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?


1 Answer 1


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 $1-a$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.