现在有位于第四象限的数据:
a=[1,-1;2,-2;3,-3;4,-4;5,-5];
画出这个图像:

预测的结局是:
坐标轴是顺时针旋转α角度和逆时针旋转2*pi-α,结果是一样的。
代码是:
a=[0,0;1,-1;2,-2;3,-3;4,-4;5,-5]; % 画出散点a的二维图像 close(figure(60000)); figure(60000);hold on;plot(a(:,1),a(:,2),'b.','markersize',11);axis equal;%axis off; figure(60000);hold on;plot(a(:,1),a(:,2),'r-','linewidth',1.1);axis equal;%axis off; figure(60000);hold on;set(gca,'XAxisLocation','origin','YAxisLocation','origin'); figure(60000);hold on;xlabel('x','FontSize',16); figure(60000);hold on;ylabel('y','FontSize',16); % 旋转矩阵 顺时针旋转3*pi/4 rotate_a1=[cos(3*pi/4) -sin(3*pi/4) sin(3*pi/4) cos(3*pi/4) ]; % 旋转矩阵 逆时针旋转7*pi/4 rotate_a2=[cos(7*pi/4) sin(7*pi/4) -sin(7*pi/4) cos(7*pi/4) ]; % 旋转矩阵 逆时针旋转5*pi/4 rotate_a3=[cos(5*pi/4) sin(5*pi/4) -sin(5*pi/4) cos(5*pi/4) ]; a_after_rotate_clock=rotate_a1*a'; a_after_rotate_clock_1=a_after_rotate_clock'; a_after_rotate_clockwise=rotate_a2*a'; a_after_rotate_clockwise_1=a_after_rotate_clockwise'; a_after_rotate_clockwise_2=rotate_a3*a'; a_after_rotate_clockwise_3=a_after_rotate_clockwise_2';
结果是:
