Range space $mathcal{R}(textbf{A})$ the same as $mathcal{R}(textbf{AA}^H)$?
$begingroup$
I'm working on a problem as follows:
Given $textbf{A}inmathbb{C}^{Mtimes N}$, show that $mathcal{R}(textbf{A})=mathcal{R}(textbf{AA}^H)$
where $mathcal{R}()$ denotes the range space of a matrix/transformation and $*^H$ denotes the conjugate transpose of a matrix.
I'm trying to show this by using SVD, but I get stuck relating $textbf{A}$ to $textbf{AA}^H$. So far, I have:
$textbf{A}=textbf{U}Sigmatextbf{V}^H$
$textbf{AA}^H=textbf{U}Sigmatextbf{V}^Htextbf{A}^H$
$textbf{AA}^H=textbf{U}textbf{U}^HSigmaSigma^Htextbf{V}^Htextbf{V}$
Note that $textbf{U}$ is the unitary $mtimes m$ matrix in SVD, $textbf{V}^H$ is the unitary $ntimes n$, and $Sigma$ is the matrix of singular values.
I know that the columns of $textbf{U}$ form a basis for $mathcal{R}(textbf{A})$, but when I use the SVD of A to try to do SVD on AA^H, I get to $textbf{UU}^H$, which is just the identity matrix and thus can't be a basis for a complex space?
Am I going about this all wrong, or is there some misstep I've made that's preventing me from solving the problem?
linear-algebra svd
edited Dec 5 '18 at 20:17
gangrene
905514
asked Dec 5 '18 at 20:12
W. MacTurkW. MacTurk
105
$endgroup$
add a comment |
$begingroup$
I'm working on a problem as follows:
Given $textbf{A}inmathbb{C}^{Mtimes N}$, show that $mathcal{R}(textbf{A})=mathcal{R}(textbf{AA}^H)$
where $mathcal{R}()$ denotes the range space of a matrix/transformation and $*^H$ denotes the conjugate transpose of a matrix.
I'm trying to show this by using SVD, but I get stuck relating $textbf{A}$ to $textbf{AA}^H$. So far, I have:
$textbf{A}=textbf{U}Sigmatextbf{V}^H$
$textbf{AA}^H=textbf{U}Sigmatextbf{V}^Htextbf{A}^H$
$textbf{AA}^H=textbf{U}textbf{U}^HSigmaSigma^Htextbf{V}^Htextbf{V}$
Note that $textbf{U}$ is the unitary $mtimes m$ matrix in SVD, $textbf{V}^H$ is the unitary $ntimes n$, and $Sigma$ is the matrix of singular values.
I know that the columns of $textbf{U}$ form a basis for $mathcal{R}(textbf{A})$, but when I use the SVD of A to try to do SVD on AA^H, I get to $textbf{UU}^H$, which is just the identity matrix and thus can't be a basis for a complex space?
Am I going about this all wrong, or is there some misstep I've made that's preventing me from solving the problem?
linear-algebra svd
edited Dec 5 '18 at 20:17
gangrene
905514
asked Dec 5 '18 at 20:12
W. MacTurkW. MacTurk
105
$endgroup$
$begingroup$
Is there something that breaks down in this arguement for complex? math.stackexchange.com/q/349742 (sorry I got the wrong link initially)
$endgroup$
– TrostAft
Dec 5 '18 at 20:17
$begingroup$
Also shouldnt $AA^H = U Sigma V^H V Sigma U^H = U Sigma^2 U^H$? Then using this it's clear that they have the same rank.
$endgroup$
– TrostAft
Dec 5 '18 at 20:20
$begingroup$
Ah, the first resource was very helpful, I forgot to search for the question using transpose instead of conjugate transpose (to include real vector spaces and find an analog). Your second answer I think is what I was looking for - sometimes I miss the obvious. Thank you so much!
$endgroup$
– W. MacTurk
Dec 5 '18 at 20:26
1
$begingroup$
@TrostAft In fact $ Sigma^2$ corresponds to $Sigma Sigma^H$. $Sigma$ is not a square matrix. I agree this notation is often used, but can be misleading in some occasions. In particular, we don't get the same $Sigma^2$ when calculating $AA^H$ or $A^HA$
$endgroup$
– Damien
Dec 5 '18 at 20:42
add a comment |
$begingroup$
I'm working on a problem as follows:
Given $textbf{A}inmathbb{C}^{Mtimes N}$, show that $mathcal{R}(textbf{A})=mathcal{R}(textbf{AA}^H)$
where $mathcal{R}()$ denotes the range space of a matrix/transformation and $*^H$ denotes the conjugate transpose of a matrix.
