Mixing
An inequality for a quadratic form and an inner product, and its relationship to the singular value...
$begingroup$
Let $A$ be a $p times q$ real matrix of full column rank $q$, and let $u, v$ be two real vectors of (euclidean) norm 1.
I want to know whether the following inequality holds:
$$
v^TA^TAv geq v^T A^T u u^T A v,
$$
with equality if and only if $u$ ($v$) is the left (right) singular vector associated with any singular value of $A$.
As trivial as I think it must be, I can't wrap my head around this.
[Edit]
I could come up with a solution using the Cauchy-Schwarz inequality. Letting $y= Av$, we have
$$
v^T A^T A v = y^T y = <y, y> <v, v> quadgeqquad <y, v>^2 = y^Tvv^Ty = v^TA^Tv v^TA v
$$
So while I could not prove the "if and only if" part, it seems I could at least prove the inequality.
matrix-decomposition quadratic-forms svd
asked Jan 30 at 14:36
wiwhwiwh
11
$endgroup$
add a comment |
$begingroup$
Let $A$ be a $p times q$ real matrix of full column rank $q$, and let $u, v$ be two real vectors of (euclidean) norm 1.
I want to know whether the following inequality holds:
$$
v^TA^TAv geq v^T A^T u u^T A v,
$$
with equality if and only if $u$ ($v$) is the left (right) singular vector associated with any singular value of $A$.
As trivial as I think it must be, I can't wrap my head around this.
[Edit]
I could come up with a solution using the Cauchy-Schwarz inequality. Letting $y= Av$, we have
$$
v^T A^T A v = y^T y = <y, y> <v, v> quadgeqquad <y, v>^2 = y^Tvv^Ty = v^TA^Tv v^TA v
$$
So while I could not prove the "if and only if" part, it seems I could at least prove the inequality.
matrix-decomposition quadratic-forms svd
asked Jan 30 at 14:36
wiwhwiwh
11
$endgroup$
$begingroup$
In terms of the SVD of $A$, what's the maximum possible value of $u^{T}Av$? What about $v^{T}A^{T}Av$ and its eigenvalues?
$endgroup$
– Brian Borchers
Jan 30 at 15:20
$begingroup$
The maximum possible value of $u^T Av$ is the spectral norm of $A$ (the largest of its singular value). As for $v^TA^Tv$, its maximum eigenvalue is the square the spectral norm of $A$ (by definintion). Is it not? I'm not so much stuck for the equality part, rather for the inequality. I'm sure you're hinting at that, but I fail to see it.
$endgroup$
– wiwh
Jan 30 at 16:13
$begingroup$
You might try a change of basis to expand $u$ in terms of the columns of the $U$ matrix of $A$ and $v$ in terms of the columns of the $V$ matrix of the SVD of $A$. The coordinates of $u$ and $v$ with respect to these bases will then be combined with the singular values of $A$ in the inequality.
$endgroup$
– Brian Borchers
Jan 30 at 16:24
$begingroup$
I edited my question with a beginning of an answer. This is enough for my use, but it is not as strong a result as I was expecting to find. Thank you for your hints @BrianBorchers. Should I write my edit as an answer or close this question somehow? (please bear with me as I am -obviously- new here, trying to do the right thing).
$endgroup$
– wiwh
Feb 1 at 13:45
$begingroup$
Do you know the "if and only if" version of Cuachy-Schwarz? When is there equality?
$endgroup$
– Brian Borchers
Feb 1 at 14:20
add a comment |
$begingroup$
Let $A$ be a $p times q$ real matrix of full column rank $q$, and let $u, v$ be two real vectors of (euclidean) norm 1.
I want to know whether the following inequality holds:
$$
v^TA^TAv geq v^T A^T u u^T A v,
$$
with equality if and only if $u$ ($v$) is the left (right) singular vector associated with any singular value of $A$.
As trivial as I think it must be, I can't wrap my head around this.
[Edit]
I could come up with a solution using the Cauchy-Schwarz inequality. Letting $y= Av$, we have
$$
v^T A^T A v = y^T y = <y, y> <v, v> quadgeqquad <y, v>^2 = y^Tvv^Ty = v^TA^Tv v^TA v
$$
So while I could not prove the "if and only if" part, it seems I could at least prove the inequality.
matrix-decomposition quadratic-forms svd
asked Jan 30 at 14:36
wiwhwiwh
11
$endgroup$
Let $A$ be a $p times q$ real matrix of full column rank $q$, and let $u, v$ be two real vectors of (euclidean) norm 1.
