. Example 2 revisited
Consider the recurrence relation ak = ak−2 with initial seeds
a0 = 4 and a1 = 0
Its corresponding characteristic polynomial is: x2 − 1
Solving the equation for x2 − 1 = 0 gives us the roots: x = −1, 1
Notice the ration of the eigenvalues’ nth powers coincide for
(−1)n = (1)n when n is even

72. Example 2 revisited
Consider the recurrence relation ak = ak−2 with initial seeds
a0 = 4 and a1 = 0
Its corresponding characteristic polynomial is: x2 − 1
Solving the equation for x2 − 1 = 0 gives us the roots: x = −1, 1
Notice the ration of the eigenvalues’ nth powers coincide for
(−1)n = (1)n when n is even
Hence, the rank will be 2 when the relation’s corresponding
recurrence matrix is an n × n matrix with odd n, and rank will be 1
when the relation’s corresponding recurrence matrix is an n × n
matrix with even n

73. Order-3 Homogeneous Relations
Theorem
An m × n recurrence matrix R whose entries come from an
order-three homogeneous recurrence relation with seeds a0, a1, a2
has rank 2 if and only if a2 = (λ1 + λ2)a1 − λ1λ2a0 where λ1 and
λ2 are (not necessarily distinct) eigenvalues of the recurrence
relation or if λn
1 = λn
2 for distinct eigenvalues of the recurrence
relation, and rank 1 if and only if, in addition, a1 = λa0 where λ is
an eigenvalue of the recurrence relation, or λn
2 = λn
3 for another
pair of distinct eigenvalues of the recurrence relation.

74. Order-3 recurrence relation with single repeated eigenvalue
Proof. To find a, b, and c in ak = aλk + bkλk + ck2λk:

75. Order-3 recurrence relation with single repeated eigenvalue
Proof. To find a, b, and c in ak = aλk + bkλk + ck2λk:


1 0 0
λ λ λ
λ2
λ2
λ2




a
b
c

 =


a0
a1
a2

 =⇒

76. Order-3 recurrence relation with single repeated eigenvalue
Proof. To find a, b, and c in ak = aλk + bkλk + ck2λk:


1 0 0
λ λ λ
λ2
λ2
λ2




a
b
c

 =


a0
a1
a2

 =⇒


a
b
c

 = 1
2λ3


2λ3
a0
−λ(3a0λ2
+ 4a1λ + a2)
λ(a0λ2
− 2a1λ + a2)



77. Order-3 recurrence relation with single repeated eigenvalue
Proof. To find a, b, and c in ak = aλk + bkλk + ck2λk:


1 0 0
λ λ λ
λ2
λ2
λ2




a
b
c

 =


a0
a1
a2

 =⇒


a
b
c

 = 1
2λ3


2λ3
a0
−λ(3a0λ2
+ 4a1λ + a2)
λ(a0λ2
− 2a1λ + a2)


So matrix has rank:
• 3 if a, b, c are non-zero.

78. Order-3 recurrence relation with single repeated eigenvalue
Proof. To find a, b, and c in ak = aλk + bkλk + ck2λk:


1 0 0
λ λ λ
λ2
λ2
λ2




a
b
c

 =


a0
a1
a2

 =⇒


a
b
c

 = 1
2λ3


2λ3
a0
−λ(3a0λ2
+ 4a1λ + a2)
λ(a0λ2
− 2a1λ + a2)


So matrix has rank:
• 3 if a, b, c are non-zero.
• 2 if a and b are non-zero, but c = 0 = a2 − 2λa1 + λ2a0
• c = 0 if a2 = 2λa1 + λ2
a0

79. Order-3 recurrence relation with single repeated eigenvalue
Proof. To find a, b, and c in ak = aλk + bkλk + ck2λk:


1 0 0
λ λ λ
λ2
λ2
λ2




a
b
c

 =


a0
a1
a2

 =⇒


a
b
c

 = 1
2λ3


2λ3
a0
−λ(3a0λ2
+ 4a1λ + a2)
λ(a0λ2
− 2a1λ + a2)


