# Computing $\operatorname{Tr} \bigl( \bigl( (A+I )^{-1} \bigr)^2\bigr)$

Suppose that $A \in \mathbb{R}^{n \times n}$ is a symmetric positive semi-definite matrix such that $\operatorname{Tr}(A)\le n$. I want a lower bound on the following quantity $$\operatorname{Tr} \bigl( \bigl( (A+I )^{-1} \bigr)^2\bigr)$$ where $I$ denotes the identity matrix.

My intuition tells me it should be

$$\operatorname{Tr} \bigl( \bigl( (A+I )^{-1} \bigr)^2\bigr) \ge 1/4$$

However, I'm not sure how to show it. Also my intuition might be wrong?

• Could be helpful: math.stackexchange.com/questions/391128/… – Squirtle Jul 31 '15 at 21:10
• The expression is invariant under conjugation, and $A$ is diagonalizable. – anomaly Jul 31 '15 at 21:11
• @Squirtle thank. But the square inside is hard to deal with – Boby Jul 31 '15 at 21:13
• @anomaly Is that a suggestion how to solve it? – Boby Jul 31 '15 at 21:24
• @Boby: Well, yes. – anomaly Jul 31 '15 at 21:27

By spectral theorem, there is invertible matrix $P$ such that $$P^{-1}(I+A)P=\pmatrix{1+\mu_1 \\ & \ddots & \\ & & 1+\mu_n}$$ where $\mu_i$ is eigenvalue of $A$ and $\mu_i\geqslant0$. So $$(I+A)^{-1}=P\pmatrix{\dfrac1{1+\mu_1} \\ & \ddots & \\ & & \dfrac1{1+\mu_n}}P^{-1}$$ And $$((I+A)^{-1})^2=P\pmatrix{\dfrac1{(1+\mu_1)^2} \\ & \ddots & \\ & & \dfrac1{(1+\mu_n)^2}}P^{-1}$$ So $$Tr(((I+A)^{-1})^2)=\sum_{i=1}^n\dfrac1{(1+\mu_i)^2}\leqslant n$$ Then by $\sum_{i=1}^n\mu_i\leqslant n$, we have $\mu_i\leqslant n$ for all $i$. So a lower bound is $$Tr(((I+A)^{-1})^2)\geqslant\frac{n}{(1+n)^2}$$

• @Boby, you should convince yourself (google or prove it yourself) that adding the identity to a matrix will shift the eigenspectrum in the manner of this proof (on line one). – Squirtle Jul 31 '15 at 21:33
• @hermes Sorry, but it is not clear to me how does this help to find the lower bound. Could you be more explicit? – Start wearing purple Jul 31 '15 at 21:48
• @hermes I mean this calculation only implies an obvious lower bound $n\cdot \frac{1}{(1+n)^2}$. But this is very far from the conjectured one and definitely not optimal. – Start wearing purple Jul 31 '15 at 22:10
• @L.G. Not sure but maybe we can use concavity of $\frac{1}{(1+x)^2}$ to show that $\sum _{i=1} \frac{1}{(1+\mu_i)^2} \ge \frac{1}{(1+\sum _{i=1}\mu_i)^2} =\frac{1}{4}$ – Boby Jul 31 '15 at 22:29
• The conjecture may not be true.I mean we can get a lower bound but it may not be $1/4$. The lower bound can be obtained by Lagrangian multiplier. – Math Wizard Jul 31 '15 at 22:35

Since $A$ is real symmetric, there exists a matrix $P$ such that

$$PAP^{-1} = D =\verb/diag/(\lambda_1,\lambda_2,\ldots,\lambda_n)$$ is diagonal with eigenvalues $\lambda_1,\ldots,\lambda_n$. This leads to

$$P (I_n+A)^{-2}P^{-1} = (I_n + PAP^{-1})^{-2} = \verb/diag/\left(\frac{1}{(1+\lambda_1)^2},\ldots,\frac{1}{(1+\lambda_n)^2}\right)\\ \implies \verb/Tr/((I_n + A)^{-2}) = \sum_{i=1}^{n}\frac{1}{(1+\lambda_i)^2}$$

Consider the function $g(x) = \frac{1}{(1+x)^2}$. It is easy to check for $x \in (0,\infty)$, $$g(x) > 0, \quad g'(x) = -\frac{2}{(1+x)^3} < 0 \quad\text{ and }\quad g''(x) = \frac{6}{(1+x)^4} > 0$$ This means $g(x)$ is a positive, monotonic decreasing and convex function over $(0,\infty)$.
By Jensen's inequality, we obtain following lower bound:

$$\verb/Tr/((I_n + A)^{-2}) = \sum_{i=1}^{n} g(\lambda_i) \underbrace{\ge}_{\text{Jensen}} n g\left(\frac1n\sum_{i=1}^n \lambda_i\right) \underbrace{\ge}_{g\text{ decreasing}} n g(1) = \frac{n}{4}$$

Since $\verb/Tr/(I_n) \le n$ and $\verb/Tr/((I_n + I_n)^{-2}) = \frac{n}{4}$, above lower bound $\frac{n}{4}$ is achievable and hence the optimal one.

As a consequence, your intution $\verb/Tr/((I_n + A)^{-1}) \ge \frac14$ is true (but far from the optimal).

• +1 Yet $\frac14\ge\frac{n}{(1+n)^2}$, so the OP's bound is at least better than the one given in the other answer ;D – user1551 Aug 24 '15 at 11:44