I'm trying to show this by using SVD, but I get stuck relating $textbf{A}$ to $textbf{AA}^H$. So far, I have:
$textbf{A}=textbf{U}Sigmatextbf{V}^H$
$textbf{AA}^H=textbf{U}Sigmatextbf{V}^Htextbf{A}^H$
$textbf{AA}^H=textbf{U}textbf{U}^HSigmaSigma^Htextbf{V}^Htextbf{V}$
Note that $textbf{U}$ is the unitary $mtimes m$ matrix in SVD, $textbf{V}^H$ is the unitary $ntimes n$, and $Sigma$ is the matrix of singular values.
I know that the columns of $textbf{U}$ form a basis for $mathcal{R}(textbf{A})$, but when I use the SVD of A to try to do SVD on AA^H, I get to $textbf{UU}^H$, which is just the identity matrix and thus can't be a basis for a complex space?
Am I going about this all wrong, or is there some misstep I've made that's preventing me from solving the problem?
linear-algebra svd
edited Dec 5 '18 at 20:17
gangrene
905514
asked Dec 5 '18 at 20:12
W. MacTurkW. MacTurk
105
$endgroup$
I'm working on a problem as follows:
Given $textbf{A}inmathbb{C}^{Mtimes N}$, show that $mathcal{R}(textbf{A})=mathcal{R}(textbf{AA}^H)$
where $mathcal{R}()$ denotes the range space of a matrix/transformation and $*^H$ denotes the conjugate transpose of a matrix.
I'm trying to show this by using SVD, but I get stuck relating $textbf{A}$ to $textbf{AA}^H$. So far, I have:
$textbf{A}=textbf{U}Sigmatextbf{V}^H$
$textbf{AA}^H=textbf{U}Sigmatextbf{V}^Htextbf{A}^H$
$textbf{AA}^H=textbf{U}textbf{U}^HSigmaSigma^Htextbf{V}^Htextbf{V}$
Note that $textbf{U}$ is the unitary $mtimes m$ matrix in SVD, $textbf{V}^H$ is the unitary $ntimes n$, and $Sigma$ is the matrix of singular values.
I know that the columns of $textbf{U}$ form a basis for $mathcal{R}(textbf{A})$, but when I use the SVD of A to try to do SVD on AA^H, I get to $textbf{UU}^H$, which is just the identity matrix and thus can't be a basis for a complex space?
Am I going about this all wrong, or is there some misstep I've made that's preventing me from solving the problem?
linear-algebra svd
linear-algebra svd
edited Dec 5 '18 at 20:17
gangrene
905514
asked Dec 5 '18 at 20:12
W. MacTurkW. MacTurk
105
edited Dec 5 '18 at 20:17
gangrene
905514
asked Dec 5 '18 at 20:12
W. MacTurkW. MacTurk
105
edited Dec 5 '18 at 20:17
gangrene
905514
edited Dec 5 '18 at 20:17
gangrene
905514
edited Dec 5 '18 at 20:17
gangrene
905514
905514
asked Dec 5 '18 at 20:12
W. MacTurkW. MacTurk
105
asked Dec 5 '18 at 20:12
W. MacTurkW. MacTurk
105
asked Dec 5 '18 at 20:12
W. MacTurkW. MacTurk
105
105
$begingroup$
Is there something that breaks down in this arguement for complex? math.stackexchange.com/q/349742 (sorry I got the wrong link initially)
$endgroup$
– TrostAft
Dec 5 '18 at 20:17
$begingroup$
Also shouldnt $AA^H = U Sigma V^H V Sigma U^H = U Sigma^2 U^H$? Then using this it's clear that they have the same rank.
$endgroup$
– TrostAft
Dec 5 '18 at 20:20
$begingroup$
Ah, the first resource was very helpful, I forgot to search for the question using transpose instead of conjugate transpose (to include real vector spaces and find an analog). Your second answer I think is what I was looking for - sometimes I miss the obvious. Thank you so much!
$endgroup$
– W. MacTurk
Dec 5 '18 at 20:26
1
$begingroup$
@TrostAft In fact $ Sigma^2$ corresponds to $Sigma Sigma^H$. $Sigma$ is not a square matrix. I agree this notation is often used, but can be misleading in some occasions. In particular, we don't get the same $Sigma^2$ when calculating $AA^H$ or $A^HA$
$endgroup$
– Damien
Dec 5 '18 at 20:42
add a comment |
$begingroup$
Is there something that breaks down in this arguement for complex? math.stackexchange.com/q/349742 (sorry I got the wrong link initially)
$endgroup$
– TrostAft
Dec 5 '18 at 20:17
$begingroup$
Also shouldnt $AA^H = U Sigma V^H V Sigma U^H = U Sigma^2 U^H$? Then using this it's clear that they have the same rank.
