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Assume that we have a positive definite matrix $C$, and a positive definite diagonal matrix $\Lambda$.

Are all the diagonal entries of $(C + \Lambda)^{-1}$ smaller than those of $C^{-1}$? In other words, if we increase the diagonal entries of a positive definite matrix, will the diagonal entries of its inverse decrease? If so, how to prove it?

Thanks a lot

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The proof is as follows:

Consider $C^{-1}-(C+\Lambda)^{-1} $ $$ C^{-1}-(C+\Lambda)^{-1} = C^{-1} \left( [C+\Lambda] - C\right) (C+\Lambda)^{-1} =C^{-1} \Lambda (C+\Lambda)^{-1} = C^{-1} \Lambda (I+C^{-1}\Lambda)^{-1} C^{-1}= C^{-1} (\Lambda^{-1}+C^{-1})^{-1} C^{-1} $$ The last matrix is clearly positive definite. Hence, its diagonals must be positive. This shows that $$\left[C^{-1}\right]_{ii} >\left[(C+\Lambda)^{-1}\right]_{ii}$$

Added in response to OP's question

If $\Lambda$ is only positive semi definite, then $\Lambda + \epsilon I$ is positive definite for any $\epsilon >0$. Hence, by virtue of previous argument $$\left[C^{-1}\right]_{ii} >\left[(C+\Lambda+\epsilon I)^{-1}\right]_{ii}, \forall \epsilon > 0$$ By continuity argument, as $\epsilon \to 0$ $$\left[C^{-1}\right]_{ii} \ge \left[(C+\Lambda)^{-1}\right]_{ii}$$

Note that $\Lambda + \epsilon I$ is positive definite for any $\epsilon >0$ because $$ x^T (\Lambda + \epsilon I) x = x^T \Lambda x + \epsilon x^T x \ge \epsilon x^T x \gt 0 \text{ if $x \neq 0$}$$

The trick of adding $\epsilon I$ to a semi-definite matrix to make is positive definite and then taking the limit as $\epsilon \to 0^+$ is the standard way to extend results from definite to semidefinite.

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  • $\begingroup$ I just come up with the same way to prove it. $\endgroup$
    – Charles Zheng
    Jan 18, 2014 at 5:45
  • $\begingroup$ what if $\Lambda$ is positive semidifinite, then $\Lambda$ can be non-invertible. $\endgroup$
    – Charles Zheng
    Jan 18, 2014 at 6:37
  • $\begingroup$ If $\Lambda$ is only positive semi-definite, then you use continuity argument to show that the result is true except you have to replace $\gt$ by $\ge$. The argument is to replace $\Lambda$ by $\Lambda + \epsilon I$ and take the limit as $\epsilon \to 0^+$ $\endgroup$
    – user44197
    Jan 18, 2014 at 6:47
  • $\begingroup$ Why can we replace $\Lambda$ by $\Lambda + \epsilon I$? Besides, if $\Lambda$ is non-invertible, then the last step is not right. $\endgroup$
    – Charles Zheng
    Jan 18, 2014 at 6:54
  • $\begingroup$ If $\Lambda \ge 0$ then $\Lambda + \epsilon I \gt 0$ for all $\epsilon >0$ and hence is invertible. $\endgroup$
    – user44197
    Jan 18, 2014 at 6:56

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