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Upper bound for smallest eigenvalue for the matrix $mathbfS (mathbfS^T mathbfS + lambda mathbfA)^-2 mathbfS^T$
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My book claims that the smallest eigenvalue of the matrix
$$mathbfS (mathbfS^T mathbfS + lambda mathbfA)^-2 mathbfS^T,$$
where $mathbfS$ is an $n times k $ matrix and $mathbfA$ is a $k times k$ positive semi-definite matrix and $lambda>0$, is proportional to $1/lambda$. Could you please provide an argument for this? It's not immediately obvious to me.
Thank you in advance.
inequality eigenvalues-eigenvectors
asked Jul 31 at 14:35
JohnK
2,70011432
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up vote
1
down vote
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My book claims that the smallest eigenvalue of the matrix
$$mathbfS (mathbfS^T mathbfS + lambda mathbfA)^-2 mathbfS^T,$$
where $mathbfS$ is an $n times k $ matrix and $mathbfA$ is a $k times k$ positive semi-definite matrix and $lambda>0$, is proportional to $1/lambda$. Could you please provide an argument for this? It's not immediately obvious to me.
Thank you in advance.
inequality eigenvalues-eigenvectors
asked Jul 31 at 14:35
JohnK
2,70011432
The thing that strikes me as odd about this assertion is that - if true - it implies that the eigenvalues must diverge as $lambda rightarrow 0$, i.e. that the eigenvalues/vectors of the matrix $bf S(bf S^Tbf S)^-2bf S^T$ do not exist. I don't think that's true in general. Do we have any other information or restrictions on these matrices?
– John Barber
Jul 31 at 15:00
For instance, there's nothing in the statement of the problem that prevents having $n = k$ and $bf S = bf I$, where $bf I$ is the $ntimes n$ identity matrix.
– John Barber
Jul 31 at 15:06
@JohnBarber Thank you for your comments. Perhaps it will help to know that $mathbfA$ can be taken to be positive definite, that is, with strictly positive eigenvalues.
– JohnK
Jul 31 at 15:20
1
If that's the only restriction, then the assertion is false. Counterexample: Take $n = k$ and $bf S = bf I$ the identity matrix. Then it's pretty straightforward to show that if an eigenvalue of $bf A$ is $alpha$, then $1/(1 + lambdaalpha)^2$ is an eigenvalue of $bf S(bf S^T bf S + lambda bf A)^-2 bf S^T$. The latter expression is not proportional to $1/lambda$. Is there some additional context? For example, is this by any chance from (say) a physics text?
– John Barber
Jul 31 at 15:25
@JohnBarber Could they mean the largest eigenvalue? I can see what they want to do is establish a bound on the expression $mathbfb^T mathbfS (mathbfS^T mathbfS + lambda mathbfA)^-2 mathbfS^T mathbfb$ off the form $C mathbfb^T mathbfb$ for some $C>0$.
– JohnK
Jul 31 at 15:39
 |Â
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
My book claims that the smallest eigenvalue of the matrix
$$mathbfS (mathbfS^T mathbfS + lambda mathbfA)^-2 mathbfS^T,$$
where $mathbfS$ is an $n times k $ matrix and $mathbfA$ is a $k times k$ positive semi-definite matrix and $lambda>0$, is proportional to $1/lambda$. Could you please provide an argument for this? It's not immediately obvious to me.
Thank you in advance.
inequality eigenvalues-eigenvectors
asked Jul 31 at 14:35
JohnK
2,70011432
My book claims that the smallest eigenvalue of the matrix
$$mathbfS (mathbfS^T mathbfS + lambda mathbfA)^-2 mathbfS^T,$$
where $mathbfS$ is an $n times k $ matrix and $mathbfA$ is a $k times k$ positive semi-definite matrix and $lambda>0$, is proportional to $1/lambda$. Could you please provide an argument for this? It's not immediately obvious to me.
Thank you in advance.
inequality eigenvalues-eigenvectors
asked Jul 31 at 14:35
JohnK
2,70011432
asked Jul 31 at 14:35
JohnK
2,70011432
asked Jul 31 at 14:35
JohnK
2,70011432
asked Jul 31 at 14:35
JohnK
2,70011432
2,70011432
The thing that strikes me as odd about this assertion is that - if true - it implies that the eigenvalues must diverge as $lambda rightarrow 0$, i.e. that the eigenvalues/vectors of the matrix $bf S(bf S^Tbf S)^-2bf S^T$ do not exist. I don't think that's true in general. Do we have any other information or restrictions on these matrices?
