Non-obvious basis

Algebra Level 3

The real vector space of " Fibonacci -like" sequences contains all sequences $\{a_n\}_{n \ge 0}$ satisfying $a_n + a_{n+1} = a_{n+2}$ for all natural numbers $n$ . Two sequences can be added by adding their corresponding terms, and a sequence can be multiplied by a scalar by multiplying all elements in the sequence by a real number. (Note that the resulting sequence in each case still satisfies the Fibonacci property and is therefore still in the vector space.)

The Fibonacci numbers $F = \{0,\, 1,\, 1,\, 2,\, 3,\, 5,\, \ldots\}$ and the Lucas number $L = \{2,\,1,\,3,\,4,\,7,\,\ldots\}$ are both elements of this sequence. The two geometric series $G_+ = \left\{1,\,\varphi,\,\varphi^2,\,\varphi^3,\,\dots\right\}$ and $G_- = \left\{1,\,-\varphi^{-1},\,\varphi^{-2},\,-\varphi^{-3},\,\ldots\right\}$ , where $\varphi^2 = \varphi + 1$ and $\varphi > 0$ are also elements of this sequence.

What makes a basis for this vector space?

\{F,\,L\}

and

\{G_+,\,G_-\}

are both bases A basis contains

F

L

G_+

, and

G_-

but also other sequences This vector space has no basis

\{F,\,L,\,G_+,\,G_-\}

is a basis

1 solution

Henry Maltby
Jul 7, 2016

In fact, both $\{F,\,L\}$ and $\{G_+,\,G_-\}$ are bases of this vector space! A sequence $\{a_n\}_{n \ge 0}$ of this form is determined entirely by its initial values $a_0$ and $a_1$ . Since $F$ and $L$ are linearly independent, they form a basis; similarly, since $G_+$ and $G_-$ are linearly independent, they form a basis.

Since any element of a vector space is expressible as a linear combination of basis elements, we could go on to write the Fibonacci sequence as a sum of $G_+$ and $G_-$ . Each element of $G_\pm$ is easily expressible, so this should give us a closed form for the Fibonacci numbers. Can you complete this reasoning?

Non-obvious basis

1 solution

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