# Determinant of Matrix Exponential is Non-Zero

## Theorem

Let $\mathbf A$ be a square matrix.

Let $t \in \R$ be a real number.

Let $e^{\mathbf A t}$ denote the matrix exponential of $\mathbf A$.

Then:

- $\det e^{\mathbf A t} \ne 0$

where $\det$ denotes the determinant.

## Proof

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The linear system $x' = \mathbf A x$ has $n$ linearly independent solutions.

Putting together these solutions as columns in a matrix creates a matrix solution to the differential equation, considering the initial conditions for the matrix exponential.

From Existence and Uniqueness Theorem for 1st Order IVPs, this solution is unique.

By linear independence of its columns:

- $\det e^{\mathbf A t} \ne 0$

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The nonzero determinant property also follows as a corollary to Liouville's Theorem (Differential Equations).