What is an application of diagonalization?

\[ 0, 1, 1, 2, 3, 5, 8, \cdots \]

\(F_k\) is the \(k^{th}\) term of the sequence

\[ F_{k + 2} \]

\[ F_{k + 2} = F_{k + 1} + F_{k} \]

\[ \begin{aligned} F_{k + 2} &= F_{k + 1} + F_{k}\\\\ F_{k + 1} &= F_{k + 1} \end{aligned} \]

\[ \begin{aligned} F_{k + 2} &= 1 \cdot F_{k + 1} + 1 \cdot F_{k}\\\\ F_{k + 1} &= 1 \cdot F_{k + 1} + 0 \cdot F_{k} \end{aligned} \]

\[ \begin{bmatrix} F_{k + 2}\\ F_{k + 1} \end{bmatrix} = \begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} F_{k + 1}\\ F_{k} \end{bmatrix} \]

\[ \begin{bmatrix} F_{k + 2}\\ F_{k + 1} \end{bmatrix} = \begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} F_{k + 1}\\ F_{k} \end{bmatrix} \]

\[ \begin{bmatrix} F_{k + 2}\\ F_{k + 1} \end{bmatrix} = \begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} F_{k + 1}\\ F_{k} \end{bmatrix} \]

\[ A = \begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix}, u_k = \begin{bmatrix} F_{k + 1}\\ F_{k} \end{bmatrix} \]

\[ \begin{bmatrix} F_{k + 2}\\ F_{k + 1} \end{bmatrix} = \begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} F_{k + 1}\\ F_{k} \end{bmatrix} \]

\[ A = \begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix}, u_k = \begin{bmatrix} F_{k + 1}\\ F_{k} \end{bmatrix} \]

\[ u_{k + 1} = A u_k \]

\[ u_{k + 1} = A u_k \]

\[ u_{k + 1} = A u_k \]

\[ u_0 = \begin{bmatrix} 1\\ 0 \end{bmatrix} \]

\[ u_{k + 1} = A u_k \]

\[ u_0 = \begin{bmatrix} 1\\ 0 \end{bmatrix} \]
\[ u_1 = Au_0 \]

\[ u_{k + 1} = A u_k \]

\[ u_0 = \begin{bmatrix} 1\\ 0 \end{bmatrix} \]
\[ u_1 = Au_0 \]
\[ \begin{aligned} u_2 &= Au_1 \end{aligned} \]

\[ u_{k + 1} = A u_k \]

\[ u_0 = \begin{bmatrix} 1\\ 0 \end{bmatrix} \]
\[ u_1 = Au_0 \]
\[ \begin{aligned} u_2 &= Au_1\\ &= A^2 u_0 \end{aligned} \]

\[ u_{k + 1} = A u_k \]

\[ u_0 = \begin{bmatrix} 1\\ 0 \end{bmatrix} \]
\[ u_1 = Au_0 \]
\[ \begin{aligned} u_2 &= A^2u_0 \end{aligned} \]

\[ u_{k + 1} = A u_k \]

\[ u_0 = \begin{bmatrix} 1\\ 0 \end{bmatrix} \]
\[ u_1 = Au_0 \]
\[ \begin{aligned} u_k &= A^ku_0 \end{aligned} \]

\[ \huge{u_k = A^k u_0} \]

\[ \huge{u_k = A^k u_0} \]

How do we compute \(A^k\)?

\[ \huge{u_k = A^k u_0} \]

How do we compute \(A^k\)?

Diagonalization

\[ A = \begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix} \]

Is \(A\) diagonalizable?

\[ \begin{aligned} \quad \ \ \begin{vmatrix} 1 - \lambda & 1\\ 1 & -\lambda \end{vmatrix} &= 0 \end{aligned} \]

\[ \begin{aligned} \begin{vmatrix} 1 - \lambda & 1\\ 1 & -\lambda \end{vmatrix} &= 0\\\\\\ (1 - \lambda)(-\lambda) - 1 &= 0 \end{aligned} \]

\[ \begin{aligned} \begin{vmatrix} 1 - \lambda & 1\\ 1 & -\lambda \end{vmatrix} &= 0\\\\\\ (1 - \lambda)(-\lambda) - 1 &= 0\\\\\\ \lambda^2 - \lambda - 1 &= 0 \end{aligned} \]

\[ \begin{aligned} \begin{vmatrix} 1 - \lambda & 1\\ 1 & -\lambda \end{vmatrix} &= 0\\\\\\ (1 - \lambda)(-\lambda) - 1 &= 0\\\\\\ \lambda^2 - \lambda - 1 &= 0\\\\ \end{aligned} \]

\[ \lambda_1 = \cfrac{1 - \sqrt{5}}{2}, \lambda_2 = \cfrac{1 + \sqrt{5}}{2} \]

\[ \begin{aligned} \quad \quad\ \ (A - \lambda I)v &= 0\\\\ \end{aligned} \]

\[ \begin{aligned} (A - \lambda I)v &= 0\\\\ \begin{bmatrix} 1 - \lambda & 1\\ 1 & -\lambda \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \end{bmatrix} &= \begin{bmatrix} 0\\ 0 \end{bmatrix}\\\\ \end{aligned} \]

\[ \begin{aligned} (A - \lambda I)v &= 0\\\\ \begin{bmatrix} 1 - \lambda & 1\\ 1 & -\lambda \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \end{bmatrix} &= \begin{bmatrix} 0\\ 0 \end{bmatrix}\\\\ \cfrac{v_1}{v_2} &= \lambda \end{aligned} \]

