If I have an $n \times n$ positive definite symmetric matrix $A$, with eigenvalues $\lambda_{1}>\lambda_{2}\cdots>\lambda_{n}$, can I claim that the highest value which matrix $A^{-1}$ can have will be $\lambda_{n}^{-1}$ (where $\lambda_{n}$ is the minimum eigenvalue of the matrix $A$)

If I have an $n \times n$ positive definite symmetric matrix $A$, with eigenvalues $\lambda_{1}>\lambda_{2}\cdots>\lambda_{n}$, can I claim that the highest value which matrix $A^{-1}$ can have will be $\lambda_{n}^{-1}$ (where $\lambda_{n}$ is the minimum eigenvalue of the matrix $A$). All the values of $A$ are real.