I believe there should be tag badges for questions as well as answers.   While it is true that SX sites are, in the bottom line, about answers, there are still badges for asking questions with many votes or many views. I believe there should be question-based tag-specific badges: In my opinion, users asking many good questions in a certain tag contribute to the "development" of the tag on the site, and this merits recognition.
Why not make  but for questions too?  I love that everything is well though behind SO but I can't see no reason as to why great questions can't be prized with "Badges by subject".
Pagination not working with custom loop Maximize $\mathrm{tr}(Q^TCQ)$ subject to $Q^TQ=I$ Let $C \in \mathbb{R}^{d \times d}$ be symmetric, and   $$Q = \begin{bmatrix}       \vert &amp; \vert &amp;    &amp; \vert \\       q_1   &amp; q_2   &amp; \dots &amp; q_K  \\       \vert &amp; \vert &amp;    &amp; \vert       \end{bmatrix} \in \mathbb{R}^{d\times K}$$  where $d \geq K$. Using Lagrange multipliers,   $$\begin{array}{ll} \text{maximize} &amp; \mbox{tr} \left( Q^T C Q \right)\\ \text{subject to} &amp; Q^T Q = I\end{array}$$    I am unfamiliar with these kind of constraints with this method, and after reading  I believe the same specific and simple result given is also applicable, and therefore the lagrangian would be:  $$\mathcal{L}(Q,\lambda)=\mathrm{tr}(Q^TCQ)-\left&lt;\lambda,Q^TQ-I\right&gt;$$  where $\lambda\in\mathbb{R}^{K\times K}$, and $\left&lt;\cdot,\cdot\right&gt;$ is the element wise inner product (what kind of makes sense to me since we're actually adding as many constraints as there are elements in these matrices.  In attempting to do that I start taking $\frac{\partial \mathcal{L}}{\partial Q}=O\in\mathbb{R}^{d\times K}$, and compute that LHS element by element; for the $(l,m)$ one:  \begin{equation} 0=\frac{\partial \mathcal{L}}{\partial Q_{lm}}=(CQ+C^TQ)_{lm}-\underbrace{\frac{\partial}{\partial Q_{lm}}\sum_{i,j}\lambda_{i,j}(Q^TQ-I)_{ij}}_{=\lambda_{lm}\frac{\partial (Q^TQ)_{lm}}{\partial Q_{lm}}}=2(CQ)_{lm}-\lambda_{lm}\frac{\partial (q_l^Tq_m)}{\partial q_m(l)} \tag{1}\end{equation}  where in the last step I've used the definition I made at the beginning for $Q$, and $q_m(l)$ denotes the $l$-th component of the column vector $q_m$.  In trying to compute the very last term:  $$\frac{\partial (q_l^Tq_m)}{\partial q_m(l)}=\frac{\partial \left[q_l(1)q_m(1)+ \ldots + q_l(d)q_m(d)\right]}{\partial q_m(l)}= \begin{cases}     q_l(l)\equiv Q_{ll}    &amp; \text{if } l\neq m\\     2q_l(l)\equiv 2Q_{ll}  &amp; \text{if} l=m \end{cases}$$  The whole equality (1) then can be written:  $$0=2(CQ)_{lm}-\lambda_{lm}Q_{ll}(1+\delta_{lm})$$  where $\delta_{lm}$ is the Kronecker delta.  The equation for the other stationary point of the lagrangian, $\frac{\partial \mathcal{L}}{\partial \lambda}=O\in\mathbb{R}^{K\times K}$, for the $(l,m)$ element as well:  $$ 0=\frac{\partial \mathcal L}{\partial \lambda_{lm}}= \frac{\partial }{\partial \lambda_{lm}}\sum_{i,j}\lambda_{i,j}(Q^TQ-I)_{ij}=(Q^TQ-I)_{lm}\tag{2}$$  what obviously leads to $(Q^TQ)_{lm}=\delta_{lm}$.  All this should tell that the columns of $Q$ are eventually the $K$ first eigenvectors of $C$, but I don't know how to continue from here to prove that, supposing I didn't make a mistake. Please I would sincerely appreciate any help.    Edit:  I have rewritten the inner product as a trace of a product of matrices (after seeing ):  $$\left&lt;\lambda,Q^TQ-I\right&gt;=\sum_{i,j}\lambda_{i,j}(Q^TQ-I)_{ij}=\mathrm{tr}(\lambda^TQ^TQ) $$  and have thus managed to do the derivative without losing the matrix format (using formulas from the ):  \begin{align} O=&amp;\frac{\partial \mathcal{L}}{\partial Q}=\frac{\partial}{\partial Q}\mathrm{tr}(Q^TCQ)-\frac{\partial}{\partial Q}\underbrace{\mathrm{tr}(\lambda^T(Q^TQ-I))}_{\mathrm{tr}(\lambda^TQ^TQ)-\mathrm{tr}(\lambda^T)}\\=&amp;(CQ+C^TQ)-(Q(\lambda^T)^T+Q\lambda^T)=2CQ+Q(\lambda+\lambda^T) \end{align}  And this leads to:  $$CQ=Q\underbrace{\left(-\frac{\lambda+\lambda^T}{2}\right)}_{:=\widetilde{\lambda}};\quad CQ=Q$$  If the defined matrix $\widetilde{\lambda}=Q^TCQ$ were diagonal we would already have the result. finding rational roots of polynomials Could someone please explain me how to apply the Descartes's Criterion? 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