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

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.

user2457324
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2 Answers2

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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 $\Delta$.

Community
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michalOut
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  • Not necessarily all of them need to be negative, we are interested in the sum. But if it were to happen that this is the case and all of them are non-positive, the statement follows. – michalOut Aug 06 '16 at 19:13
  • $UE_{ii}U^T$ can be further simplyfied. Note that $E_{ii}U^T$ is a matrix with columns zero except $i$-th column being $u_i$. Also try to use that $\forall i\neq j : u_i^Tu_j=0$, while computing this product. – michalOut Aug 06 '16 at 20:04
  • Regarding the multiplication with the diagonal matric, just try realize what does multiplication with diagonal matrix mean. It corresponds to stretching the vectors. Hence, geometrically it should be acceptable that the product $UDe_i^Tu_i$ is equal to $d_{ii}E_{ii}$, thanks to the fact that $U$ is unitary and has orthonormal columns and the stretching does not change the orthogonality. Regarding the $F$, I have paid that matrix no mind, since the multiplication by $F$ should not change the sign of the trace. But of that I am not so sure, be sure to double check it. – michalOut Aug 07 '16 at 01:13
  • thanks for that. based on your suggestion, I was thinking to do something like (with $u_i$ being the $i$'th row of $U$): $U D u_i^T = (U D^{1/2})(D^{1/2}u_i^T) = \bar{U} \bar{u}i^T$, but this does not give a column vector of zeros with $1$ at the $i$'th position since $D^{1/2}$ only scales the columns of $U$ and $u_i$ in my notation is the $i$'th row of $U$ (I had to choose it to be the rows in order to simplify $UE{ii}U^T = u_i^T u_i$). Is it possible that I can substitute this back in trace (and use the cyclic property) and simplify while ignoring the $F$ since its elements are positive? – user2457324 Aug 07 '16 at 04:19
  • Hi, I have also posted the main problem here with a working code, if you may have a chance to comment on this: http://math.stackexchange.com/questions/1885411/will-the-following-trace-inequality-be-true it would be appreciated. I don't know if there may be a much simpler way to prove this since I have not applied the Sherman-Morrison formula to get the above? thank you. – user2457324 Aug 08 '16 at 00:44
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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 $\operatorname{diag}(0,1,0)$.

user1551
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  • thank you for this. Just a quick Q: when you say that it is equivalent to $\Delta$ have non-positive diagonal entries -- is this really the case since I was thinking that there is a balance between the eigenvalues of $\Delta F$ that causes the inequality to follow? – user2457324 Aug 07 '16 at 18:16
  • also, I applied the Sherman-Morrison formula to the first term of Δ here http://math.stackexchange.com/questions/1885411/will-the-following-trace-inequality-be-true to get to the above. I have also included a small code to verify the numerics. Please take a look at your convenience. thanks. – user2457324 Aug 07 '16 at 21:30
  • @user2457324 The trace of $\Delta F$ is merely the sum of all elements on the diagonal of $\Delta F$. If $\Delta$ has a nonpositive diagonal, the diagonal of $\Delta F$ and in turn the trace are nonpositive too. Conversely, if the trace of $\Delta F$ is nonpositive for every nonpositive diagonal $F$, then in particular the trace is nonpositive when $F$ has only one positive diagonal element. So, the corresponding diagonal element of $\Delta$ must be nonpositive. Since $F$ is arbitrary, it follows that the diagonal of $\Delta$ is entrywise nonpositive. – user1551 Aug 08 '16 at 17:07
  • Thank you for this. Just had one more question if you can kindly help with this...I would really appreciate it http://math.stackexchange.com/questions/1887554/find-2-norm-conditions-on-a-matrix-to-make-the-following-true – user2457324 Aug 09 '16 at 21:51