lower bound for determinant of $(X^TY)$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)are there any bounds on the eigenvalues of products of positive-semidefinite matrices?Lower bound on Gram determinantLower bound on the smallest eigenvalueProve an upper bound for the determinant of a matrix ALower Bound for the induced matrix norm $||A||_infty,1$Simple lower bound for a determinantLower bound on absolute value of determinant of sum of matricesLower bounds for determinant of $AA^T$The Hadamard determinant problem: understanding a proof for the upper bound on matrix determinantsGreatest lower bound
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lower bound for determinant of $(X^TY)$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)are there any bounds on the eigenvalues of products of positive-semidefinite matrices?Lower bound on Gram determinantLower bound on the smallest eigenvalueProve an upper bound for the determinant of a matrix ALower Bound for the induced matrix norm $||A||_infty,1$Simple lower bound for a determinantLower bound on absolute value of determinant of sum of matricesLower bounds for determinant of $AA^T$The Hadamard determinant problem: understanding a proof for the upper bound on matrix determinantsGreatest lower bound
$begingroup$
Am looking for a lower bound for determinant, $det(X^TY)$ where $X^T$ is $p times n$ and $Y$ is $n times p$. Is it $Tr(X^TY)^-1$? Regardless, what are other lower bounds for this? $X,Y$ are real-valued matrices.
linear-algebra matrices determinant inner-product-space upper-lower-bounds
asked Mar 27 at 16:42
hearsehearse
278
$endgroup$
add a comment |
$begingroup$
Am looking for a lower bound for determinant, $det(X^TY)$ where $X^T$ is $p times n$ and $Y$ is $n times p$. Is it $Tr(X^TY)^-1$? Regardless, what are other lower bounds for this? $X,Y$ are real-valued matrices.
linear-algebra matrices determinant inner-product-space upper-lower-bounds
asked Mar 27 at 16:42
hearsehearse
278
$endgroup$
$begingroup$
What is $Z$ in this context?
$endgroup$
– tch
Mar 27 at 17:30
$begingroup$
sry, typo. changed it to $Y$
$endgroup$
– hearse
Mar 27 at 17:31
$begingroup$
If $X = -cI$, $c>0$ and $Y=I$ when $c$ is negative and $n$ is odd then the $det(X^T)$ will be $-c^n$ while $1/Tr(X^TY)$ will be $-1/(cn)$. For $c$ large enough the determinant would be arbitrarily smaller than the bound from the trace.
$endgroup$
– tch
Mar 27 at 19:54
$begingroup$
You might want to check out en.wikipedia.org/wiki/Cauchy%E2%80%93Binet_formula
$endgroup$
– daw
Mar 27 at 20:36
$begingroup$
If $n < p$, then the determinant is always $0$, For $n ge p$, the lower bound of the absolute value of the determinant is $0$. The lower bound for the determinant itself is $-infty$. It is independent of the value of the trace for $p > 1$.
$endgroup$
– Paul Sinclair
Mar 28 at 0:38
add a comment |
$begingroup$
Am looking for a lower bound for determinant, $det(X^TY)$ where $X^T$ is $p times n$ and $Y$ is $n times p$. Is it $Tr(X^TY)^-1$? Regardless, what are other lower bounds for this? $X,Y$ are real-valued matrices.
linear-algebra matrices determinant inner-product-space upper-lower-bounds
asked Mar 27 at 16:42
hearsehearse
278
$endgroup$
Am looking for a lower bound for determinant, $det(X^TY)$ where $X^T$ is $p times n$ and $Y$ is $n times p$. Is it $Tr(X^TY)^-1$? Regardless, what are other lower bounds for this? $X,Y$ are real-valued matrices.
linear-algebra matrices determinant inner-product-space upper-lower-bounds
linear-algebra matrices determinant inner-product-space upper-lower-bounds
asked Mar 27 at 16:42
hearsehearse
278
asked Mar 27 at 16:42
hearsehearse
278
edited Mar 27 at 17:31
hearse
asked Mar 27 at 16:42
hearsehearse
278
asked Mar 27 at 16:42
hearsehearse
278
asked Mar 27 at 16:42
hearsehearse
278
278
$begingroup$
What is $Z$ in this context?
$endgroup$
– tch
Mar 27 at 17:30
$begingroup$
sry, typo. changed it to $Y$
$endgroup$
– hearse
Mar 27 at 17:31
$begingroup$
If $X = -cI$, $c>0$ and $Y=I$ when $c$ is negative and $n$ is odd then the $det(X^T)$ will be $-c^n$ while $1/Tr(X^TY)$ will be $-1/(cn)$. For $c$ large enough the determinant would be arbitrarily smaller than the bound from the trace.
