Assume an N*N covariance matrix (Q) which is a positive definite matrix. The decoder X is assumed to be N*s, where s<=N. X is calculated to be s eigenvectors corresponding to s minimum eigenvalues. Is this solution for X unique to achieve minimum of Tr(X'QX)? if no, what is the other solutions?
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1$\begingroup$ isn't the minimum of ${\rm Tr}(X^{T}QX)$ for positive definite $Q$ reached for $X=0$? $\endgroup$– Carlo BeenakkerCommented Sep 13, 2013 at 17:17
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$\begingroup$ I can't write comments, but you want some constraint on X like columns of X have norm 1 and are orthogonal to each other? This is stronger than the constraint suggested by loup blanc. X could be unique only up to permutations of its columns. $\endgroup$– lmgCommented Feb 4, 2014 at 13:59
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If you consider eigenvectors associated to distinct eigenvalues, then you don't obtain an extremum. In general, the minimum is not reached if $X$ is assumed to have maximal rank !
Thus I assume that you search the minimum of $f(X)=tr(X^TQX)$ under (for instance) the condition $g(X)=tr(X^TX)-s=0$. One has $Df_X:H\rightarrow 2tr(X^TQH)$ and $Dg_X:H\rightarrow 2tr(X^TH)$. We use the Lagrange's method: there is a real number $\lambda$ s.t. $Df_x-\lambda Dg_x=0$, that is, for every $H$, $tr((X^TQ-\lambda X^T)H)=0$. That implies $X^TQ-\lambda X^T=0$ or $QX=\lambda X$. Thus, $\lambda$ is an eigenvalue of $Q$ and $X=[u_1,\cdots,u_s]$ where $u_i$ is an eigenvector associated to $\lambda$ s.t. $u_i^Tu_i=1$. For such a $X$, $f(X)=s\lambda$.
Finally the minimum of $f$ is $s\lambda_m$, where $\lambda_m$ is the minimum of the spectrum of $Q$. Of course, if the condition is $tr(X^TX)-1=0$, then the minimum is $\lambda_m$.
