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0 $begingroup$ I have a vector $mathbf{v}=begin{pmatrix} v_1 & v_2 & v_3 end{pmatrix}^top$ and the matrix $G$ . $$G = begin{pmatrix} partial_1^2v_1 & partial_2^2v_1 & partial_3^2v_1 \ partial_1^2v_2 & partial_2^2v_2 & partial_3^2v_2 \ partial_1^2v_3 & partial_2^2v_3 & partial_3^2v_3end{pmatrix}$$ $G$ is not the same as the gradient of $mathbf{v}$ $$nablamathbf{v}=begin{pmatrix} partial_1v_1 & partial_2v_1 & partial_3v_1 \ partial_1v_2 & partial_2v_2 & partial_3v_2 \ partial_1v_3 & partial_2v_3 & partial_3v_3end{pmatrix}$$ nor is it the Laplacian of $mathbf{v}$ . $$nabla^2mathbf{v}=Deltamathbf{v}=begin{pmatrix} partial_1^2v_1 + partial_2^2v_1 + partial_3^2v_1\ partial_1^2v_2 + partial_2^2v_2 + partial_3^2v_2\partial_1^2v_3 + partial_2^2v_3 + partial_3^2v

2 $begingroup$ I was reading this topic to show that $(x-a,y-b)$ is an ideal of $K[x,y]$ , where $K$ is a field. The answer suggests to show that $(x-a,y-b)$ is the kernel of the evaluation in $ev_{(a,b)} : K[x,y] rightarrow K$ . I understand that $ev_{(a,b)}(x-a)=0$ and $ev_{(a,b)}(y-b)=0$ so we have the inclusion $(x-a,y-b) subset Ker(ev_{(a,b)})$ but how to show the other inclusion? ideals share | cite | improve this question asked Dec 20 '18 at 16:59 roi_saumon roi_saumon 634 3 8 $endgroup$