– John Barber
Jul 31 at 15:00
For instance, there's nothing in the statement of the problem that prevents having $n = k$ and $bf S = bf I$, where $bf I$ is the $ntimes n$ identity matrix.
– John Barber
Jul 31 at 15:06
@JohnBarber Thank you for your comments. Perhaps it will help to know that $mathbfA$ can be taken to be positive definite, that is, with strictly positive eigenvalues.
– JohnK
Jul 31 at 15:20
1
If that's the only restriction, then the assertion is false. Counterexample: Take $n = k$ and $bf S = bf I$ the identity matrix. Then it's pretty straightforward to show that if an eigenvalue of $bf A$ is $alpha$, then $1/(1 + lambdaalpha)^2$ is an eigenvalue of $bf S(bf S^T bf S + lambda bf A)^-2 bf S^T$. The latter expression is not proportional to $1/lambda$. Is there some additional context? For example, is this by any chance from (say) a physics text?
– John Barber
Jul 31 at 15:25
@JohnBarber Could they mean the largest eigenvalue? I can see what they want to do is establish a bound on the expression $mathbfb^T mathbfS (mathbfS^T mathbfS + lambda mathbfA)^-2 mathbfS^T mathbfb$ off the form $C mathbfb^T mathbfb$ for some $C>0$.
– JohnK
Jul 31 at 15:39
 |Â
show 2 more comments
The thing that strikes me as odd about this assertion is that - if true - it implies that the eigenvalues must diverge as $lambda rightarrow 0$, i.e. that the eigenvalues/vectors of the matrix $bf S(bf S^Tbf S)^-2bf S^T$ do not exist. I don't think that's true in general. Do we have any other information or restrictions on these matrices?
– John Barber
Jul 31 at 15:00
For instance, there's nothing in the statement of the problem that prevents having $n = k$ and $bf S = bf I$, where $bf I$ is the $ntimes n$ identity matrix.
– John Barber
Jul 31 at 15:06
@JohnBarber Thank you for your comments. Perhaps it will help to know that $mathbfA$ can be taken to be positive definite, that is, with strictly positive eigenvalues.
– JohnK
Jul 31 at 15:20
1
If that's the only restriction, then the assertion is false. Counterexample: Take $n = k$ and $bf S = bf I$ the identity matrix. Then it's pretty straightforward to show that if an eigenvalue of $bf A$ is $alpha$, then $1/(1 + lambdaalpha)^2$ is an eigenvalue of $bf S(bf S^T bf S + lambda bf A)^-2 bf S^T$. The latter expression is not proportional to $1/lambda$. Is there some additional context? For example, is this by any chance from (say) a physics text?
– John Barber
Jul 31 at 15:25
@JohnBarber Could they mean the largest eigenvalue? I can see what they want to do is establish a bound on the expression $mathbfb^T mathbfS (mathbfS^T mathbfS + lambda mathbfA)^-2 mathbfS^T mathbfb$ off the form $C mathbfb^T mathbfb$ for some $C>0$.
– JohnK
Jul 31 at 15:39
The thing that strikes me as odd about this assertion is that - if true - it implies that the eigenvalues must diverge as $lambda rightarrow 0$, i.e. that the eigenvalues/vectors of the matrix $bf S(bf S^Tbf S)^-2bf S^T$ do not exist. I don't think that's true in general. Do we have any other information or restrictions on these matrices?
– John Barber
Jul 31 at 15:00
The thing that strikes me as odd about this assertion is that - if true - it implies that the eigenvalues must diverge as $lambda rightarrow 0$, i.e. that the eigenvalues/vectors of the matrix $bf S(bf S^Tbf S)^-2bf S^T$ do not exist. I don't think that's true in general. Do we have any other information or restrictions on these matrices?
– John Barber
Jul 31 at 15:00
For instance, there's nothing in the statement of the problem that prevents having $n = k$ and $bf S = bf I$, where $bf I$ is the $ntimes n$ identity matrix.
