Solution: Given that the matrix $A= \begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix}$, we want to compute $A^{2023}$. At first we will check whether $A$ is diagonalizable or not, if it is diagonalizable, then the computation of the power will be very easy. Let us find out the eigenvalues of the matrix $A$. Note that the determinant of $A$ is zero, so one of the eigenvalues will be $0$. Since the sum of all the eigenvalues is the trace of the matrix, so the other eigenvalue will be $2$. We now will find out the corresponding eigenvectors.

Consider the eigenvalue $\lambda _1 = 0$. We have \begin{align*} A \mathbf{v} = \lambda _1 \mathbf{v} & \implies \begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \end{pmatrix} \\ & \implies v_1 + v_2 = 0 \\ & \implies v_2 = -v_1. \end{align*} Thus one of the eigenvector will be \[ \mathbf{v} = \begin{pmatrix} 1 \\ -1 \\ \end{pmatrix}. \]

Similarly for the eigenvalue $\lambda _2 = 1$. We have \begin{align*} A \mathbf{v} = \lambda _2 \mathbf{v} & \implies (A-I \lambda _2)\mathbf{v} = \begin{pmatrix} 0 \\ 0 \\ \end{pmatrix} \\ & \implies \begin{pmatrix} -1 & 1 \\ 1 & -1 \\ \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ & \implies v_1 - v_2 = 0 \\ & \implies v_2 = v_1. \end{align*} Thus one of the eigenvector will be \[ \mathbf{v} = \begin{pmatrix} 1 \\ 1 \\ \end{pmatrix}. \]

If we write \[ P = \begin{pmatrix} 1 & 1 \\ -1 & 1 \\ \end{pmatrix} \text{ and } D = \begin{pmatrix} 0 & 0 \\ 0 & 2 \\ \end{pmatrix}, \] then, \[ A = PDP^{-1} \implies A^{2023} = PD^{2023}P^{-1} \] We have \[ P^{-1} = \frac{1}{2} \begin{pmatrix} 1 & -1 \\ 1 & 1 \\ \end{pmatrix}. \] Therefore, \begin{align*} A^{2023} & = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ - & 1 \\ \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 2^{2023} \\ \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 1 & 1 \\ \end{pmatrix} \\ & = \frac{1}{2}\begin{pmatrix} 2^{2023} & 2^{2023} \\ 2^{2023} & 2^{2023} \\ \end{pmatrix} \\ & = 2^{2022} \begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix}. \end{align*}