Exercise 1.19

The "clever algorithm" used in Exercise 1.19 for computing Fibonacci numbers should be familiar to students of linear algebra. Referring, for example, to Exercise 4.2.9 in the text Linear Algebra Done Wrong, available from the Math Department at Brown University, we encounter a very similar problem statement in relation to computing matrix eigenvalues.

Linear Algebra Done Wrong - Exercise 4.2.9

Let us recall that the famous Fibonacci sequence:

[ 0, 1, 1, 2, 3, 5, 8, 13, 21, ... ]

is defined as follows:

We want a formula for the n-th Fibonacci number:

(a) Find a 2 x 2 matrix A such that:

(b) Diagonalize A and find a formula for the n-th power of A.

(c) Noticing that

find a formula for the n-th Fibonacci number.

(d) Show that the vector

converges to an eigenvector of A.

Linear Algebra Done Wrong - Solution 4.2.9

It's worthwhile to work through this example, both for its own merits, and also to understand in more depth the procedure we designed in SICP Exercise 1.19.

The linear operator A can easily be derived as follows:

To diagonalize A, we must calculate its eigenvalues, and verify that the algebraic multiplicity of each eigenvalue is equal to its geometric multiplicity. The characteristic equation of A is given by:

and hence we determine that the two distinct eigenvalues of A are:

The algebraic and geometric multiplicity of each eigenvalue is 1, and so the matrix can be diagonalized.

Knowing this, we know that the operator A can be expressed as a diagonal matrix as follows:

where the column vectors of P are the eigenvectors of the linear operator A.

If we wish to obtain a representation for the inversion matrix P, we may proceed as follows:

A representative eigenvector, chosen from the eigenspace corresponding to the eigenvalue lambda(1), is:

Similarly, a representative eigenvector chosen from the eigenspace corresponding to the eigenvalue lambda(2) is:

So that the inversion matrix we are seeking can be expressed succinctly as:

so that:

From here, it may be easily verified that:

Since the eigenvalues lambda are solutions to the characteristic equation, the off-diagonal elements reduce to 0.

Similarly, bearing in mind that:

we have:

so that:

and likewise

so that:

Hence we have that:

To obtain an expression for A^n, we write:

Furthermore, since:

We can now obtain our expression for A^n as follows:

Since we showed above that:

we can now use this expression to find an expression for the n-th Fibonacci number as follows:

From which we obtain the following expression for the n-th Fibonacci number:

Let's try out this expression for a few Fibonacci numbers, to see if it holds up:

Clearly, the expression holds up.

If we "re-index" the n-th Fibonacci number to begin with 0 rather than 1, we obtain the following expression for the n-th Fibonacci number:

Compare this with the expression utilized in Exercise 1.13.

Finally, let us consider the following expression:

and consider further that:

for all positive n, so that:

and so that:

Hence the vector

converges to the vector

which we already proved was an eigenvector of the operator A.

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