# Linear Algebra Project

```Linear Algebra Project
Randy, Stephanie, Caleb
April 23, 2020
Part I - Matrix Exponents
Recall from Calculus that a Maclaurin Series is an infinite series with real coefficients of the form
f (x) =
∞
X
an xn
n=0
These series are used in analysis, combinatorics (as generating functions) and
in electrical engineering (under the name Z-transforms). In Calculus, you learn
how to add, subtract, and multiply series and how to find the radius and interval
of convergence. You also calculate examples of series for common functions. One
of the first examples you see of a convergent Maclaurin series defined for all real
numbers is the series
∞
X
x2
x3
xn
=1+x+
+
+ ...
n!
2!
3!
n=0
Our goal in this part of the project is to define this series with a square matrix
A in place of the variable x. These matrix exponentials have applications in
higher branches of mathematics including differential equations and Lie theory,
a theory of continuous symmetries named after mathematician Sophus Lie.
Let A be an n x n matrix. Define eA = limm→∞ Bm , where
Am
A2
+ ...
2!
m!
This limit always converges for all A and therefore
Bm = In + A +
eA =
∞
X
Am
A2
A3
=I +A+
+
+ ...
m!
2!
3!
n=0
and Bm is the mth partial sum. Notice that if A is a 1 &times; 1 matrix, then above
coincides with the power series for f (x) = ex from Calculus.
1
Assignment (Part 1)
1. (2 points) Compute e0n , where the n &times; n zero matrix.
e0n = I + 0 + 02 + 03 + ...
=I
2. (2 points) Computer eIn , where In represents the n &times; n identity matrix
I3
+ ...
2!
eI = I + I +
1
+ ...)
2!
= I ∗ (1 + 1 +
=I ∗e
3. (4 points) In general, it is not true that eAe B = eA+B which differs when
compared to the exponential function for real numbers. (However, if A
and B commute, that is, if AB = BA, then the above relation does hold.)
Show, using the matrices

0
A = 0
0

7
2
0
1
0
0

0
B = 0
0
1
0
0

−1
3
0
that eA eB 6= eA+B .To calculate eA and eB , use the fact that A3 = 0/3
and B 3 = 03 .


0 1 7
A = 0 0 2
0 0 0

0
A2 = 0
0
1
0
0

7
2
0

0
x 0
0
1
0
0

7
2
0

0
= . 0
0
0
0
0

2
0
0

0
A3 = 0
0
0
0
0

2
0
0

0
x 0
0
1
0
0

7
2
0

0
= 0
0
0
0
0

0
0
0

1
eA = 0
0
0
1
0

0
0
1

0
+0
0
1
0
0

7
2
0
2

0
1
0
+
2
0
0
0
0

2
0
0

1
= 0
0
1
1
0

8
2
1

0 1
B = 0 0
0 0

0
2

B = 0
0
1
0
0

0
B 3 = 0
0

1
eB = 0
0
0
1
0
0
0
0

−1
3
0

3
0
0

0
x 0
0

0
+0
0
1
0
0

0
0
1

1
eA ∗ eB = 0
0
1
1
0

0
A + B = 0
0
eA+B

1
= 0
0
0
1
0

1
0
0

8
2
1
1
0
0

−1
3
0
1
0
0
1
0
0

−1
3
0
1
2

1 1
x 0 1
0 0


10
0
0

0
(A + B)3 = 0
0
0
0
0

0
0
0

11
5
1
2
0
0

6
5
0

0
1
+ 0
2
0

1
doesN OT equal 0
0
3

3
0
0
 
−1
0
3  = 0
0
0
0
0
0

0
+0
0

0
0
0
0
0
0

1
= 0
0
23
2

1 2
= 0 1
0 0
3
1

0 1
+ 0 0
0 0
0
0
0

3
0
0
0
0
0

0
= 0
0

0
1
0
+
2
0

−1
3
0

7
2
0

0

= 0
0

0
(A + B)2 = 0
0

0
0
1
2
1
0

0

x 0
0

−1
3
0
0
0
0
2
1
0

10
0
0
23
2

5
1
1
2
1
1
0

3
1

5
1
2
0
0

6
5
0

1
= 0
0
2
1
0

11
5
1

d1
0

4. (6 points) Show that if D =  .
 ..
0
d2
..
.
...
...
..
.

0
0

..  is a diagonal matrix,
.
0

0
...
dn
then eD
 d1
e
 0

= .
 ..
0
0
ed2
..
.
...
...
..
.
0
0
..
.
0
...
edn




eD = 1 + D +
eD

1
0

= .
 ..
0
1
..
.
...
...
..
.
0
0
...
 
d1
0
0
0
 
.. + ..
.  .
0
1
0
d2
..
.
...
...
..
.
0
...
1 2
D + ...
