t**o
发帖数: 2618 | 1 Ａ是对称半正定矩阵，特征值【０，１】之间，
求证ＡＱ的特征值也在【０，１】之间，Ｑ是个正交矩阵 | t**o
发帖数: 2618 | | d****r
发帖数: 1017 | | n******d
发帖数: 244 | 4 You need to require that Q be special 正交矩阵, i.e. det Q=1.
Otherewise you only get ＡＱ的特征值在【-1，１】之间.
You can prove it easily with the observation of its geometry.
【在 t**o 的大作中提到】
: Ａ是对称半正定矩阵，特征值【０，１】之间，
: 求证ＡＱ的特征值也在【０，１】之间，Ｑ是个正交矩阵
| t**o
发帖数: 2618 | 5 兄台能否详细一点....
我没发现如何use its geometry.... | n******d
发帖数: 244 | 6 I was wrong. I think you can only get eigenvalue between -1 and 1 even if
det Q=1.
For example, if Q =-I in even dimension, you get negative eigenvalues.
As to the proof, matrix A maps unit ball into itself, and so matrix AQ has
the same property since Q is only a rotation or reflection. And this
property implies that all eigenvalues are between -1 and 1.
If you want to prove it algebraically, you could write Qx into linear
combination of eigenvectors of A and argue that AQx=rx implies that r^2\
【在 t**o 的大作中提到】
: 兄台能否详细一点....
: 我没发现如何use its geometry....
| t**o
发帖数: 2618 | 7 我想符号 是一定的
由于rotation是单联通的， 所以 特征值不会穿越0这个点。（A正定是对的，如果A半
正定，估计也对）
只是 此题我始终没看出来
如何用几何 直接把它说清楚 | s**e
发帖数: 1834 | 8 reflection is not 单联通的.
I think netnomad's "geometric interpretation" is fairly clear, I just restate it here:
Since A is 半正定 and has all eigenvalues between [0,1], so for any vector x inside a unit ball, Ax is also inside a unit ball. So A maps a unit ball into itself.
Now, because Q is a rotation/reflection, so for any x, we have |Qx|=|x|. So AQ also maps a unit ball into itself. So for any REAL eigenvalue of AQ, it must be between [-1,1]. Since we can prove AQ only has real eigenvalue (becaus
【在 t**o 的大作中提到】
: 我想符号 是一定的
: 由于rotation是单联通的， 所以 特征值不会穿越0这个点。（A正定是对的，如果A半
: 正定，估计也对）
: 只是 此题我始终没看出来
: 如何用几何 直接把它说清楚
|
| t**o
发帖数: 2618 | | s**e
发帖数: 1834 | 10 You are right. e.g. A = [1,0;0,1], Q=[cos(t),sin(t);-sin(t),cos(t)].
Then AQ=Q has eigenvalues exp(i*t) and exp(-i*t).
But you can still get that "AQ's eightvalue's absolute value is between [0,1]".
【在 t**o 的大作中提到】
: 对了，AQ特征值可能不是实数。。。。
| t**o
发帖数: 2618 | 11 was wrong. I think you can only get eigenvalue between -1 and 1 even if
det Q=1.
如果是rotation，我想AQ的特征值的符号不会变 |
|
