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Nov 12, 2017 at 21:48 comment added Igor Rivin @ChristianRemling Re your first comment, I was tired :( Re the second, I am specifically looking at the OP's lagrange multiplier equation.
Nov 12, 2017 at 21:44 comment added Christian Remling I don't understand your second sentence: $x^tBx=0$ does not imply that $x\in N(B)$ (for example take $B=\textrm{diag}(1,-1)$).
Nov 12, 2017 at 21:36 comment added Yang Thank you for the comments. In viewing of your comments, I removed the "full rank" condition to accept a broader class of solutions. One obviously can use the vectors in the null space of $B$ to minimize $A$. If $B$ is not symmetric, are there any additional solutions beyond the null-space? I appreciate your answer.
Nov 12, 2017 at 14:51 comment added Igor Rivin @FedericoPoloni I said "rarely" not never.
Nov 12, 2017 at 10:04 comment added Federico Poloni Why "$x^TBx$ is rarely zero for full-rank matrices"? For instance, the block matrix $\begin{bmatrix}0 & I\\I & 0\end{bmatrix}$ has two dim/2-dimensional subspaces of vectors such that $x^TBx=0$.
Nov 12, 2017 at 5:04 history answered Igor Rivin CC BY-SA 3.0