\[ \begin{aligned} (A - \lambda I)v &= 0\\\\ \begin{bmatrix} 1 - \lambda & 1\\ 1 & -\lambda \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \end{bmatrix} &= \begin{bmatrix} 0\\ 0 \end{bmatrix}\\\\ \cfrac{v_1}{v_2} &= \lambda \end{aligned} \]

\[ \begin{bmatrix} \lambda_1\\ 1 \end{bmatrix}, \begin{bmatrix} \lambda_2\\ 1 \end{bmatrix} \]

\[ \begin{aligned} (A - \lambda I)v &= 0\\\\ \begin{bmatrix} 1 - \lambda & 1\\ 1 & -\lambda \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \end{bmatrix} &= \begin{bmatrix} 0\\ 0 \end{bmatrix}\\\\ \cfrac{v_1}{v_2} &= \lambda \end{aligned} \]

\[ \begin{bmatrix} \lambda_1\\ 1 \end{bmatrix}, \begin{bmatrix} \lambda_2\\ 1 \end{bmatrix} \]

We have two linearly independent eigenvectors.

\[ A = \begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix} \]

\[ Q = \begin{bmatrix} \lambda_1 & \lambda_2\\ 1 & 1 \end{bmatrix} \]
\[ D = \begin{bmatrix} \lambda_1 & 0\\ 0 & \lambda_2 \end{bmatrix} \]

\[ A = \begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix} \]

\[ Q = \begin{bmatrix} \lambda_1 & \lambda_2\\ 1 & 1 \end{bmatrix} \]
\[ D = \begin{bmatrix} \lambda_1 & 0\\ 0 & \lambda_2 \end{bmatrix} \]

\[ A = QDQ^{-1} \]

\[ A = \begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix} \]

\[ Q = \begin{bmatrix} \lambda_1 & \lambda_2\\ 1 & 1 \end{bmatrix}, Q^{-1} = \cfrac{1}{\lambda_1 - \lambda_2}\begin{bmatrix} 1 & -\lambda_2\\ -1 & \lambda_1 \end{bmatrix} \]
\[ D = \begin{bmatrix} \lambda_1 & 0\\ 0 & \lambda_2 \end{bmatrix} \]

\[ A = QDQ^{-1} \]

\[ u_k = A^k u_0 \]

\[ u_k = A^k u_0 \]

\[ \begin{aligned} u_k &= (QDQ^{-1})^k u_0 \end{aligned} \]

\[ u_k = A^k u_0 \]

\[ \begin{aligned} u_k &= (QDQ^{-1})^k u_0\\\\ &= Q D^k Q^{-1} u_0 \end{aligned} \]

\[ \begin{aligned} u_k &= Q D^k Q^{-1} u_0 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\ \ \end{aligned} \]

\[ \begin{aligned} u_k &= Q D^k Q^{-1} u_0\\\\ &= \cfrac{1}{\lambda_1 - \lambda_2} \begin{bmatrix} \lambda_1 & \lambda_2\\ 1 & 1 \end{bmatrix} \begin{bmatrix} \lambda_1^k & 0\\ 0 & \lambda_2^k \end{bmatrix} \begin{bmatrix} 1 & -\lambda_2\\ -1 & \lambda_1 \end{bmatrix} \begin{bmatrix} 1\\ 0 \end{bmatrix} \end{aligned} \]

\[ \begin{aligned} u_k &= Q D^k Q^{-1} u_0\\\\ &= \cfrac{1}{\lambda_1 - \lambda_2} \begin{bmatrix} \lambda_1 & \lambda_2\\ 1 & 1 \end{bmatrix} \begin{bmatrix} \lambda_1^k & 0\\ 0 & \lambda_2^k \end{bmatrix} \begin{bmatrix} 1 & -\lambda_2\\ -1 & \lambda_1 \end{bmatrix} \begin{bmatrix} 1\\ 0 \end{bmatrix}\\\\ &= \cfrac{1}{\lambda_1 - \lambda_2} \begin{bmatrix} \lambda_1^{k + 1} - \lambda_2^{k + 1}\\ \lambda_1^{k} - \lambda_2^{k} \end{bmatrix} \end{aligned} \]

\[ u_k = \cfrac{1}{\lambda_1 - \lambda_2} \begin{bmatrix} \lambda_1^{k + 1} - \lambda_2^{k + 1}\\ \lambda_1^{k} - \lambda_2^{k} \end{bmatrix} \]

\[ u_k = \begin{bmatrix} F_{k + 1}\\ F_{k} \end{bmatrix} \]

\[ F_k = \cfrac{1}{\lambda_1 - \lambda_2} \left(\lambda_1^k - \lambda_2^k \right) \]

\[ \lambda_1 = \cfrac{1 - \sqrt{5}}{2}, \lambda_2 = \cfrac{1 + \sqrt{5}}{2} \]

\[ F_k = \cfrac{1}{\lambda_1 - \lambda_2} \left(\lambda_1^k - \lambda_2^k \right) \]

\[ \boxed{F_k = \cfrac{1}{\sqrt{5}} \left[ \left(\cfrac{1 + \sqrt{5}}{2} \right)^k - \left(\cfrac{1 - \sqrt{5}}{2} \right)^k \right]} \]