i*******d
发帖数: 6 | 1 Can anyone prove that the 2-norm of the following matrix is equal to 1?
i.e. || I - Mxx'M/(x'M'Mx) ||2 = 1
M is an any n by n matrix and x is a vector having n elements; ' indicats
transpose.
Thank you very much! |
B********e
发帖数: 10014 | 2
~~here mush be M'?
the problem is actually :
prove for any vector x
||I-xx'/x'x||_2=1
represent xx' and x'x in form of summation you'll see it
【在 i*******d 的大作中提到】
: Can anyone prove that the 2-norm of the following matrix is equal to 1?
: i.e. || I - Mxx'M/(x'M'Mx) ||2 = 1
: M is an any n by n matrix and x is a vector having n elements; ' indicats
: transpose.
: Thank you very much!
|
r*******y
发帖数: 1081 | 3 I also think so. And I-xx'/x'x is symmetric and 2-幂等矩阵 which has
only 1 and 0 as eigenvalues.
【在 B********e 的大作中提到】
:
: ~~here mush be M'?
: the problem is actually :
: prove for any vector x
: ||I-xx'/x'x||_2=1
: represent xx' and x'x in form of summation you'll see it
|
i*******d
发帖数: 6 | 4 Sorry, the correct one is || I - Mxx'M'/(x'M'Mx) ||2 = 1. By the way, I can
check it with any numeral values showing that it is true, but i couldn't
prove it. Thanks again! |
B********e
发帖数: 10014 | 5 why don't you just write down the summation form and the matrix?
let S=sum_1^n (v_i)^2=x'*x, ... ,go ahead
can
【在 i*******d 的大作中提到】
: Sorry, the correct one is || I - Mxx'M'/(x'M'Mx) ||2 = 1. By the way, I can
: check it with any numeral values showing that it is true, but i couldn't
: prove it. Thanks again!
|
i*******d
发帖数: 6 | 6 I think it is easy for me to prove that the matrix is idempotent and an
idempotent matrix has only 1 and 0 as eigenvalues.
Thank you both! |
c*******h
发帖数: 1096 | 7 just to remind you that 2-norm of a matrix A does not relate to the
eigenvalues
of A. you are just lucky here that the matrix in question is symmetric and
diagonalizable.
【在 i*******d 的大作中提到】
: I think it is easy for me to prove that the matrix is idempotent and an
: idempotent matrix has only 1 and 0 as eigenvalues.
: Thank you both!
|
i*******d
发帖数: 6 | 8 I am confused. I found that ||A||_2 = sqrt(max(eigenvalues of A)) on
Internet.
【在 c*******h 的大作中提到】
: just to remind you that 2-norm of a matrix A does not relate to the
: eigenvalues
: of A. you are just lucky here that the matrix in question is symmetric and
: diagonalizable.
|
B********e
发帖数: 10014 | 9 come on baby，who said that?
~~~A'*A吧
【在 i*******d 的大作中提到】
: I am confused. I found that ||A||_2 = sqrt(max(eigenvalues of A)) on
: Internet.
|
r*******y
发帖数: 1081 | 10 How can you calculate 2-norm if you don't know the definition ?
【在 i*******d 的大作中提到】
: I am confused. I found that ||A||_2 = sqrt(max(eigenvalues of A)) on
: Internet.
|
i*******d
发帖数: 6 | 11 Yes, you are right for general matrix A. But for idempotent matrix A, i.e. A
^2=A, the eigenvalues of A is the same as ones of A'*A, right?
【在 B********e 的大作中提到】
: come on baby，who said that?
:
: ~~~A'*A吧
|
