= Solution
An <invertible matrix> is a <square matrix> $X$ admitting a <matrix inverse> $X^{-1}$ with $XX^{-1}=X^{-1}X=I$, where $I$ is the <identity matrix>. Equivalently, its associated <linear map> is bijective.
Let $X\mathbf v=\lambda\mathbf v$ with <eigenvector> $\mathbf v\ne0$. The <eigenvalue> cannot be zero: otherwise applying $X^{-1}$ gives $\mathbf v=0$. Apply $X^{-1}$ to the <eigenvector> equation:
$$
\mathbf v=\lambda X^{-1}\mathbf v,
\qquad\boxed{X^{-1}\mathbf v=\frac1\lambda\mathbf v.}
$$
Thus $1/\lambda$ is an <eigenvalue> of $X^{-1}$ with the same <eigenvector>.
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