# Fibonacci Matrices

Posted by **Cleve Moler**,

### Contents

#### A bug report

MathWorks recently received a bug report involving the matrix

X = [ 63245986, 102334155 102334155, 165580141]

X = 63245986 102334155 102334155 165580141

The report claimed that MATLAB computed an inaccurate result for the determinant of `X`.

```
format long
detX = det(X)
```

detX = 1.524897739291191

The familiar high school formula gives a significantly different result.

detX = X(1,1)*X(2,2) - X(1,2)*X(2,1)

detX = 2

While checking out the report, I computed the elementwise ratio of the rows of `X` and discovered an old friend.

ratio = X(2,:)./X(1,:)

ratio = 1.618033988749895 1.618033988749895

This is $\phi$, the Golden ratio.

phi = (1 + sqrt(5))/2

phi = 1.618033988749895

So, I decided to investigate further.

#### Powers of a Fibonacci matrix

Let's call the following 2-by-2 matrix the Fibonacci matrix.

$$F = \pmatrix{0 & 1 \cr 1 & 1}$$

Generate the matrix in MATLAB

F = [0 1; 1 1]

F = 0 1 1 1

The elements of powers of $F$ are Fibonacci numbers. Let $f_n$ be the $n$-th Fibanacci number. Then the $n$-th power of $F$ contains three successive $f_n$.

$$F^n = \pmatrix{f_{n-1} & f_n \cr f_n & f_{n+1}}$$

For example

F2 = F*F F3 = F*F2 F4 = F*F3 F5 = F*F4

F2 = 1 1 1 2 F3 = 1 2 2 3 F4 = 2 3 3 5 F5 = 3 5 5 8

The matrix in the bug report is $F^{40}$.

X = F^40

X = 63245986 102334155 102334155 165580141

These matrix powers are computed without any roundoff error. Their elements are "flints", floating point numbers whose values are integers.

#### Determinants

The determinant of $F$ is clearly $-1$. Hence

$$\mbox{det}(F^n) = (-1)^n$$

So `det(F^40)` should be `1`, not `1.5249`, and not `2`.

Let's examine the computation of determinants using floating point arithmetic. We expect these to be +1 or -1.

det1 = det(F) det2 = det(F^2) det3 = det(F^3)

det1 = -1 det2 = 1 det3 = -1

So far, so good. However,

det40 = det(F^40)

det40 = 1.524897739291191

This has barely one digit of accuracy.

#### What happened?

It is instructive to look at the accuracy of all the determinants.

d = zeros(40,1); for n = 1:40 d(n) = det(F^n); end format long d

d = -1.000000000000000 1.000000000000000 -1.000000000000000 0.999999999999999 -0.999999999999996 1.000000000000000 -0.999999999999996 0.999999999999996 -1.000000000000057 0.999999999999872 -0.999999999999943 0.999999999997726 -0.999999999998209 1.000000000000227 -1.000000000029786 1.000000000138243 -0.999999999990905 1.000000000261935 -1.000000000591172 0.999999999999091 -0.999999996060069 1.000000003768946 -0.999999926301825 0.999999793712050 -1.000000685962732 0.999998350307578 -0.999995890248101 0.999983831774443 -0.999970503151417 1.000005397945643 -0.999914042185992 0.999138860497624 -0.997885793447495 0.998510751873255 -0.996874589473009 0.973348170518875 -0.989943694323301 0.873688660562038 -0.471219316124916 1.524897739291191

We see that the accuracy deteriorates as `n` increases. In fact, the number of correct digits is roughly a linear function of `n`, reaching zero around n = 40.

#### A log plot of the error

semilogy(abs(abs(d)-1),'.') set(gca,'ydir','rev') title('error in det(F^n)') xlabel('n') ylabel('error') axis([0 41 eps 1])

#### Computing determinants

MATLAB computes determinants as the product of the diagonal elements of the triangular factors resulting from Gaussian elimination.

Let's look at $F^{12}$.

format long e F12 = F^12 [L,U] = lu(F12)

F12 = 89 144 144 233 L = 6.180555555555555e-01 1.000000000000000e+00 1.000000000000000e+00 0 U = 1.440000000000000e+02 2.330000000000000e+02 0 -6.944444444428655e-03

Since $det(L) = -1$,

det(L)

ans = -1

we have

det12 = -prod(diag(U))

det12 = 9.999999999977263e-01

We can see that, for $F^{12}$, the crucial element `U(2,2)` has lost about 5 significant digits in the elimination and that this is reflected in the accuracy of the determinant.

How about $F^{40}$?

[L,U] = lu(F^40) det40 = -prod(diag(U))

L = 6.180339887498949e-01 1.000000000000000e+00 1.000000000000000e+00 0 U = 1.023341550000000e+08 1.655801410000000e+08 0 -1.490116119384766e-08 det40 = 1.524897739291191e+00

The troubling value of `1.5249` from the bug report is a direct result of the subtractive cancellation involved in computing `U(2,2)`. In order for the computed determinant to be `1`, the value of `U(2,2)` should have been

U22 = -1/U(1,1)

U22 = -9.771908508943080e-09

Compare `U22` with `U(2,2)`. The value of `U(2,2)` resulting from elimination has lost almost all its accuracy.

This prompts us to check the condition of $F^{40}$.

cond40 = cond(F^40)

cond40 = Inf

Well, that's not much help. Something is going wrong in the computation of `cond(F^40)`.

