Every Invertible Matrix Can be Expressed as a Matrix Exponential
Theorem
Every invertible matrix $A \in M_{n \times n}(\mathbb{C})$ can be expressed in the form of a matrix exponential of some $X \in M_{n \times n}(\mathbb{C})$.
$$ A = e^{X} \quad \text{for some } X \in M_{n \times n}(\mathbb{C}) $$
Proof
Let $A$ be an invertible matrix. Then none of the eigenvalues of $A$ is $0$. (See ../3004.) Moreover, the matrix $A$ can be decomposed into the following block‑diagonal matrix: (See ../3770.)
$$ A = \begin{bmatrix} \lambda_{1} I + N_{1} & O & \cdots & O \\ O & \lambda_{2} I + N_{2} & \cdots & O \\ \vdots & \vdots & \ddots & \vdots \\ O & O & \cdots & \lambda_{k} I + N_{k} \end{bmatrix} $$
Here $N_{i}$ is a nilpotent matrix. Since $\lambda_{i} \ne 0$, write each diagonal block as $\lambda_{i}(I + \lambda_{i}^{-1}N_{i})$. The matrix inside the parentheses is unipotent, so $I + \lambda_{i}^{-1}N_{i} = e^{\log (I + \lambda_{i}^{-1}N_{i})}$ holds. Also, if we set $\lambda_{i} = e^{\mu_{i}}$, then $\lambda_{i} I = e^{\mu_{i}I}$ holds. (See ../3758.) Therefore $A$ is given by
$$ \begin{align*} A &= \begin{bmatrix} e^{\mu_{1}I} e^{\log (I + \lambda_{1}^{-1}N_{1})} & O & \cdots & O \\ O & e^{\mu_{2}I} e^{\log (I + \lambda_{2}^{-1}N_{2})} & \cdots & O \\ \vdots & \vdots & \ddots & \vdots \\ O & O & \cdots & e^{\mu_{k}I} e^{\log (I + \lambda_{k}^{-1}N_{k})} \end{bmatrix} \\[3em] &= \begin{bmatrix} e^{\mu_{i}I + \log (I + \lambda_{i}^{-1}N_{i})} & O & \cdots & O \\ O & e^{\mu_{i}I + \log (I + \lambda_{i}^{-1}N_{i})} & \cdots & O \\ \vdots & \vdots & \ddots & \vdots \\ O & O & \cdots & e^{\mu_{k}I + \log (I + \lambda_{k}^{-1}N_{k})} \end{bmatrix} \end{align*} $$
Now let $X_{i} = \mu_{i}I + \log (I + \lambda_{i}^{-1}N_{i})$. Then the following holds. (See ../3758.)
$$ A = \diag(e^{X_{1}}, e^{X_{2}}, \dots, e^{X_{k}}) = e^{\diag(X_{1}, X_{2}, \dots, X_{k})} $$
Consequently, if we put $X = \diag(X_{1}, X_{2}, \dots, X_{k})$, we obtain
$$ A = e^{X} $$
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