I want to know whether the following inequality holds:
$$
v^TA^TAv geq v^T A^T u u^T A v,
$$
with equality if and only if $u$ ($v$) is the left (right) singular vector associated with any singular value of $A$.
As trivial as I think it must be, I can't wrap my head around this.
[Edit]
I could come up with a solution using the Cauchy-Schwarz inequality. Letting $y= Av$, we have
$$
v^T A^T A v = y^T y = <y, y> <v, v> quadgeqquad <y, v>^2 = y^Tvv^Ty = v^TA^Tv v^TA v
$$
So while I could not prove the "if and only if" part, it seems I could at least prove the inequality.
matrix-decomposition quadratic-forms svd
matrix-decomposition quadratic-forms svd
asked Jan 30 at 14:36
wiwhwiwh
11
asked Jan 30 at 14:36
wiwhwiwh
11
edited Feb 1 at 13:44
wiwh
asked Jan 30 at 14:36
wiwhwiwh
11
asked Jan 30 at 14:36
wiwhwiwh
11
asked Jan 30 at 14:36
wiwhwiwh
11
11
$begingroup$
In terms of the SVD of $A$, what's the maximum possible value of $u^{T}Av$? What about $v^{T}A^{T}Av$ and its eigenvalues?
$endgroup$
– Brian Borchers
Jan 30 at 15:20
$begingroup$
The maximum possible value of $u^T Av$ is the spectral norm of $A$ (the largest of its singular value). As for $v^TA^Tv$, its maximum eigenvalue is the square the spectral norm of $A$ (by definintion). Is it not? I'm not so much stuck for the equality part, rather for the inequality. I'm sure you're hinting at that, but I fail to see it.
$endgroup$
– wiwh
Jan 30 at 16:13
$begingroup$
You might try a change of basis to expand $u$ in terms of the columns of the $U$ matrix of $A$ and $v$ in terms of the columns of the $V$ matrix of the SVD of $A$. The coordinates of $u$ and $v$ with respect to these bases will then be combined with the singular values of $A$ in the inequality.
$endgroup$
– Brian Borchers
Jan 30 at 16:24
$begingroup$
I edited my question with a beginning of an answer. This is enough for my use, but it is not as strong a result as I was expecting to find. Thank you for your hints @BrianBorchers. Should I write my edit as an answer or close this question somehow? (please bear with me as I am -obviously- new here, trying to do the right thing).
$endgroup$
– wiwh
Feb 1 at 13:45
$begingroup$
Do you know the "if and only if" version of Cuachy-Schwarz? When is there equality?
$endgroup$
– Brian Borchers
Feb 1 at 14:20
add a comment |
$begingroup$
In terms of the SVD of $A$, what's the maximum possible value of $u^{T}Av$? What about $v^{T}A^{T}Av$ and its eigenvalues?
$endgroup$
– Brian Borchers
Jan 30 at 15:20
$begingroup$
The maximum possible value of $u^T Av$ is the spectral norm of $A$ (the largest of its singular value). As for $v^TA^Tv$, its maximum eigenvalue is the square the spectral norm of $A$ (by definintion). Is it not? I'm not so much stuck for the equality part, rather for the inequality. I'm sure you're hinting at that, but I fail to see it.
$endgroup$
– wiwh
Jan 30 at 16:13
$begingroup$
You might try a change of basis to expand $u$ in terms of the columns of the $U$ matrix of $A$ and $v$ in terms of the columns of the $V$ matrix of the SVD of $A$. The coordinates of $u$ and $v$ with respect to these bases will then be combined with the singular values of $A$ in the inequality.
$endgroup$
– Brian Borchers
Jan 30 at 16:24
$begingroup$
I edited my question with a beginning of an answer. This is enough for my use, but it is not as strong a result as I was expecting to find. Thank you for your hints @BrianBorchers. Should I write my edit as an answer or close this question somehow? (please bear with me as I am -obviously- new here, trying to do the right thing).
$endgroup$
– wiwh
Feb 1 at 13:45
$begingroup$
Do you know the "if and only if" version of Cuachy-Schwarz? When is there equality?
$endgroup$
– Brian Borchers
Feb 1 at 14:20
$begingroup$
In terms of the SVD of $A$, what's the maximum possible value of $u^{T}Av$? What about $v^{T}A^{T}Av$ and its eigenvalues?
