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Recitation 9A

Problem 1: Define the Fibonacci sequence by F0=0F_0=0, F1=1F_1=1 and Fn+2=Fn+1+FnF_{n+2}=F_{n+1}+F_n. Denote [1110]\begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix} as MM. Prove that M[Fn+1Fn]=[Fn+2Fn+1]M \begin{bmatrix}F_{n+1} \\ F_{n}\end{bmatrix} = \begin{bmatrix}F_{n+2} \\ F_{n+1}\end{bmatrix} and Mn[10]=[Fn+1Fn]M^n \begin{bmatrix}1 \\ 0\end{bmatrix}=\begin{bmatrix}F_{n+1} \\ F_n\end{bmatrix}. Use these results to find a closed formula for FnF_n.

Proof: Check that M[Fn+1Fn]=[1110][Fn+1Fn]=[Fn+1+FnFn+1]=[Fn+2Fn+1]M \begin{bmatrix}F_{n+1} \\ F_{n}\end{bmatrix}= \begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix} \begin{bmatrix}F_{n+1} \\ F_{n}\end{bmatrix} = \begin{bmatrix}F_{n+1} + F_n \\ F_{n+1}\end{bmatrix} = \begin{bmatrix}F_{n+2} \\ F_{n+1}\end{bmatrix}. In other words, if we multiply [Fn+1Fn]\begin{bmatrix}F_{n+1} \\ F_n\end{bmatrix} with matrix MM, both indices increase by 1. Therefore if we multiply [F1F0]\begin{bmatrix}F_1 \\ F_0\end{bmatrix} with matrix MM for nn times, both indices increase by nn. That is Mn[10]=Mn[F1F0]=[Fn+1Fn]M^n \begin{bmatrix}1 \\ 0\end{bmatrix} = M^n \begin{bmatrix}F_1 \\ F_0\end{bmatrix} = \begin{bmatrix}F_{n+1} \\ F_n\end{bmatrix}. In order to find a closed formula for FnF_n, we have to compute MnM^n using diagonalization. Set λ1=1+52,λ2=1−52\lambda_1 = \frac{1+\sqrt{5}}{2}, \lambda_2 = \frac{1-\sqrt{5}}{2}. The diagonalization of MM is PDP−1PDP^{-1}, where P=[λ1λ211]P= \begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 1\end{bmatrix} and D=[λ100λ2]D= \begin{bmatrix}\lambda_1 & 0 \\ 0 & \lambda_2\end{bmatrix}. Hence [Fn+1Fn]=Mn[10]=PDnP−1[10]=[λ1λ211][λ1n00λ2n]15[1−λ2−1λ1][10]\begin{bmatrix}F_{n+1} \\ F_n\end{bmatrix}= M^n \begin{bmatrix}1 \\ 0\end{bmatrix}=PD^nP^{-1}\begin{bmatrix}1 \\ 0\end{bmatrix}=\begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 1\end{bmatrix}\begin{bmatrix}\lambda_1^n & 0 \\ 0 & \lambda_2^n\end{bmatrix}\frac{1}{\sqrt{5}}\begin{bmatrix}1 & -\lambda_2 \\ -1 & \lambda_1\end{bmatrix}\begin{bmatrix}1 \\ 0\end{bmatrix}. Expand the equation and obtain Fn=15(λ1n−λ2n)F_n = \frac{1}{\sqrt{5}}(\lambda_1^n-\lambda_2^n).

Problem 2: Let u=[2−5−1]u=\begin{bmatrix}2\\-5\\-1\end{bmatrix} and v=[−7−46]v= \begin{bmatrix}-7\\-4\\6\end{bmatrix}. Compute u⋅vu\cdot v, ∣∣u∣∣2||u||^2, ∣∣v∣∣2||v||^2, ∣∣u+v∣∣2||u+v||^2 and ∣∣u−v∣∣2||u-v||^2. Check the Pythagorean theorem and the parallelogram law hold true.

Solution: Plug in uu and vv and obtain u⋅v=0u\cdot v=0, ∣∣u∣∣2=30||u||^2=30, ∣∣v∣∣2=101||v||^2=101, ∣∣u+v∣∣2=∣∣u−v∣∣2=131||u+v||^2=||u-v||^2=131. The Pythagorean theorem holds because u⋅v=0u\cdot v=0 and ∣∣u∣∣2+∣∣v∣∣2=∣∣u+v∣∣2||u||^2+||v||^2=||u+v||^2. The parallelogram law holds as well because ∣∣u+v∣∣2+∣∣u−v∣∣2=2∣∣u∣∣2+2∣∣v∣∣2||u+v||^2+||u-v||^2=2||u||^2+2||v||^2.