$endgroup$
– tch
Mar 27 at 19:54
$begingroup$
You might want to check out en.wikipedia.org/wiki/Cauchy%E2%80%93Binet_formula
$endgroup$
– daw
Mar 27 at 20:36
$begingroup$
If $n < p$, then the determinant is always $0$, For $n ge p$, the lower bound of the absolute value of the determinant is $0$. The lower bound for the determinant itself is $-infty$. It is independent of the value of the trace for $p > 1$.
$endgroup$
– Paul Sinclair
Mar 28 at 0:38
add a comment |
$begingroup$
What is $Z$ in this context?
$endgroup$
– tch
Mar 27 at 17:30
$begingroup$
sry, typo. changed it to $Y$
$endgroup$
– hearse
Mar 27 at 17:31
$begingroup$
If $X = -cI$, $c>0$ and $Y=I$ when $c$ is negative and $n$ is odd then the $det(X^T)$ will be $-c^n$ while $1/Tr(X^TY)$ will be $-1/(cn)$. For $c$ large enough the determinant would be arbitrarily smaller than the bound from the trace.
$endgroup$
– tch
Mar 27 at 19:54
$begingroup$
You might want to check out en.wikipedia.org/wiki/Cauchy%E2%80%93Binet_formula
$endgroup$
– daw
Mar 27 at 20:36
$begingroup$
If $n < p$, then the determinant is always $0$, For $n ge p$, the lower bound of the absolute value of the determinant is $0$. The lower bound for the determinant itself is $-infty$. It is independent of the value of the trace for $p > 1$.
$endgroup$
– Paul Sinclair
Mar 28 at 0:38
$begingroup$
What is $Z$ in this context?
$endgroup$
– tch
Mar 27 at 17:30
$begingroup$
What is $Z$ in this context?
$endgroup$
– tch
Mar 27 at 17:30
$begingroup$
sry, typo. changed it to $Y$
$endgroup$
– hearse
Mar 27 at 17:31
$begingroup$
sry, typo. changed it to $Y$
$endgroup$
– hearse
Mar 27 at 17:31
$begingroup$
If $X = -cI$, $c>0$ and $Y=I$ when $c$ is negative and $n$ is odd then the $det(X^T)$ will be $-c^n$ while $1/Tr(X^TY)$ will be $-1/(cn)$. For $c$ large enough the determinant would be arbitrarily smaller than the bound from the trace.
$endgroup$
– tch
Mar 27 at 19:54
$begingroup$
If $X = -cI$, $c>0$ and $Y=I$ when $c$ is negative and $n$ is odd then the $det(X^T)$ will be $-c^n$ while $1/Tr(X^TY)$ will be $-1/(cn)$. For $c$ large enough the determinant would be arbitrarily smaller than the bound from the trace.
$endgroup$
– tch
Mar 27 at 19:54
$begingroup$
You might want to check out en.wikipedia.org/wiki/Cauchy%E2%80%93Binet_formula
$endgroup$
– daw
Mar 27 at 20:36
$begingroup$
You might want to check out en.wikipedia.org/wiki/Cauchy%E2%80%93Binet_formula
$endgroup$
– daw
Mar 27 at 20:36
$begingroup$
If $n < p$, then the determinant is always $0$, For $n ge p$, the lower bound of the absolute value of the determinant is $0$. The lower bound for the determinant itself is $-infty$. It is independent of the value of the trace for $p > 1$.
$endgroup$
– Paul Sinclair
Mar 28 at 0:38
$begingroup$
If $n < p$, then the determinant is always $0$, For $n ge p$, the lower bound of the absolute value of the determinant is $0$. The lower bound for the determinant itself is $-infty$. It is independent of the value of the trace for $p > 1$.
$endgroup$
– Paul Sinclair
Mar 28 at 0:38
add a comment |
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UmA7A5CuNAAqGQtVHNFh680 WhLhofmZia,AfvybOOv F,B,re
$begingroup$
What is $Z$ in this context?
$endgroup$
– tch
Mar 27 at 17:30
$begingroup$
sry, typo. changed it to $Y$
$endgroup$
– hearse
Mar 27 at 17:31
$begingroup$
If $X = -cI$, $c>0$ and $Y=I$ when $c$ is negative and $n$ is odd then the $det(X^T)$ will be $-c^n$ while $1/Tr(X^TY)$ will be $-1/(cn)$. For $c$ large enough the determinant would be arbitrarily smaller than the bound from the trace.
$endgroup$
– tch
Mar 27 at 19:54
$begingroup$
You might want to check out en.wikipedia.org/wiki/Cauchy%E2%80%93Binet_formula
$endgroup$
– daw
Mar 27 at 20:36
$begingroup$
If $n < p$, then the determinant is always $0$, For $n ge p$, the lower bound of the absolute value of the determinant is $0$. The lower bound for the determinant itself is $-infty$. It is independent of the value of the trace for $p > 1$.
$endgroup$
– Paul Sinclair
Mar 28 at 0:38