Proving the following trace inequality?

Question

$$\mbox{trace} (\Delta F) \leq 0 $$

where

$$\Delta = \frac{A^{-1}}{\alpha} \bigg( \Lambda \Lambda^T - 2 \alpha \Lambda_{symm} \bigg) \frac{A^{-1}}{\alpha} $$ $$ \Lambda = (c^T c) A^{-1}$$ $$ 0 \neq \alpha = 1 + c A^{-1} c^T \in \mathbb{R}$$

$A$ is a (symmetric) positive definite matrix and $c$ is a row vector with only one non-zero entry. And $F$ is a diagonal matrix with non-negative entries. $\Lambda_{symm}$ is the symmetric part of $\Lambda$

Testing this via numerics verify this claim (though $\Delta$ has positive eigenvalues at times); however, I can't prove it. Any help/advice on how to go about proving this would be much appreciated.

Answer 1

The inequality is false. It is equivalent to the statement that $\Delta$ has a nonpositive diagonal. Here is a counterexample. Consider $$ A^{-1}=\pmatrix{3&-1&3\\ -1&5&4\\ 3&4&9}, \ c^T=\pmatrix{1\\ 0\\ 0}, \ \Lambda=\pmatrix{3&-1&3\\ 0&0&0\\ 0&0&0}. $$ Then $\alpha=4$ and $\det A^{-1}=9$. Sylvester's criterion shows that $A^{-1}$ and in turn $A$ are positive definite. Straightforward calculations show that $$ \Delta = \frac1{16}\pmatrix{-285&-29&-441\\ -29&51&23\\ -441&23&-597}. $$ Therefore the trace of $\Delta F$ is positive when $F$ is equal to or close to $\operatorname{diag}(0,1,0)$.

Answer 2

It is awful amount of work, but simply writing everything in one term, using spectral decomposition for the matrix $A$ and the product rule for the trace. Taking $c$ of the form $$c=(0,\dotsc,0, c_i,0,\dotsc ,0)$$

one can observe that the matrix $cc^T$ has actually only one non-zero entry, i.e., the entry $i,i$. Working with that and at the same time the fact that $A=UDU^T$ for some unitary matrix $U$ and diagonal matrix $D$ with positive entries on the diagonal, it is possible to show the result.

Since it is rather long, I won't go into further details.

HINT Let us start with writting $$\Delta =A^{-1}\left[(1+cA^{-1}c^T) (c^Tc)A^{-1}A^{-T}(c^Tc)^T - (c^T c)A^{-1} - A^{-T}(c^Tc)^T \right]A^{-1}. $$ Try to use that $$A=UDU^T$$ and hence $A^{-1}=U^TD^{-1}U$ and since $A$ is symmetric and commutativness of inverse and transpose, one has $A^{-T}=A^{-1}$. Try to start making $\Delta$ simpler using $UU^T=U^TU=I$ and the fact that $c^Tc=c_i^2 \cdot E_{ii}$, where $E_{ii}$ is the matrix filled with zeros and with the onlynonzero entry $e_{ii}=1$. Try to compute $E_{ii}U$ and use it in further simplying the $\Delta$.