So matrix has rank:
• 3 if a, b, c are non-zero.
• 2 if a and b are non-zero, but c = 0 = a2 − 2λa1 + λ2a0
• c = 0 if a2 = 2λa1 + λ2
a0
• 1 if a is non-zero, but b = 4λa1 − 3λ2a0 − a2 = c = 0
• b = c = 0 if 4λa1 − 3λ2
a0 − 2λa1 + λ2
a0=2λ(a1 − λa0) = 0
• 2λ(a1 − λa0) = 0 if a1=λa0

80. Order-3 recurrence relation with single repeated eigenvalue
Proof. To find a, b, and c in ak = aλk + bkλk + ck2λk:


1 0 0
λ λ λ
λ2
λ2
λ2




a
b
c

 =


a0
a1
a2

 =⇒


a
b
c

 = 1
2λ3


2λ3
a0
−λ(3a0λ2
+ 4a1λ + a2)
λ(a0λ2
− 2a1λ + a2)


So matrix has rank:
• 3 if a, b, c are non-zero.
• 2 if a and b are non-zero, but c = 0 = a2 − 2λa1 + λ2a0
• c = 0 if a2 = 2λa1 + λ2
a0
• 1 if a is non-zero, but b = 4λa1 − 3λ2a0 − a2 = c = 0
• b = c = 0 if 4λa1 − 3λ2
a0 − 2λa1 + λ2
a0=2λ(a1 − λa0) = 0
• 2λ(a1 − λa0) = 0 if a1=λa0
• 0 if a = b = c = 0

81. Order-3 recurrence relation with single repeated eigenvalue
Proof. To find a, b, and c in ak = aλk + bkλk + ck2λk:


1 0 0
λ λ λ
λ2
λ2
λ2




a
b
c

 =


a0
a1
a2

 =⇒


a
b
c

 = 1
2λ3


2λ3
a0
−λ(3a0λ2
+ 4a1λ + a2)
λ(a0λ2
− 2a1λ + a2)


So matrix has rank:
• 3 if a, b, c are non-zero.
• 2 if a and b are non-zero, but c = 0 = a2 − 2λa1 + λ2a0
• c = 0 if a2 = 2λa1 + λ2
a0
• 1 if a is non-zero, but b = 4λa1 − 3λ2a0 − a2 = c = 0
• b = c = 0 if 4λa1 − 3λ2
a0 − 2λa1 + λ2
a0=2λ(a1 − λa0) = 0
• 2λ(a1 − λa0) = 0 if a1=λa0
• 0 if a = b = c = 0
• There are no width rank drops present in this case since there
exists only one eigenvalue

82. Order-3 recurrence relation with three distinct eigenvalues
To find a, b, and c in ak = aλk
1 + bλk
2 + cλk
3:

83. Order-3 recurrence relation with three distinct eigenvalues
To find a, b, and c in ak = aλk
1 + bλk
2 + cλk
3:


1 1 1
λ1 λ2 λ3
λ2
1 λ2
2 λ2
3




a
b
c

 =


a0
a1
a2

 =⇒

84. Order-3 recurrence relation with three distinct eigenvalues
To find a, b, and c in ak = aλk
1 + bλk
2 + cλk
3:


1 1 1
λ1 λ2 λ3
λ2
1 λ2
2 λ2
3




a
b
c

 =


a0
a1
a2

 =⇒


a
b
c

=




a0λ1λ2−a1λ1−a1λ2+a2
λ1λ2−λ1λ3−λ1λ2−λ2
3
−a0λ1λ3+a1λ1+a1λ3−a2
λ1λ2−λ1λ3+λ1λ2−λ2
2
a0λ2λ3−a1λ2−a1λ3+a2
λ2λ3−λ1λ3λ1λ2+λ2
1





85. Order-3 recurrence relation with three distinct eigenvalues
To find a, b, and c in ak = aλk
1 + bλk
2 + cλk
3:


1 1 1
λ1 λ2 λ3
λ2
1 λ2
2 λ2
3




a
b
c

 =


a0
a1
a2

 =⇒


a
b
c

=




a0λ1λ2−a1λ1−a1λ2+a2
λ1λ2−λ1λ3−λ1λ2−λ2
3
−a0λ1λ3+a1λ1+a1λ3−a2
λ1λ2−λ1λ3+λ1λ2−λ2
2
a0λ2λ3−a1λ2−a1λ3+a2
λ2λ3−λ1λ3λ1λ2+λ2
1




So matrix has rank:
• 3 if λn
1 = λn
2, λn
1 = λn
3, or λn
2 = λn
3, and a, b, c are non-zero.

