# matrices and fibonacci .pdf

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Using Matrices to Count: Binet’s Formula and

Characteristic Polynomials

redacted

June 12, 2017

1 Introduction

Today, we’ll be looking at how to use matrices to solve certain counting problems,

including how to use matrices to find a formula for Fibonacci numbers! To get the most

out of this handout, you should be fairly comfortable with matrices. This will be fairly

low on explanations and hints as I had to make it fast, so come to me if you get stuck.

2 Eigenvectors

Let M be a two by two matrix. We say that a vector v 6= 0 is an eigenvector of M if

and only if

M v = λv,

for some constant λ. λ is called an eigenvalue of M. We can similarly define eigenvectors

and values for higher dimensional matrices, but for this paper we’ll just need two by two

matrices.

Why on earth does anybody care about eigenvectors and eigenvalues? We’ll see soon!

But first, some practice:

Problem 1. Find the eigenvectors, and their corresponding eigenvalues, of the

matrices

1 0

M=

0 1

and

N=

1 2

.

3 4

As you’ve seen, it’s fairly simple to find eigenvalues using systems of equations. But

those are long and tedious. Is there any way to generate eigenvalues without systems of

equations?

Well, let’s look back at the definition M v = λv. Note that λ(Iv) = λv, where I is

the identity matrix. So, we’re now solving

M v = λ(Iv),

1

(M − λI)v = 0.

Lemma. The above implies that det(M − λI) = 0.

Proof of Lemma. Use contradiction to show M − λI is not invertible as v 6= 0, then

the claim follows.

Using this, we start to get an idea. Can we go backwards? That is, show that if

det(M − λI) = 0, then λ is an eigenvalue?

Other Lemma. det X = 0 implies there exists some v 6= 0 such that Xv = 0.

Proof of Other Lemma. Let v, w be two vectors such that v 6= w and Xv = Xw

(why can we always pick two such vectors if det X = 0?). Then, X(v − w) = 0, and

v − w 6= 0.

So yes, we can go backwards.

3 Characteristic Polynomials

We call the polynomial det(M − xI) the characteristic polynomial of matrix M. The

roots of the characteristic polynomial are the eigenvalues of M.

Problem 2. Find the eigenvalues of matrices M and N from problem 1 using

characteristic polynomials.

Problem 3. (hard) Find a quick way to compute M n v, given that v = ae1 + be2 ,

and e1 and e2 are two distinct eigenvectors (that are linearly independent) of M with

eigenvalues λ1 and λ2 , respectively.

4 Fibonacci Numbers

Fibonacci numbers are the sequence of numbers defined by the recurrence F0 = F1 = 1,

and Fn+1 = Fn + Fn−1 .

Problem 4. (hard) Find a closed formula for the Fibonacci numbers using characteristic polynomials. Hint: Consider a two by two matrix M such that

Fn

Fn+1

M

=

.

Fn−1

Fn

We can think of M as a ‘generator matrix’ of the Fibonacci numbers.

This formula is usually called Binet’s Formula.

5 More Recurrence Problems.

‘ These might take awhile, but you’ll have all summer to work on them!

Problem 5. Find a closed form for the recurrence given by a0 = 2, a1 = 4, an =

an−1 − 2an .

Problem 6. Find a closed form for the recurrence given by a0 = 2, a1 = 4, a2 = 8,

and an = 3an−3 + 2an−2 + an−1 .

2

Problem 7. The formula for Fibonacci numbers uncovered in the last section is

pretty sucky. There’s a cooler formula using the floor function. Try to find it.

Problem 8. Try finding a matrix M whose eigenvectors are all linearly dependent.

What kind of interesting properties does it have? Can you find any recurrences that

relate to it?

6 Conclusion

Enjoy your summers! Learning about this in high school was what made me want to

major in math, so I really hope this interested you.

3

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