Mixing
Prove $I-YY^dagger$ is a projection operator for $N(Y)$
$begingroup$
Let $Yin mathbb{R}^{m*m}$ be a symmetric matrix and how to prove $I-YY^dagger$ is a projection operator for $N(Y)$? I know $I-Y^dagger Y$ is a projection operator for $N(Y)$.
Thanks in advance!
matrices
asked Feb 1 at 8:15
Jingyang LiJingyang Li
113
$endgroup$
add a comment |
$begingroup$
Let $Yin mathbb{R}^{m*m}$ be a symmetric matrix and how to prove $I-YY^dagger$ is a projection operator for $N(Y)$? I know $I-Y^dagger Y$ is a projection operator for $N(Y)$.
Thanks in advance!
matrices
asked Feb 1 at 8:15
Jingyang LiJingyang Li
113
$endgroup$
$begingroup$
Does $N(Y) = ker Y = { y mid Yy = 0}$?
$endgroup$
– Robert Lewis
Feb 1 at 8:18
$begingroup$
@RobertLewis yes
$endgroup$
– Jingyang Li
Feb 1 at 8:44
add a comment |
$begingroup$
Let $Yin mathbb{R}^{m*m}$ be a symmetric matrix and how to prove $I-YY^dagger$ is a projection operator for $N(Y)$? I know $I-Y^dagger Y$ is a projection operator for $N(Y)$.
Thanks in advance!
matrices
asked Feb 1 at 8:15
Jingyang LiJingyang Li
113
$endgroup$
Let $Yin mathbb{R}^{m*m}$ be a symmetric matrix and how to prove $I-YY^dagger$ is a projection operator for $N(Y)$? I know $I-Y^dagger Y$ is a projection operator for $N(Y)$.
Thanks in advance!
matrices
matrices
asked Feb 1 at 8:15
Jingyang LiJingyang Li
113
asked Feb 1 at 8:15
Jingyang LiJingyang Li
113
asked Feb 1 at 8:15
Jingyang LiJingyang Li
113
asked Feb 1 at 8:15
Jingyang LiJingyang Li
113
asked Feb 1 at 8:15
Jingyang LiJingyang Li
113
113
$begingroup$
Does $N(Y) = ker Y = { y mid Yy = 0}$?
$endgroup$
– Robert Lewis
Feb 1 at 8:18
$begingroup$
@RobertLewis yes
$endgroup$
– Jingyang Li
Feb 1 at 8:44
add a comment |
$begingroup$
Does $N(Y) = ker Y = { y mid Yy = 0}$?
$endgroup$
– Robert Lewis
Feb 1 at 8:18
$begingroup$
@RobertLewis yes
$endgroup$
– Jingyang Li
Feb 1 at 8:44
$begingroup$
Does $N(Y) = ker Y = { y mid Yy = 0}$?
$endgroup$
– Robert Lewis
Feb 1 at 8:18
$begingroup$
Does $N(Y) = ker Y = { y mid Yy = 0}$?
$endgroup$
– Robert Lewis
Feb 1 at 8:18
$begingroup$
@RobertLewis yes
$endgroup$
– Jingyang Li
Feb 1 at 8:44
$begingroup$
@RobertLewis yes
$endgroup$
– Jingyang Li
Feb 1 at 8:44
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
How can you rewrite $I-YY^dagger$ if you know Y is symmetric and you already know that $I-Y^dagger Y$ is a projection operator?
answered Feb 1 at 8:55
GenericUsrnmeGenericUsrnme
34017
$endgroup$
add a comment |
$begingroup$
Since $Y$ is real symmetric,
$Y^dagger = Y^T = Y; tag 1$
thus
$Y^dagger Y = YY^dagger = Y^2, tag 2$
whence
$I - Y^dagger Y = I - YY^dagger = I - Y^2; tag 3$
it therefore follows that $I - YY^dagger$ is a projection onto $N(Y)$ if and only if $I - Y^dagger Y$ is . . .
answered Feb 2 at 5:34
Robert LewisRobert Lewis
48.9k23168
$endgroup$
$begingroup$
why is $Y^dagger = Y^T$ true?
$endgroup$
– Jingyang Li
Feb 3 at 8:44
$begingroup$
@Jingyang: Because $Y$ is real and symmetric.
