Prove matrix identity by induction

Question

Let $\lambda_1=\frac{1-\sqrt{5}}{2}$ og $\lambda_2=\frac{1+\sqrt{5}}{2}$.

Show by induction in $k$ that \begin{equation*} A^k=\frac{1}{\sqrt{5}} \begin{pmatrix} \lambda_2^{k-1}-\lambda_1^{k-1} & \lambda_2^k-\lambda_1^k\\ \lambda_2^k-\lambda_1^k & \lambda_2^{k+1}-\lambda_1^{k+1} \end{pmatrix} , k>0 \end{equation*} when \begin{equation*} A= \begin{pmatrix} 0 & 1\\ 1 & 1 \end{pmatrix} \in\mathrm{Mat}_2(\mathbb{R}). \end{equation*}

My attempt

When $k=1$ I get: \begin{equation*} A^1= \begin{pmatrix} 0 & 1\\ 1 & 1 \end{pmatrix} \end{equation*} and \begin{align*} A^1&=\frac{1}{\sqrt{5}} \begin{pmatrix} \lambda_2^0-\lambda_1^0 & \lambda_2-\lambda_1\\ \lambda_2-\lambda_1 & \lambda_2^2-\lambda_1^2 \end{pmatrix}\\ &=\frac{1}{\sqrt{5}} \begin{pmatrix} 0 & \frac{1+\sqrt{5}}{2}-\frac{1-\sqrt{5}}{2}\\ \frac{1+\sqrt{5}}{2}-\frac{1-\sqrt{5}}{2} & 1+\frac{1+\sqrt{5}}{2}-\left(1+\frac{1-\sqrt{5}}{2}\right) \end{pmatrix}\\ &= \begin{pmatrix} 0 & 1\\ 1 & 1 \end{pmatrix}. \end{align*}

Therefore the statement is true for $k=1$. Now I assume that the statement is true for $k=n$. Then I have to prove that it is true for $k=n+1$, but I can't get it to work for $k=n+1$. Can you help?

Answer 1

Hint: Just use what induction means, write $A^{n+1} = A^n\times A$, you know both terms in the product, plug them in, find $A^{n+1}$. Also, observe that $\lambda_i$ are the roots of $x^2-x-1 = 0$. Thus, $$\lambda_i^n+\lambda_i^{n-1} = \lambda_i^{n+1}$$

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$$A^{n+1} =A^n\times A = \frac{1}{\sqrt{5}} \begin{pmatrix} \lambda_2^{n-1}-\lambda_1^{n-1} & \lambda_2^n-\lambda_1^n\\ \lambda_2^n-\lambda_1^n & \lambda_2^{n+1}-\lambda_1^{n+1} \end{pmatrix} \begin{pmatrix} 0&1\\ 1&1 \end{pmatrix} =$$ $$= \frac{1}{\sqrt{5}} \begin{pmatrix} \lambda_2^n-\lambda_1^n& \lambda_2^{n-1}-\lambda_1^{n-1} + \lambda_2^n-\lambda_1^n \\ \lambda_2^{n+1}-\lambda_1^{n+1}& \lambda_2^n-\lambda_1^n +\lambda_2^{n+1}-\lambda_1^{n+1} \end{pmatrix} $$ $$= \frac{1}{\sqrt{5}} \begin{pmatrix} \lambda_2^n-\lambda_1^n& \left(\lambda_2^{n-1} + \lambda_2^n\right)-\left(\lambda_1^{n-1}+\lambda_1^n\right) \\ \lambda_2^{n+1}-\lambda_1^{n+1}& \left(\lambda_2^n+\lambda_2^{n+1}\right)-\left(\lambda_1^n +\lambda_1^{n+1}\right) \end{pmatrix} $$ $$= \frac{1}{\sqrt{5}} \begin{pmatrix} \lambda_2^n-\lambda_1^n& \lambda_2^{n+1}-\lambda_1^{n+1} \\ \lambda_2^{n+1}-\lambda_1^{n+1}& \lambda_2^{n+2}-\lambda_2^{n+2} \end{pmatrix} $$