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Let $U$ be a non-zero $(q\times q)$ matrix, and assume this matrix is normalized such that $tr(U^{\intercal}U)=1$. Let $R$ be a symmetric $(q\times q)$ matrix, and $N>0$ be a positive integer. Define the following two scalar qunatities

$Q_{1}=tr(URRU^{\intercal})-N^{-1}\left(tr(UR)\right)^{2}$,

and

$Q_{2}=tr(URU^{\intercal}R)-N^{-1}\left(tr(UR)\right)^{2}$.

I am trying to follow a proof in a book which states that $Q_{1}>0\Rightarrow Q_{2}>0$.

Despite the many trace identities that can be used, I cannot derive the obvious equality $tr(URRU^{\intercal})=tr(URU^{\intercal}R)$, and so I am pretty certain this does not hold in general. Furthermore I cannot derive the inequality $tr(URRU^{\intercal})<=tr(URU^{\intercal}R)$ which would also do. I suspect the condition $tr(U^{\intercal}U)=1$ is the key to cracking this, yet I cannot bring this to bear on the problem. The only inequality I can find related to this condition is

$\left[tr(U^{\intercal}R)\right]^{2}\leq\left[tr(U^{\intercal}U)\right]\left[tr(R^{\intercal}R)\right]$,

where the above inequality actually holds for any two matrices of the same size. For the matrices $U$, and $R$ this becomes $\left[tr(U^{\intercal}R)\right]^{2}\leq tr(R^{2})$, but I cannot see how this helps. Finally I would also be interested in solutions to this problem, even if it means demanding $R$ is positive definite, although leaving $R$ to be general is preferred.

Any help would be greatly appreciated.

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up vote 2 down vote accepted

Unless some contextual details are missing, your book is wrong. For a counterexample, consider $R=\begin{pmatrix}1&1\\1&2\end{pmatrix}$ and $U=\begin{pmatrix}0&0\\0&1\end{pmatrix}$. Clearly $R$ is symmetric and $tr(U^TU)=1$. You may verify that \begin{align*} Q_1&=tr(URRU^T)-N^{-1}\left(tr(UR)\right)^2 =tr\begin{pmatrix}0&0\\0&5\end{pmatrix} - N^{-1}tr^2\begin{pmatrix}0&0\\1&2\end{pmatrix}=5-\frac4N,\\ Q_2&=tr(URU^TR)-N^{-1}\left(tr(UR)\right)^2 =tr\begin{pmatrix}0&0\\2&4\end{pmatrix} - N^{-1}tr^2\begin{pmatrix}0&0\\1&2\end{pmatrix}=4-\frac4N. \end{align*} When $N=1$, we have $Q_1>0$ but $Q_2\le0$.

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This is a very useful answer, however I must apologise that you I should have said $N$ must be an integer not a real number so in this case your counter-example will not work. –  dandar Jan 22 '13 at 22:08
    
Thank-you so much for this. Although I did not state this in my question (for fear of putting people off), I have become increasingly convinced something is wrong with the proof I am following, and it is good to get independent verification. I did not think to try and devise counter examples - I will try and remember this technique. I will take another look at the proof to find the problem. –  dandar Jan 22 '13 at 22:50
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