$endgroup$
– TrostAft
Dec 5 '18 at 20:20
$begingroup$
Ah, the first resource was very helpful, I forgot to search for the question using transpose instead of conjugate transpose (to include real vector spaces and find an analog). Your second answer I think is what I was looking for - sometimes I miss the obvious. Thank you so much!
$endgroup$
– W. MacTurk
Dec 5 '18 at 20:26
1
$begingroup$
@TrostAft In fact $ Sigma^2$ corresponds to $Sigma Sigma^H$. $Sigma$ is not a square matrix. I agree this notation is often used, but can be misleading in some occasions. In particular, we don't get the same $Sigma^2$ when calculating $AA^H$ or $A^HA$
$endgroup$
– Damien
Dec 5 '18 at 20:42
$begingroup$
Is there something that breaks down in this arguement for complex? math.stackexchange.com/q/349742 (sorry I got the wrong link initially)
$endgroup$
– TrostAft
Dec 5 '18 at 20:17
$begingroup$
Is there something that breaks down in this arguement for complex? math.stackexchange.com/q/349742 (sorry I got the wrong link initially)
$endgroup$
– TrostAft
Dec 5 '18 at 20:17
$begingroup$
Also shouldnt $AA^H = U Sigma V^H V Sigma U^H = U Sigma^2 U^H$? Then using this it's clear that they have the same rank.
$endgroup$
– TrostAft
Dec 5 '18 at 20:20
$begingroup$
Also shouldnt $AA^H = U Sigma V^H V Sigma U^H = U Sigma^2 U^H$? Then using this it's clear that they have the same rank.
$endgroup$
– TrostAft
Dec 5 '18 at 20:20
$begingroup$
Ah, the first resource was very helpful, I forgot to search for the question using transpose instead of conjugate transpose (to include real vector spaces and find an analog). Your second answer I think is what I was looking for - sometimes I miss the obvious. Thank you so much!
$endgroup$
– W. MacTurk
Dec 5 '18 at 20:26
$begingroup$
Ah, the first resource was very helpful, I forgot to search for the question using transpose instead of conjugate transpose (to include real vector spaces and find an analog). Your second answer I think is what I was looking for - sometimes I miss the obvious. Thank you so much!
$endgroup$
– W. MacTurk
Dec 5 '18 at 20:26
1
1
$begingroup$
@TrostAft In fact $ Sigma^2$ corresponds to $Sigma Sigma^H$. $Sigma$ is not a square matrix. I agree this notation is often used, but can be misleading in some occasions. In particular, we don't get the same $Sigma^2$ when calculating $AA^H$ or $A^HA$
$endgroup$
– Damien
Dec 5 '18 at 20:42
$begingroup$
@TrostAft In fact $ Sigma^2$ corresponds to $Sigma Sigma^H$. $Sigma$ is not a square matrix. I agree this notation is often used, but can be misleading in some occasions. In particular, we don't get the same $Sigma^2$ when calculating $AA^H$ or $A^HA$
$endgroup$
– Damien
Dec 5 '18 at 20:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For this kind of problem, it is convenient to use the following form of SVD:
$$A = sum_{i=1}^{rank} lambda_i u_i v^H_i$$
Where the $lambda_i > 0$
Then,
$$ AA^H = sum_{i=1}^{rank} lambda^2_i u_i u^H_i$$
It follows immediately that $A$ and $AA^H$ have the same rank
answered Dec 5 '18 at 20:25
DamienDamien
58214
$endgroup$
$begingroup$
But does having the same rank here indicate that A and AA^H have the same range space?
$endgroup$
– W. MacTurk
Dec 5 '18 at 20:29
$begingroup$
Yes: the set generated by the $u_i$ for which the $lambda_i$ are greater than 0
$endgroup$
– Damien
Dec 5 '18 at 20:32
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For this kind of problem, it is convenient to use the following form of SVD:
$$A = sum_{i=1}^{rank} lambda_i u_i v^H_i$$
Where the $lambda_i > 0$
Then,
$$ AA^H = sum_{i=1}^{rank} lambda^2_i u_i u^H_i$$
It follows immediately that $A$ and $AA^H$ have the same rank
answered Dec 5 '18 at 20:25
DamienDamien
58214
$endgroup$
$begingroup$
But does having the same rank here indicate that A and AA^H have the same range space?
