Say for a real, rectangular matrix $X$ and a s.p.s.d matrix $Q$ we maximize or minimize $Tr(X^TQX)$ under the constraint $Tr(X^TM) = 1$ for some fixed real matrix $M$.

i) Would the columns of the solution $X^*$ be called 'non-orthogonal' eigenvectors of $Q$?

ii) What is the optimization problem and/or the equation to be solved to compute a non-othogonal eigenvector of $Q$? say as in the analogue to the characteristic equation or the $Ax=\lambda x$ kind in terms of orthogonal eigenvectors: what would this formulation be to compute non-orthogonal eigenvectors?

iii) If the constraint was $XX^T=I$, then the maximization corresponds to a solution of the eigenvectors corresponding to max eigen values of Q and vice versa (minimum). How does the concept of eigenvalues come into play in the case of the $Tr(X^TM) = 1$ constraint with regards to the problem being a maximum or minimum?

iv)Is it ok to say the non-orthogonal eigenvectors form a non-orthogonal basis of $Q$?

v) Can this non-orthogonal basis be subjected to a change of basis to or from the orthogonal eigenvector basis of $Q$ using the inverse operation of the matrix containing the columns being the basis (eigenvectors) vectors?

vi) if the constraint was having some positive number $p$ instead of $1$ in $Tr(X^TM) = 1$ as $Tr(X^TM) = p$, how does it effect this non-orthogonal basis?


@qlinck, a solution depends on $Q$ AND $M$. Moreover your constraint has degree $1$ and consequently $X$ has nothing to do with the eigenvectors of $Q$.

Let $X,M$ be $n\times k$ matrices and $Q$ be a $n\times n$ (sym. $\geq 0$) matrix. The Lagrange's method gives the relation: there is $\lambda\in\mathbb{R}$ s.t. $2QX+\lambda M=0$. If we assume that $Q>0$, then $X=\dfrac{-\lambda}{2}Q^{-1}M$. Since $tr(X^TM)=1$, $\lambda=-\dfrac{2}{tr(M^TQ^{-1}M)}$ and finally $X=\dfrac{1}{tr(M^TQ^{-1}M)}Q^{-1}M$ except if $tr(M^TQ^{-1}M)=0$.

EDIT: I forgot to say, that if $Q>0$, then clearly, $\sup(tr(X^TQX))=+\infty$. The $X$ above gives the $\inf$ of the considered function, that is $\dfrac{1}{tr(M^TQ^{-1}M)}$.

  • $\begingroup$ So are the columns of X as computed above called nonorthogonal eigenvectors? $\endgroup$
    – qlinck
    Apr 18 '14 at 2:14
  • $\begingroup$ I don't see why. If $U_i$ is the column of index $i$ of a matrix $U$, then $QX_i$ and $M_i$ are parallel vectors. So what ? $\endgroup$
    – user91684
    Apr 18 '14 at 14:43

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