什么是“顺时针”以及“逆时针”方向。在矩阵旋转中？

发布于 2024-10-26 18:53:16 字数 799 浏览 8 评论 0原文

已移动： 将此问题移至 math.stackexchange.com 并关闭它。

我正在学习 PCA 中涉及的数学。出于我的目的，我只是想了解 90° 旋转矩阵。我得到了旋转矩阵的概念，但是当我查看 wikipedia 时，Wolfram Mathworld 网站等。我不断看到以下定义为 90° 逆时针 旋转矩阵：

|0 -1|
|1  0|

但是当我实际进行数学计算时，我似乎得到了围绕原点顺时针旋转的点：

|1 2|     |0 -1|     |2 -1|
|3 4|  x  |1  0|  =  |4 -3|

绘制点（1,2）和（3,4），它们都在象限 1 (+,+)。绘制结果点 (2,-1) 和 (4,-3)，它们都位于四边形 4 (+,-) 中。旋转有效，但看起来是顺时针，而不是逆时针。我缺少什么？

注意一位朋友建议这是旋转的坐标系，但 Wolfram 站点（上面链接）似乎明确排除了这是我误解的原因（请参阅（ 1) 和 (3) 在 Wolfram 网站上）。

非常感谢任何帮助。

原文

MOVED: Moving this question to math.stackexchange.com and closing it.

I'm learning about the math invovled in PCA. For my purposes here, I'm just trying to understand a 90° rotation matrix. I get the concept of a rotation matrix, but when I look on wikipedia, the Wolfram Mathworld site, etc. I keep seeing the following defined as a 90° counter-clockwise rotation matrix:

|0 -1|
|1  0|

but when I actually do the math, I seem to get the point(s) rotated clockwise around the origin:

|1 2|     |0 -1|     |2 -1|
|3 4|  x  |1  0|  =  |4 -3|

Graphing the points (1,2) and (3,4), they're both in quadrant 1 (+,+). Graphing the resulting points, (2,-1) and (4,-3), they're both in quadrang 4 (+,-). The rotation works, but it seems clockwise, not counter-clockwise. What am I missing?

Note A friend suggested that it's the coordinate system that's being rotated, but the wolfram site (linked above) seems to explicitly exclude that from being the cause of my misunderstanding (see (1) and (3) on the wolfram site).

Any help greatly appreciated.

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

你好，陌生人 2024-11-02 18:53:16

x'  =  | 0  -1 | * | x |
y'     | 1   0 |   | y |

所以你想要的是

x'  =  | 0  -1 | * | 1 |
y'     | 1   0 |   | 2 |  =  (-2, 1)

x'  =  | 0  -1 | * | 3 |
y'     | 1   0 |   | 4 |  =  (-4, 3)

如果你在图表上绘制它并在两个原始点和两个新点之间画一条线，然后从原点到每条线的第一个点画一条虚线，然后画一个 90两条虚线之间的度角标记，您将看到 90 度角已从原点逆时针旋转。

x'  =  | 0  -1 | * | x |
y'     | 1   0 |   | y |

So what you want is

x'  =  | 0  -1 | * | 1 |
y'     | 1   0 |   | 2 |  =  (-2, 1)

x'  =  | 0  -1 | * | 3 |
y'     | 1   0 |   | 4 |  =  (-4, 3)

If you plot that on a graph and draw a line between the two original points, and the two new points, and then draw a dotted line from the origin to the first point of each line, and then draw a 90 degree angle marker between the two dotted lines you will see that the 90 degree angle has rotated counter clockwise from the origin.

回复收藏 0 原文