86. Order-3 recurrence relation with three distinct eigenvalues
To find a, b, and c in ak = aλk
1 + bλk
2 + cλk
3:


1 1 1
λ1 λ2 λ3
λ2
1 λ2
2 λ2
3




a
b
c

 =


a0
a1
a2

 =⇒


a
b
c

=




a0λ1λ2−a1λ1−a1λ2+a2
λ1λ2−λ1λ3−λ1λ2−λ2
3
−a0λ1λ3+a1λ1+a1λ3−a2
λ1λ2−λ1λ3+λ1λ2−λ2
2
a0λ2λ3−a1λ2−a1λ3+a2
λ2λ3−λ1λ3λ1λ2+λ2
1




So matrix has rank:
• 3 if λn
1 = λn
2, λn
1 = λn
3, or λn
2 = λn
3, and a, b, c are non-zero.
• 2 if λn
1 = λn
2, λn
1 = λn
3, or λn
2 = λn
3

87. Order-3 recurrence relation with three distinct eigenvalues
To find a, b, and c in ak = aλk
1 + bλk
2 + cλk
3:


1 1 1
λ1 λ2 λ3
λ2
1 λ2
2 λ2
3




a
b
c

 =


a0
a1
a2

 =⇒


a
b
c

=




a0λ1λ2−a1λ1−a1λ2+a2
λ1λ2−λ1λ3−λ1λ2−λ2
3
−a0λ1λ3+a1λ1+a1λ3−a2
λ1λ2−λ1λ3+λ1λ2−λ2
2
a0λ2λ3−a1λ2−a1λ3+a2
λ2λ3−λ1λ3λ1λ2+λ2
1




So matrix has rank:
• 3 if λn
1 = λn
2, λn
1 = λn
3, or λn
2 = λn
3, and a, b, c are non-zero.
• 2 if λn
1 = λn
2, λn
1 = λn
3, or λn
2 = λn
3
• 2 if a and b are non-zero, but c = 0
• c = 0 if a2 = (λ2 + λ3)a1 − λ2λ3a0

88. Order-3 recurrence relation with three distinct eigenvalues
To find a, b, and c in ak = aλk
1 + bλk
2 + cλk
3:


1 1 1
λ1 λ2 λ3
λ2
1 λ2
2 λ2
3




a
b
c

 =


a0
a1
a2

 =⇒


a
b
c

=




a0λ1λ2−a1λ1−a1λ2+a2
λ1λ2−λ1λ3−λ1λ2−λ2
3
−a0λ1λ3+a1λ1+a1λ3−a2
λ1λ2−λ1λ3+λ1λ2−λ2
2
a0λ2λ3−a1λ2−a1λ3+a2
λ2λ3−λ1λ3λ1λ2+λ2
1




So matrix has rank:
• 3 if λn
1 = λn
2, λn
1 = λn
3, or λn
2 = λn
3, and a, b, c are non-zero.
• 2 if λn
1 = λn
2, λn
1 = λn
3, or λn
2 = λn
3
• 2 if a and b are non-zero, but c = 0
• c = 0 if a2 = (λ2 + λ3)a1 − λ2λ3a0
• 1 if another pair of eigenvalues’ nth powers coincide (different
pair from rank dropping to 2)

89. Order-3 recurrence relation with three distinct eigenvalues
To find a, b, and c in ak = aλk
1 + bλk
2 + cλk
3:


1 1 1
λ1 λ2 λ3
λ2
1 λ2
2 λ2
3




a
b
c

 =


a0
a1
a2

 =⇒


a
b
c

=




a0λ1λ2−a1λ1−a1λ2+a2
λ1λ2−λ1λ3−λ1λ2−λ2
3
−a0λ1λ3+a1λ1+a1λ3−a2
λ1λ2−λ1λ3+λ1λ2−λ2
2
a0λ2λ3−a1λ2−a1λ3+a2
λ2λ3−λ1λ3λ1λ2+λ2
1




So matrix has rank:
• 3 if λn
1 = λn
2, λn
1 = λn
3, or λn
2 = λn
3, and a, b, c are non-zero.
• 2 if λn
1 = λn
2, λn
1 = λn
3, or λn
2 = λn
3
• 2 if a and b are non-zero, but c = 0
• c = 0 if a2 = (λ2 + λ3)a1 − λ2λ3a0
• 1 if another pair of eigenvalues’ nth powers coincide (different
pair from rank dropping to 2)
• 1 if a is non-zero, but b = a2 −λ1a1 −λ3a1 +λ1λ3a0 = c = 0
• b = 0 if λ1a1 + λ2a1 − λ1λ2a0 − λ1a1 − λ3a1 + λ1λ3a0 = c = 0
• c = a1(λ2 − λ3) − λ1(λ2 − λ3)a0
• c = 0 if a1 = λ1a0

90. Order-3 recurrence relation with three distinct eigenvalues
To find a, b, and c in ak = aλk
1 + bλk
2 + cλk
3:


1 1 1
λ1 λ2 λ3
λ2
1 λ2
2 λ2
3




a
b
c

 =


a0
a1
a2

 =⇒


a
b
c

=




a0λ1λ2−a1λ1−a1λ2+a2
λ1λ2−λ1λ3−λ1λ2−λ2
3
−a0λ1λ3+a1λ1+a1λ3−a2
λ1λ2−λ1λ3+λ1λ2−λ2
2
a0λ2λ3−a1λ2−a1λ3+a2
λ2λ3−λ1λ3λ1λ2+λ2
1




So matrix has rank:
• 3 if λn
1 = λn
2, λn
1 = λn
3, or λn
2 = λn
3, and a, b, c are non-zero.
• 2 if λn
1 = λn
2, λn
1 = λn
3, or λn
2 = λn
3
• 2 if a and b are non-zero, but c = 0
• c = 0 if a2 = (λ2 + λ3)a1 − λ2λ3a0
• 1 if another pair of eigenvalues’ nth powers coincide (different
pair from rank dropping to 2)
• 1 if a is non-zero, but b = a2 −λ1a1 −λ3a1 +λ1λ3a0 = c = 0
• b = 0 if λ1a1 + λ2a1 − λ1λ2a0 − λ1a1 − λ3a1 + λ1λ3a0 = c = 0
• c = a1(λ2 − λ3) − λ1(λ2 − λ3)a0
• c = 0 if a1 = λ1a0
• 0 if a = b = c = 0

91. Order-3 recurrence relation with one repeated and one
distinct eigenvalue
To find a, b, and c in ak = aλk
1 + bλk
2 + ckλk
2:

92. Order-3 recurrence relation with one repeated and one
distinct eigenvalue
To find a, b, and c in ak = aλk
1 + bλk
2 + ckλk
2:


1 1 0
λ1 λ2 λ2
λ2
1 λ2
2 2λ2
2




a
b
c

 =


a0
a1
a2

 =⇒

93. Order-3 recurrence relation with one repeated and one
distinct eigenvalue
To find a, b, and c in ak = aλk
1 + bλk
2 + ckλk
2:


1 1 0
λ1 λ2 λ2
λ2
1 λ2
2 2λ2
2




a
b
c

 =


a0
a1
a2

 =⇒


a
b
c

=




λ2
2a0−2λ2a1+a2
(λ1−λ2)2
λ2
1a0−2λ1λ2a0+2λ2a1−a2
(λ1−λ2)2
λ1a1−λ1λ2a0+λ2a1−a2
λ2(λ1−λ2)





94. Order-3 recurrence relation with one repeated and one
distinct eigenvalue
To find a, b, and c in ak = aλk
1 + bλk
2 + ckλk
2:


1 1 0
λ1 λ2 λ2
λ2
1 λ2
2 2λ2
2




a
b
c

 =


a0
a1
a2

 =⇒


a
b
c

=




λ2
2a0−2λ2a1+a2
(λ1−λ2)2
λ2
1a0−2λ1λ2a0+2λ2a1−a2
(λ1−λ2)2
λ1a1−λ1λ2a0+λ2a1−a2
λ2(λ1−λ2)




