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Is there a $2 \times 2$ matrix that can be raised to any power to obtain the Lucas Numbers? If so, what is that matrix? I've looked around on this website and other sites and am not able to find the solution.

VividD
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guest111
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3 Answers3

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Use the same matrix you would use for Fibonacci numbers, then apply to the initial state vector for Lucas numbers. The matrix captures the recurrence relation. So, same recurrence relation, same matrix. The differences between the Lucas numbers and the Fibonacci numbers lies entirely in the initial values, which do not come into the matrix, but rather an initial state vector to apply the matrix to.

Using one option for orienting the vectors: $$\begin{bmatrix}L_{n}\\L_{n+1}\end{bmatrix}=\begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}^n\begin{bmatrix}L_0\\L_1\end{bmatrix}=\begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}^n\begin{bmatrix}2\\1\end{bmatrix}$$

If you are using powers of $\begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}$ to identify Fibonacci numbers, then maybe you aren't thinking about the method in the best way. $$\begin{bmatrix}F_{n}\\F_{n+1}\end{bmatrix}=\begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}^n\begin{bmatrix}F_0\\F_1\end{bmatrix}=\begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}^n\begin{bmatrix}0\\1\end{bmatrix}$$ happens to just pick out the second column of $\begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}^n$, so there is some misleading coincidence there that makes the actual entries of $\begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}^n$ be Fibonacci numbers.

2'5 9'2
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Let $$Q=\pmatrix{1&1\\1&0}\;,$$ so that $$\pmatrix{F_{n+1}&F_n\\F_n&F_{n-1}}=Q^n$$ for $n\ge 0$. I don’t know that you can find a matrix that does exactly the same thing for the Lucas numbers, but it is true that

$$\pmatrix{L_{n+1}&L_n\\L_n&L_{n-1}}=Q^n\pmatrix{1&2\\2&-1}\;.$$

Another matrix that’s not quite what you want is $$Q_L=\pmatrix{3&1\\1&2}\;,$$ since

$$Q_L^n=\begin{cases} 5^{n/2}\pmatrix{F_{n+1}&F_n\\F_n&F_{n-1}},&\text{if }n\text{ is even}\\\\ 5^{(n-1)/2}\pmatrix{L_{n+1}&L_n\\L_n&L_{n-1}},&\text{if }n\text{ is odd} \end{cases}$$

for $n\ge 1$. (This last result is taken from this paper.)

Brian M. Scott
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The Lucas number $L_n$ is $\text{tr} \left[ \begin{array}{cc} 1 & 1 \\\ 1 & 0 \end{array} \right]^n$.

Qiaochu Yuan
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