# Lower bound on smallest eigenvalue of covariance matrices

Assume that a class of $p\times p$ covariance matrices is characterized by a parameter $\theta$, i.e,

$$\mathbb{F} = \left\{\Sigma(\theta), \theta\in R\right\}$$

Also suppose we know the following

$$||\Sigma(\theta) - I_{p}||_{\text{op}} \leq ||\Sigma(\theta) - I_{p}||_{l_1} \leq a$$ where $a$ is a constant. This implies that

$$\lambda_{\text{max}}(\Sigma(\theta)) \leq 1+a$$

Then we know that the smallest eigenvalue of $\Sigma(\theta)$ is lower bounded by the following

$$\lambda_{\text{min}}(\Sigma(\theta)) > 1-a$$ However, notice that the bound $1-a$ needs not to be positive. How would one get from the upper bound on the largest eigenvalue to the lower bound argument?

The condition on the norm $$||\Sigma(\theta) - I_{p}||_{1} = ||\Sigma(\theta) - I_{p}||_{\infty} \leq a$$ implies that $$|\sigma_{ii} - 1| + \sum_{j\neq i}|\sigma_{ij}| \leq a \quad \forall i.\qquad (1)$$
If $\sigma_{ii} \geq 1$, then (1) implies $\sum_{j\neq i}|\sigma_{ij}| \leq a$, and therefore: $$\sigma_{ii} - \sum_{j\neq i}|\sigma_{ij}| \geq 1 - a.\qquad (2)$$
If $\sigma_{ii} \leq 1$, then (1) becomes $1 - \sigma_{ii} + \sum_{j\neq i}|\sigma_{ij}| \leq a$. Rearranging gives (2) again. Applying Gershgorin circle thorem, we have that all eigenvalues must be greater or equal than $1-a$.