I am trying to bound the following expression:
$$\det\left(I+\frac{1}{\lambda_{\min}\left(\Lambda+A\right)}C\right)$$
Where $\Lambda$ is a PD matrix and $A,C$ are PSD matrices.
I know that:
$$\lambda_{\min}\left(\Lambda+A\right)\geq\lambda_{\min}\left(\Lambda\right)+\lambda_{\min}\left(A\right)=\lambda_{\min}\left(\Lambda_{k}\right)+0=\lambda_{\min}\left(\Lambda_{k}\right)$$
The question is, does this bound the original expression, using the above argument? That is:
$$\det\left(I+\frac{1}{\lambda_{\min}\left(\Lambda+A\right)}C\right)\leq\det\left(I+\frac{1}{\lambda_{\min}\left(\Lambda\right)}C\right)$$
I assume that, in this context, positive (semi)-definite matrices are necessarily symmetric.
The answer is yes. Because we are dealing with positive semidefinite matrices, we can use the Loewner order. We have $$ \left(I + \frac{C}{\lambda_{\min}(\Lambda)}\right) - \left(I + \frac{C}{\lambda_{\min}(\Lambda + A)}\right) = \left( \frac{1}{\lambda_{\min}(\Lambda)} -\frac{1}{\lambda_{\min}(\Lambda + A)}\right)C \succeq 0, $$ which is to say that $$ \left(I + \frac{C}{\lambda_{\min}(\Lambda)}\right) \succeq \left(I + \frac{C}{\lambda_{\min}(\Lambda + A)}\right). $$ It follows that $$ \det\left(I + \frac{C}{\lambda_{\min}(\Lambda)}\right) \geq \det \left(I + \frac{C}{\lambda_{\min}(\Lambda + A)}\right). $$