2!
  d21
0
 2!
0
0
 
.. +
 ..

.
.
dn
0
0
d21
2!
..
.
0
...
...
..
.
...

 d1
0
e


0
 0

.
.. +... = 
 ..
.
d21
0
2!
5. (6 points) Let A be diagonalizable, so that A = P DP − 1 for some diagonal
matrix D and invertible matrix P. Show that eA = P eD P − 1
A = P DP − 1
A2 = P D 2 P − 1
A3 = P D 3 P − 1
eA = 1 +
=I+
A3
A A2
+
+
+ ...
1
2
3
P DP − 1 P D2 P − 1 P D3 P − 1
+
+
+ ...
1
2
3
= P ∗ (1 +
D D2
D3
+
+
+ ...)P − 1
1
2
3
= P eD P − 1
4
0
ed 2
..
.
...
...
..
.
0
0
..
.
0
...
edn





Part II - The Fibonacci Sequence
Recall that we can define the Fibonacci sequence recursively by the following:
a1 = 1, a2 = 1, an+1 = an + an−1
This recursive sequence is named after Leonardo Fibonacci, who wrote about
the sequence in 1202. The first few terms of this sequence are:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, . . .
These numbers have appeared in popular culture because they relate mathematics to seemingly unrelated things in science and nature. In this part of the
project, we will see how linear algebra can be used to study this famous sequence.
If we wanted to find the 1000t h Fibonacci number with the recursive definition, we would first have to find the first 999 terms of the Fibonacci sequence.
We will apply the techniques of this
unit
to find an explicit
formula for an , the
1 1
an+1
−
→
nth Fibonacci number. Let A =
and let xn =
1 0
an
Notice that we have:
1 1 an+1
an + an+1
an + 2
−
→
−
Ax n =
=
=
=→
x n+1 w
1 0
an
an
an + 1
So multiplying by the matrix A gives the next number in the Fibonacci
sequence. In general:
−
→, with−
→= 1
n−
x→
=
A
x
x
n
0
0
0
5
Assignment (Part II)
1. (3 points) Verify that the eigenvalues of A are:
√
√
1+ 5
1− 5
λ1 =
,
λ2 =
.
2
2
If these numbers look familiar, its because λ1 is called the The Golden
Ratio, often denoted as ϕ, and λ2 is 1 − ϕ, or ϕ
1 1
1−λ 1
A−
x→
=
→
− λ2 − λ − 1 = 0 →
−λ =
→
− det(A−λ)=0 →
−
n
1 0
1
−λ
√
1&plusmn; 1+4
2
√
1&plusmn; 5
2
√
√
Eigenvalues: λ1 = 1+2 5 and λ2 = 1−2 5
ϕ
1−ϕ
2. Verify that
and
are eigenvectors for the eigenvalues λ1 and
1
1
λ2 , respectively.
=
#
&quot;
√
√
1√
ϕ 1−ϕ
1 − ( 1+2 5 )
→
− x2 IsF ree−
→( 1+2 5 )x1 +x2 =
→
− F orλ1 →
−
1+
5
1
1
1
−( 2 )
√
√
2
2
2
− a − b2 = (a − b)(a + b)−
→ √5−1
&middot; √5+1
=
0−
→(1 − 5)x1 = −2−
→x1 = √5−1 →
5+1
√ √ √
√
1− 5
1+ 5
2( 5+1
2
2
→
− λ2 = v2 =
→
− verif ied
= 5+1
− v1 =
5−1
2 →
1
1
3. (6 points) Diagonalize A by writing A = P DP −1 for
and an invertible matrix P.
√
&quot;
#
√ √
5−1
√
√1
1+ 5
1− 5
1 1
−1
5
2
A=
P= 2
,P = −15 √25+1
1 0
√
√
1
1
5
2 5
&quot; √
#
1+ 5
0
2
√
D=
1− 5
0
2
a diagonal matrix D
&quot; 5+√5 #
1
= 5−10√5 , and
1
10
→
n−
4. (6 points) Use the fact that −
x→
part to verify the formula:
√
√
1 1+ 5 n
1 1− 5 n
an = √ (
) −√ (
)
2
2
5
5
From the previous part, if we use A = P DP −1 it would look like this:
√
#&quot; √
#k &quot; √ #
&quot; √1
5−1
1+ 5
5+ 5
√
Fk+3
0
5
2
10
√
√
If you multiply this out,
= −15 √25+1
1− 5
5− 5
Fk+2
√
√
0
2
10
5
2 5
√
√ √
√
√
√
1+ 5
1− 5
1+
5
5+
5
k
2
an extensive form would look like this:
( 2 ) ( 10 )+ 2
( 1−2 5 )k ( 5−10 5 )
1
1
which corresponds to the given formula
6
5. (2 points) Use this formula and a calculator to find the 45th term of the
Fibonacci sequence.
√
√
1 1 + 5 45
1 1 − 5 45
a45 = √ (
) −√ (
) = 1134903170
2
2
5
5
7
```