$endgroup$
– Brian Borchers
Jan 30 at 15:20
$begingroup$
In terms of the SVD of $A$, what's the maximum possible value of $u^{T}Av$? What about $v^{T}A^{T}Av$ and its eigenvalues?
$endgroup$
– Brian Borchers
Jan 30 at 15:20
$begingroup$
The maximum possible value of $u^T Av$ is the spectral norm of $A$ (the largest of its singular value). As for $v^TA^Tv$, its maximum eigenvalue is the square the spectral norm of $A$ (by definintion). Is it not? I'm not so much stuck for the equality part, rather for the inequality. I'm sure you're hinting at that, but I fail to see it.
$endgroup$
– wiwh
Jan 30 at 16:13
$begingroup$
The maximum possible value of $u^T Av$ is the spectral norm of $A$ (the largest of its singular value). As for $v^TA^Tv$, its maximum eigenvalue is the square the spectral norm of $A$ (by definintion). Is it not? I'm not so much stuck for the equality part, rather for the inequality. I'm sure you're hinting at that, but I fail to see it.
$endgroup$
– wiwh
Jan 30 at 16:13
$begingroup$
You might try a change of basis to expand $u$ in terms of the columns of the $U$ matrix of $A$ and $v$ in terms of the columns of the $V$ matrix of the SVD of $A$. The coordinates of $u$ and $v$ with respect to these bases will then be combined with the singular values of $A$ in the inequality.
$endgroup$
– Brian Borchers
Jan 30 at 16:24
$begingroup$
You might try a change of basis to expand $u$ in terms of the columns of the $U$ matrix of $A$ and $v$ in terms of the columns of the $V$ matrix of the SVD of $A$. The coordinates of $u$ and $v$ with respect to these bases will then be combined with the singular values of $A$ in the inequality.
$endgroup$
– Brian Borchers
Jan 30 at 16:24
$begingroup$
I edited my question with a beginning of an answer. This is enough for my use, but it is not as strong a result as I was expecting to find. Thank you for your hints @BrianBorchers. Should I write my edit as an answer or close this question somehow? (please bear with me as I am -obviously- new here, trying to do the right thing).
$endgroup$
– wiwh
Feb 1 at 13:45
$begingroup$
I edited my question with a beginning of an answer. This is enough for my use, but it is not as strong a result as I was expecting to find. Thank you for your hints @BrianBorchers. Should I write my edit as an answer or close this question somehow? (please bear with me as I am -obviously- new here, trying to do the right thing).
$endgroup$
– wiwh
Feb 1 at 13:45
$begingroup$
Do you know the "if and only if" version of Cuachy-Schwarz? When is there equality?
$endgroup$
– Brian Borchers
Feb 1 at 14:20
$begingroup$
Do you know the "if and only if" version of Cuachy-Schwarz? When is there equality?
$endgroup$
– Brian Borchers
Feb 1 at 14:20
add a comment |
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$begingroup$
In terms of the SVD of $A$, what's the maximum possible value of $u^{T}Av$? What about $v^{T}A^{T}Av$ and its eigenvalues?
$endgroup$
– Brian Borchers
Jan 30 at 15:20
$begingroup$
The maximum possible value of $u^T Av$ is the spectral norm of $A$ (the largest of its singular value). As for $v^TA^Tv$, its maximum eigenvalue is the square the spectral norm of $A$ (by definintion). Is it not? I'm not so much stuck for the equality part, rather for the inequality. I'm sure you're hinting at that, but I fail to see it.
$endgroup$
– wiwh
Jan 30 at 16:13
$begingroup$
You might try a change of basis to expand $u$ in terms of the columns of the $U$ matrix of $A$ and $v$ in terms of the columns of the $V$ matrix of the SVD of $A$. The coordinates of $u$ and $v$ with respect to these bases will then be combined with the singular values of $A$ in the inequality.
$endgroup$
– Brian Borchers
Jan 30 at 16:24
$begingroup$
I edited my question with a beginning of an answer. This is enough for my use, but it is not as strong a result as I was expecting to find. Thank you for your hints @BrianBorchers. Should I write my edit as an answer or close this question somehow? (please bear with me as I am -obviously- new here, trying to do the right thing).
$endgroup$
– wiwh
Feb 1 at 13:45
$begingroup$
Do you know the "if and only if" version of Cuachy-Schwarz? When is there equality?
$endgroup$
– Brian Borchers
Feb 1 at 14:20