So matrix has rank:
• 3 if a, b, c are non-zero and λn
1 = λn
2

95. Order-3 recurrence relation with one repeated and one
distinct eigenvalue
To find a, b, and c in ak = aλk
1 + bλk
2 + ckλk
2:


1 1 0
λ1 λ2 λ2
λ2
1 λ2
2 2λ2
2




a
b
c

 =


a0
a1
a2

 =⇒


a
b
c

=




λ2
2a0−2λ2a1+a2
(λ1−λ2)2
λ2
1a0−2λ1λ2a0+2λ2a1−a2
(λ1−λ2)2
λ1a1−λ1λ2a0+λ2a1−a2
λ2(λ1−λ2)




So matrix has rank:
• 3 if a, b, c are non-zero and λn
1 = λn
2
• 2 if λn
1 = λn
2, or if a and b are non-zero, but
c = 0 = λ1a1 − λ1λ2a0 + λ2a1 − a2
• c = 0 if a2 = (λ1 + λ2)a1 − λ1λ2a0

96. Order-3 recurrence relation with one repeated and one
distinct eigenvalue
To find a, b, and c in ak = aλk
1 + bλk
2 + ckλk
2:


1 1 0
λ1 λ2 λ2
λ2
1 λ2
2 2λ2
2




a
b
c

 =


a0
a1
a2

 =⇒


a
b
c

=




λ2
2a0−2λ2a1+a2
(λ1−λ2)2
λ2
1a0−2λ1λ2a0+2λ2a1−a2
(λ1−λ2)2
λ1a1−λ1λ2a0+λ2a1−a2
λ2(λ1−λ2)




So matrix has rank:
• 3 if a, b, c are non-zero and λn
1 = λn
2
• 2 if λn
1 = λn
2, or if a and b are non-zero, but
c = 0 = λ1a1 − λ1λ2a0 + λ2a1 − a2
• c = 0 if a2 = (λ1 + λ2)a1 − λ1λ2a0
• 1 if a = 0, but b = λ2
1a0 − 2λ1λ2a0 + 2λ2a1 − a2 = c = 0
• b = 0 if λ2
1a0 − 2λ1λ2a0 + 2λ2a1 = c = 0
• c = −(λ1 − λ2)(a0λ1 − a1)
• c = 0 if a1 = λ1a0

97. Order-3 recurrence relation with one repeated and one
distinct eigenvalue
To find a, b, and c in ak = aλk
1 + bλk
2 + ckλk
2:


1 1 0
λ1 λ2 λ2
λ2
1 λ2
2 2λ2
2




a
b
c

 =


a0
a1
a2

 =⇒


a
b
c

=




λ2
2a0−2λ2a1+a2
(λ1−λ2)2
λ2
1a0−2λ1λ2a0+2λ2a1−a2
(λ1−λ2)2
λ1a1−λ1λ2a0+λ2a1−a2
λ2(λ1−λ2)




So matrix has rank:
• 3 if a, b, c are non-zero and λn
1 = λn
2
• 2 if λn
1 = λn
2, or if a and b are non-zero, but
c = 0 = λ1a1 − λ1λ2a0 + λ2a1 − a2
• c = 0 if a2 = (λ1 + λ2)a1 − λ1λ2a0
• 1 if a = 0, but b = λ2
1a0 − 2λ1λ2a0 + 2λ2a1 − a2 = c = 0
• b = 0 if λ2
1a0 − 2λ1λ2a0 + 2λ2a1 = c = 0
• c = −(λ1 − λ2)(a0λ1 − a1)
• c = 0 if a1 = λ1a0
• 0 if a = b = c = 0

98. Further Research
Begin the generalization for order-4 homogeneous recurrence
relations using Bozlee’s method to determine when the relations
can be rewritten to have order 3, 2, or 1 and when the rank of the
corresponding recurrence matrices drop

99. Acknowledgements
Thank you, Dr. Jason Callahan, for your guidance and the Ronald
E. McNair Scholars Program for making this research experience
possible.