– John Barber
Jul 31 at 15:06
For instance, there's nothing in the statement of the problem that prevents having $n = k$ and $bf S = bf I$, where $bf I$ is the $ntimes n$ identity matrix.
– John Barber
Jul 31 at 15:06
@JohnBarber Thank you for your comments. Perhaps it will help to know that $mathbfA$ can be taken to be positive definite, that is, with strictly positive eigenvalues.
– JohnK
Jul 31 at 15:20
@JohnBarber Thank you for your comments. Perhaps it will help to know that $mathbfA$ can be taken to be positive definite, that is, with strictly positive eigenvalues.
– JohnK
Jul 31 at 15:20
1
1
If that's the only restriction, then the assertion is false. Counterexample: Take $n = k$ and $bf S = bf I$ the identity matrix. Then it's pretty straightforward to show that if an eigenvalue of $bf A$ is $alpha$, then $1/(1 + lambdaalpha)^2$ is an eigenvalue of $bf S(bf S^T bf S + lambda bf A)^-2 bf S^T$. The latter expression is not proportional to $1/lambda$. Is there some additional context? For example, is this by any chance from (say) a physics text?
– John Barber
Jul 31 at 15:25
If that's the only restriction, then the assertion is false. Counterexample: Take $n = k$ and $bf S = bf I$ the identity matrix. Then it's pretty straightforward to show that if an eigenvalue of $bf A$ is $alpha$, then $1/(1 + lambdaalpha)^2$ is an eigenvalue of $bf S(bf S^T bf S + lambda bf A)^-2 bf S^T$. The latter expression is not proportional to $1/lambda$. Is there some additional context? For example, is this by any chance from (say) a physics text?
– John Barber
Jul 31 at 15:25
@JohnBarber Could they mean the largest eigenvalue? I can see what they want to do is establish a bound on the expression $mathbfb^T mathbfS (mathbfS^T mathbfS + lambda mathbfA)^-2 mathbfS^T mathbfb$ off the form $C mathbfb^T mathbfb$ for some $C>0$.
– JohnK
Jul 31 at 15:39
@JohnBarber Could they mean the largest eigenvalue? I can see what they want to do is establish a bound on the expression $mathbfb^T mathbfS (mathbfS^T mathbfS + lambda mathbfA)^-2 mathbfS^T mathbfb$ off the form $C mathbfb^T mathbfb$ for some $C>0$.
– JohnK
Jul 31 at 15:39
 |Â
show 2 more comments
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The thing that strikes me as odd about this assertion is that - if true - it implies that the eigenvalues must diverge as $lambda rightarrow 0$, i.e. that the eigenvalues/vectors of the matrix $bf S(bf S^Tbf S)^-2bf S^T$ do not exist. I don't think that's true in general. Do we have any other information or restrictions on these matrices?
– John Barber
Jul 31 at 15:00
For instance, there's nothing in the statement of the problem that prevents having $n = k$ and $bf S = bf I$, where $bf I$ is the $ntimes n$ identity matrix.
– John Barber
Jul 31 at 15:06
@JohnBarber Thank you for your comments. Perhaps it will help to know that $mathbfA$ can be taken to be positive definite, that is, with strictly positive eigenvalues.
– JohnK
Jul 31 at 15:20
1
If that's the only restriction, then the assertion is false. Counterexample: Take $n = k$ and $bf S = bf I$ the identity matrix. Then it's pretty straightforward to show that if an eigenvalue of $bf A$ is $alpha$, then $1/(1 + lambdaalpha)^2$ is an eigenvalue of $bf S(bf S^T bf S + lambda bf A)^-2 bf S^T$. The latter expression is not proportional to $1/lambda$. Is there some additional context? For example, is this by any chance from (say) a physics text?
– John Barber
Jul 31 at 15:25
@JohnBarber Could they mean the largest eigenvalue? I can see what they want to do is establish a bound on the expression $mathbfb^T mathbfS (mathbfS^T mathbfS + lambda mathbfA)^-2 mathbfS^T mathbfb$ off the form $C mathbfb^T mathbfb$ for some $C>0$.
– JohnK
Jul 31 at 15:39