$endgroup$
– W. MacTurk
Dec 5 '18 at 20:29
$begingroup$
Yes: the set generated by the $u_i$ for which the $lambda_i$ are greater than 0
$endgroup$
– Damien
Dec 5 '18 at 20:32
add a comment |
$begingroup$
For this kind of problem, it is convenient to use the following form of SVD:
$$A = sum_{i=1}^{rank} lambda_i u_i v^H_i$$
Where the $lambda_i > 0$
Then,
$$ AA^H = sum_{i=1}^{rank} lambda^2_i u_i u^H_i$$
It follows immediately that $A$ and $AA^H$ have the same rank
answered Dec 5 '18 at 20:25
DamienDamien
58214
$endgroup$
$begingroup$
But does having the same rank here indicate that A and AA^H have the same range space?
$endgroup$
– W. MacTurk
Dec 5 '18 at 20:29
$begingroup$
Yes: the set generated by the $u_i$ for which the $lambda_i$ are greater than 0
$endgroup$
– Damien
Dec 5 '18 at 20:32
add a comment |
$begingroup$
For this kind of problem, it is convenient to use the following form of SVD:
$$A = sum_{i=1}^{rank} lambda_i u_i v^H_i$$
Where the $lambda_i > 0$
Then,
$$ AA^H = sum_{i=1}^{rank} lambda^2_i u_i u^H_i$$
It follows immediately that $A$ and $AA^H$ have the same rank
answered Dec 5 '18 at 20:25
DamienDamien
58214
$endgroup$
For this kind of problem, it is convenient to use the following form of SVD:
$$A = sum_{i=1}^{rank} lambda_i u_i v^H_i$$
Where the $lambda_i > 0$
Then,
$$ AA^H = sum_{i=1}^{rank} lambda^2_i u_i u^H_i$$
It follows immediately that $A$ and $AA^H$ have the same rank
answered Dec 5 '18 at 20:25
DamienDamien
58214
answered Dec 5 '18 at 20:25
DamienDamien
58214
answered Dec 5 '18 at 20:25
DamienDamien
58214
answered Dec 5 '18 at 20:25
DamienDamien
58214
58214
$begingroup$
But does having the same rank here indicate that A and AA^H have the same range space?
$endgroup$
– W. MacTurk
Dec 5 '18 at 20:29
$begingroup$
Yes: the set generated by the $u_i$ for which the $lambda_i$ are greater than 0
$endgroup$
– Damien
Dec 5 '18 at 20:32
add a comment |
$begingroup$
But does having the same rank here indicate that A and AA^H have the same range space?
$endgroup$
– W. MacTurk
Dec 5 '18 at 20:29
$begingroup$
Yes: the set generated by the $u_i$ for which the $lambda_i$ are greater than 0
$endgroup$
– Damien
Dec 5 '18 at 20:32
$begingroup$
But does having the same rank here indicate that A and AA^H have the same range space?
$endgroup$
– W. MacTurk
Dec 5 '18 at 20:29
$begingroup$
But does having the same rank here indicate that A and AA^H have the same range space?
$endgroup$
– W. MacTurk
Dec 5 '18 at 20:29
$begingroup$
Yes: the set generated by the $u_i$ for which the $lambda_i$ are greater than 0
$endgroup$
– Damien
Dec 5 '18 at 20:32
$begingroup$
Yes: the set generated by the $u_i$ for which the $lambda_i$ are greater than 0
$endgroup$
– Damien
Dec 5 '18 at 20:32
add a comment |
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$begingroup$
Is there something that breaks down in this arguement for complex? math.stackexchange.com/q/349742 (sorry I got the wrong link initially)
$endgroup$
– TrostAft
Dec 5 '18 at 20:17
$begingroup$
Also shouldnt $AA^H = U Sigma V^H V Sigma U^H = U Sigma^2 U^H$? Then using this it's clear that they have the same rank.
$endgroup$
– TrostAft
Dec 5 '18 at 20:20
$begingroup$
Ah, the first resource was very helpful, I forgot to search for the question using transpose instead of conjugate transpose (to include real vector spaces and find an analog). Your second answer I think is what I was looking for - sometimes I miss the obvious. Thank you so much!
$endgroup$
– W. MacTurk
Dec 5 '18 at 20:26
1
$begingroup$
@TrostAft In fact $ Sigma^2$ corresponds to $Sigma Sigma^H$. $Sigma$ is not a square matrix. I agree this notation is often used, but can be misleading in some occasions. In particular, we don't get the same $Sigma^2$ when calculating $AA^H$ or $A^HA$
$endgroup$
– Damien
Dec 5 '18 at 20:42