Eigenvalue Diagonalization of an Invertible Matrix
Definition
For $A \in \mathbb{C}^{ m \times m }$, if there exist a unitary matrix $Q$ and a diagonal matrix $\Lambda$ satisfying $A = Q^{ \ast } \Lambda Q$, then the matrix $A$ is said to be unitarily diagonalizable.
Theorem
For the linearly independent eigenvectors $\mathbf{x}_{1}, \mathbf{x}_{2}, \cdots , \mathbf{x}_{m}$ of an invertible matrix $A \in \mathbb{R}^{m \times m}$, let $S = \begin{bmatrix} \mathbf{x}_{1}, \mathbf{x}_{2}, \cdots , \mathbf{x}_{m} \end{bmatrix}$, then $$ S^{-1} A S = \begin{bmatrix} \lambda_{1} & 0 & \cdots & 0 \\ 0 & \lambda_2 & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & \lambda_{m} \end{bmatrix} $$
Explanation
Since the assumption required eigenvectors, it is not hard to guess that $\lambda_{i}$ represents the eigenvalue of the eigenvector $\mathbf{x}_{i}$.
If such a diagonalization is possible, there is the advantage that powers of the matrix can be computed very easily. Suppose $A$ is diagonalizable but looks extremely complicated to raise to a power. Originally an enormous amount of computation would have to be performed, but if it can be expressed in the form $A = S \Lambda S^{-1}$, then it can also be expressed as $A^{n} = S \Lambda^{n} S^{-1}$, and since the power of the diagonal matrix $\Lambda = \text{diag} ( \lambda_{1} , \lambda_{2} , \cdots , \lambda_{m} )$ is $\Lambda^{n} = \text{diag} ( \lambda_{1}^{n} , \lambda_{2}^{n} , \cdots , \lambda_{m}^{n} )$, the computation is reduced dramatically. Such matrix powers are known to be actively applied in stochastic process theory, atmospheric science, and the like.
Meanwhile, note that since the eigenvector for a given eigenvalue is not unique, the matrix $S$ is likewise not unique.
Proof
Since $A$ is an invertible matrix, there exist linearly independent eigenvectors $\mathbf{x}_{1}, \mathbf{x}_{2}, \cdots , \mathbf{x}_{m}$ of $A$. $$ \begin{align*} AS =& A \begin{bmatrix} \mathbf{x}_{1}, \mathbf{x}_{2}, \cdots , \mathbf{x}_{m} \end{bmatrix} \\ =& \begin{bmatrix} A \mathbf{x}_{1}, A \mathbf{x}_{2}, \cdots , A \mathbf{x}_{m} \end{bmatrix} \\ =& \begin{bmatrix} \lambda_{1} \mathbf{x}_{1}, \lambda_{2} \mathbf{x}_{2}, \cdots , \lambda \mathbf{x}_{m} \end{bmatrix} \\ =& \begin{bmatrix} \mathbf{x}_{1}, \mathbf{x}_{2}, \cdots , \mathbf{x}_{m} \end{bmatrix} \begin{bmatrix} \lambda_{1} & 0 & \cdots & 0 \\ 0 & \lambda_2 & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & \lambda_{m} \end{bmatrix} \\ =& S \begin{bmatrix} \lambda_{1} & 0 & \cdots & 0 \\ 0 & \lambda_2 & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & \lambda_{m} \end{bmatrix} \end{align*} $$
Since $S$ is a matrix composed of linearly independent eigenvectors, $S^{-1}$ exists, and moving it to the left-hand side gives $$ S^{-1} A S = \begin{bmatrix} \lambda_{1} & 0 & \cdots & 0 \\ 0 & \lambda_2 & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & \lambda_{m} \end{bmatrix} $$
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