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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?

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1 Answer 1

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For a symmetric matrix the one norm and the infinity norm coincide. So the condition on the norm $$\Vert \Sigma(\theta) - I_{p}\Vert_{1} = \Vert\Sigma(\theta) - I_{p}\Vert_{\infty} \leq a$$ implies that $$|\sigma_{ii} - 1| + \sum_{j\neq i}|\sigma_{ij}| \leq a \quad \forall i.\qquad (1)$$

Let us consider the case $\sigma_{ii} \geq 1$. We note that (1) implies $\sum_{j\neq i}|\sigma_{ij}| \leq a$. Combining these two inequalities ($p \ge q \wedge r \le s \Rightarrow p-r \ge q-s$) we obtain: $$\sigma_{ii} - \sum_{j\neq i}|\sigma_{ij}| \geq 1 - a.\qquad (2)$$

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 $1-a$.

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