$endgroup$
– Robert Lewis
Feb 3 at 12:02
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
How can you rewrite $I-YY^dagger$ if you know Y is symmetric and you already know that $I-Y^dagger Y$ is a projection operator?
answered Feb 1 at 8:55
GenericUsrnmeGenericUsrnme
34017
$endgroup$
add a comment |
$begingroup$
How can you rewrite $I-YY^dagger$ if you know Y is symmetric and you already know that $I-Y^dagger Y$ is a projection operator?
answered Feb 1 at 8:55
GenericUsrnmeGenericUsrnme
34017
$endgroup$
add a comment |
$begingroup$
How can you rewrite $I-YY^dagger$ if you know Y is symmetric and you already know that $I-Y^dagger Y$ is a projection operator?
answered Feb 1 at 8:55
GenericUsrnmeGenericUsrnme
34017
$endgroup$
How can you rewrite $I-YY^dagger$ if you know Y is symmetric and you already know that $I-Y^dagger Y$ is a projection operator?
answered Feb 1 at 8:55
GenericUsrnmeGenericUsrnme
34017
answered Feb 1 at 8:55
GenericUsrnmeGenericUsrnme
34017
answered Feb 1 at 8:55
GenericUsrnmeGenericUsrnme
34017
answered Feb 1 at 8:55
GenericUsrnmeGenericUsrnme
34017
34017
add a comment |
add a comment |
$begingroup$
Since $Y$ is real symmetric,
$Y^dagger = Y^T = Y; tag 1$
thus
$Y^dagger Y = YY^dagger = Y^2, tag 2$
whence
$I - Y^dagger Y = I - YY^dagger = I - Y^2; tag 3$
it therefore follows that $I - YY^dagger$ is a projection onto $N(Y)$ if and only if $I - Y^dagger Y$ is . . .
answered Feb 2 at 5:34
Robert LewisRobert Lewis
48.9k23168
$endgroup$
$begingroup$
why is $Y^dagger = Y^T$ true?
$endgroup$
– Jingyang Li
Feb 3 at 8:44
$begingroup$
@Jingyang: Because $Y$ is real and symmetric.
$endgroup$
– Robert Lewis
Feb 3 at 12:02
add a comment |
$begingroup$
Since $Y$ is real symmetric,
$Y^dagger = Y^T = Y; tag 1$
thus
$Y^dagger Y = YY^dagger = Y^2, tag 2$
whence
$I - Y^dagger Y = I - YY^dagger = I - Y^2; tag 3$
it therefore follows that $I - YY^dagger$ is a projection onto $N(Y)$ if and only if $I - Y^dagger Y$ is . . .
answered Feb 2 at 5:34
Robert LewisRobert Lewis
48.9k23168
$endgroup$
$begingroup$
why is $Y^dagger = Y^T$ true?
$endgroup$
– Jingyang Li
Feb 3 at 8:44
$begingroup$
@Jingyang: Because $Y$ is real and symmetric.
$endgroup$
– Robert Lewis
Feb 3 at 12:02
add a comment |
$begingroup$
Since $Y$ is real symmetric,
$Y^dagger = Y^T = Y; tag 1$
thus
$Y^dagger Y = YY^dagger = Y^2, tag 2$
whence
$I - Y^dagger Y = I - YY^dagger = I - Y^2; tag 3$
it therefore follows that $I - YY^dagger$ is a projection onto $N(Y)$ if and only if $I - Y^dagger Y$ is . . .
answered Feb 2 at 5:34
Robert LewisRobert Lewis
48.9k23168
$endgroup$
Since $Y$ is real symmetric,
$Y^dagger = Y^T = Y; tag 1$
thus
$Y^dagger Y = YY^dagger = Y^2, tag 2$
whence
$I - Y^dagger Y = I - YY^dagger = I - Y^2; tag 3$
it therefore follows that $I - YY^dagger$ is a projection onto $N(Y)$ if and only if $I - Y^dagger Y$ is . . .
answered Feb 2 at 5:34
Robert LewisRobert Lewis
48.9k23168
answered Feb 2 at 5:34
Robert LewisRobert Lewis
48.9k23168
answered Feb 2 at 5:34
Robert LewisRobert Lewis
48.9k23168
answered Feb 2 at 5:34
Robert LewisRobert Lewis
48.9k23168
48.9k23168
$begingroup$
why is $Y^dagger = Y^T$ true?
$endgroup$
– Jingyang Li
Feb 3 at 8:44
$begingroup$
@Jingyang: Because $Y$ is real and symmetric.
$endgroup$
– Robert Lewis
Feb 3 at 12:02
add a comment |
$begingroup$
why is $Y^dagger = Y^T$ true?
$endgroup$
– Jingyang Li
Feb 3 at 8:44
$begingroup$
@Jingyang: Because $Y$ is real and symmetric.
$endgroup$
– Robert Lewis
Feb 3 at 12:02
$begingroup$
why is $Y^dagger = Y^T$ true?
$endgroup$
– Jingyang Li
Feb 3 at 8:44
$begingroup$
why is $Y^dagger = Y^T$ true?
$endgroup$
– Jingyang Li
Feb 3 at 8:44
$begingroup$
@Jingyang: Because $Y$ is real and symmetric.
$endgroup$
– Robert Lewis
Feb 3 at 12:02
$begingroup$
@Jingyang: Because $Y$ is real and symmetric.
$endgroup$
– Robert Lewis
Feb 3 at 12:02
add a comment |
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$begingroup$
Does $N(Y) = ker Y = { y mid Yy = 0}$?
$endgroup$
– Robert Lewis
Feb 1 at 8:18
$begingroup$
@RobertLewis yes
$endgroup$
– Jingyang Li
Feb 1 at 8:44