#### Golden Ratio

There is an intimate relationship between Fibonacci numbers and the Golden Ratio. The eigenvalues of $F$ are $\phi$ and its negative reciprocal.

$$\phi = (1+\sqrt{5})/2$$

$$\bar{\phi} = (1-\sqrt{5})/2$$

```
format long
phibar = (1-sqrt(5))/2
phi = (1+sqrt(5))/2
eigF = eig(F)
```

phibar = -0.618033988749895 phi = 1.618033988749895 eigF = -0.618033988749895 1.618033988749895

Because powers of $\phi$ dominate powers of $\bar{\phi}$, it is possible to generate $F^n$ by rounding a scaled matrix of powers of $\phi$ to the nearest integers.

$$F^n = \mbox{round}\left(\pmatrix{\phi^{n-1} & \phi^n \cr \phi^n & \phi^{n+1}}/\sqrt{5}\right)$$

n = 40 F40 = round( [phi^(n-1) phi^n; phi^n phi^(n+1)]/sqrt(5) )

n = 40 F40 = 63245986 102334155 102334155 165580141

Before rounding, the matrix of powers of $\phi$ is clearly singular, so the rounded matrix $F^n$ must be regarded as "close to singular", even though its determinant is $+1$. This is yet another example of the fact that the size of the determinant cannot be a reliable indication of nearness to singularity.

#### Condition

The condition number of $F$ is the ratio of its singular values, and because $F$ is symmetric, its singular values are the absolute value of its eigenvalues. So

$$\mbox{cond}(F) = \phi/|\bar{\phi}| = \phi^2 = 1+\phi$$

condF = 1+phi

condF = 2.618033988749895

This agrees with

condF = cond(F)

condF = 2.618033988749895

With the help of the Symbolic Toolbox, we can get an exact expression for $\mbox{cond}(F^{40}) = \mbox{cond}(F)^{40}$

```
Phi = sym('(1+sqrt(5))/2');
cond40 = expand((1+Phi)^40)
```

cond40 = (23416728348467685*5^(1/2))/2 + 52361396397820127/2

The numerical value is

cond40 = double(cond40)

cond40 = 5.236139639782013e+16

This is better than the `Inf` we obtain from `cond(F^40)`. Note that it is more than an order of magnitude larger than `1/eps`.

#### Backward error

J. H. Wilkinson showed us that the `L` and `U` factors computing by Gaussian elimination are the exact factors of some matrix within roundoff error of the given matrix. With hindsight and the Symbolic Toolbox, we can find that matrix.

digits(25) X = vpa(L)*vpa(U)

X = [ 63245986.00000000118877885, 102334154.9999999967942319] [ 102334155.0, 165580141.0]

Our computed `L` and `U` are the actual triangular decomposition of this extended precision matrix. And the determinant of this `X` is the shaky value that prompted this investigation.

det(X)

ans = 1.524897739291191101074219

The double precision matrix closest to `X` is precisely $F^{40}$.

double(X)

ans = 63245986 102334155 102334155 165580141

Retrospective backward error analysis confirms Wilkinson's theory in this example.

#### The high school formula

The familiar formula for 2-by-2 determinants

$$\mbox{det}(X) = x_{1,1} x_{2,2} - x_{1,2} x_{2,1}$$

gives +1 or -1 with no roundoff error for all $X = F^n$ with $n < 40$. It fails completely for $n > 40$. The behavior for exactly $n = 40$ is interesting. Set the output format to

```
format bank
```

This format allows us to see all the digits generated when converting binary floating point numbers to decimal for printing. Again, let

X = F^40

X = 63245986.00 102334155.00 102334155.00 165580141.00

Look at the last digits of these two products.

p1 = X(1,1)*X(2,2) p2 = X(2,1)*X(1,2)

p1 = 10472279279564026.00 p2 = 10472279279564024.00

The `6` at the end of `p1` is correct because `X(1,1)` ends in a `6` and `X(2,2)` ends in a `1`. But the `4` at the end of `p2` is incorrect. It should be a `5` because both `X(2,1)` and `X(1,2)` end in `5`. However, the spacing between floating point numbers of this magnitude is

```
format short
delta = eps(p2)
```

delta = 2

So near `p2` only even flints can be represented. In fact, `p1` is the next floating point number after `p2`. The true product of `X(2,1)` and `X(1,2)` falls halfway between `p1` and `p2` and must be rounded to one or the other. The familiar formula cannot possibly produce the correct result.

#### Historical irony

For many years, the `det` function in MATLAB would apply the `round` function to the computed value if all the matrix entries were integers. So, these old versions would have returned exactly `+1` or `-1` for `det(F^n)` with `n < 40`. But they would round `det(F^40)` to `2`. This kind of behavior was the reason we got rid of the rounding.

`eig` and `svd`

Let's try to compute eigenvalues and singular values of $F^{40}$.

format long e eig(F^40) svd(F^40)

ans = 0 228826127 ans = 2.288261270000000e+08 0

Unfortunately, the small eigenvalue and small singular value are completely lost. We know that the small value should be

phibar^40

ans = 4.370130339181083e-09

But this is smaller than roundoff error in the large value

eps(phi^40)

ans = 2.980232238769531e-08

Log plots of the accuracy loss in the computed small eigenvalue and the small singular value are similar to our plot for the determinant. I'm not sure how LAPACK computes eigenvalues and singular values of 2-by-2 matrices. Perhaps that can be the subject of a future blog.

Get
the MATLAB code

Published with MATLAB® 7.14