Determination of transition from non-singular matrix to singular matrix
I have the following matrix as a biproduct of inverting a matrix sum by the Woodbury matrix identity:
$$ mathbf{A} = -(gmathbf{G})^{-1} + mathbf{W}^T mathbf{K}^{-1} mathbf{W} $$
where $g$ is an arbitrary but positive scalar, $mathbf{G}$ is a $mtimes m$ diagonal matrix with positive entities, $mathbf{W}$ is an $n times m$ array with each column having a single $1$ or $-1$ entity and otherwise zeros and $mathbf{K}$ is a $n times n$ symmetric, real and positive definite matrix, representing the stiffness matrix of an elastic solid. Further more $m << n$, typically $10^4<n$ and $m < 20$.
$mathbf{K}$ has a structure such that for very small values of $g$ $mathbf{A}$ will be positive definite. When increasing $g$ towards $g_0$ ($0<gleq g_0$) the matrix $mathbf{A}$ will at some point become singular and the system loses its stability. I am to determine $g_0$ which indentifies instability.
I am trying to set up a robust way to determine $mathbf{A}$'s transition from non-singular to singular. Right now I am calculating the determinant of $mathbf{A}$ with Matlab, and it seems that as $mathbf{A}$ becomes 'less and less' positive definite, the determinant decreases and becomes $0$ when singular. For $g_0<g$ it seems that the determinant becomes negative, which I guess is a result of $mathbf{A}$ then being negative definite?
Is it correct that as $g rightarrow infty$ $mathbf{A}$ goes from being positive definite to singular to negative definite along with the determinant going from positive through $0$ to negative with $det(mathbf{A})=0$ identifying the point when $mathbf{A}$ becomes singular? Is there a precise mathematical argument for the determinant to behave in that way?
Thanks for any comments.
linear-algebra matrices determinant singularity
asked 5 hours ago
DavidH
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I have the following matrix as a biproduct of inverting a matrix sum by the Woodbury matrix identity:
$$ mathbf{A} = -(gmathbf{G})^{-1} + mathbf{W}^T mathbf{K}^{-1} mathbf{W} $$
where $g$ is an arbitrary but positive scalar, $mathbf{G}$ is a $mtimes m$ diagonal matrix with positive entities, $mathbf{W}$ is an $n times m$ array with each column having a single $1$ or $-1$ entity and otherwise zeros and $mathbf{K}$ is a $n times n$ symmetric, real and positive definite matrix, representing the stiffness matrix of an elastic solid. Further more $m << n$, typically $10^4<n$ and $m < 20$.
$mathbf{K}$ has a structure such that for very small values of $g$ $mathbf{A}$ will be positive definite. When increasing $g$ towards $g_0$ ($0<gleq g_0$) the matrix $mathbf{A}$ will at some point become singular and the system loses its stability. I am to determine $g_0$ which indentifies instability.
I am trying to set up a robust way to determine $mathbf{A}$'s transition from non-singular to singular. Right now I am calculating the determinant of $mathbf{A}$ with Matlab, and it seems that as $mathbf{A}$ becomes 'less and less' positive definite, the determinant decreases and becomes $0$ when singular. For $g_0<g$ it seems that the determinant becomes negative, which I guess is a result of $mathbf{A}$ then being negative definite?
Is it correct that as $g rightarrow infty$ $mathbf{A}$ goes from being positive definite to singular to negative definite along with the determinant going from positive through $0$ to negative with $det(mathbf{A})=0$ identifying the point when $mathbf{A}$ becomes singular? Is there a precise mathematical argument for the determinant to behave in that way?
Thanks for any comments.
linear-algebra matrices determinant singularity
asked 5 hours ago
DavidH
1
New contributor
DavidH is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Look at the corner case: $W=G=I_2$ and $K = a,I_2$. The determinant is always positive or null.
– Damien
3 hours ago
add a comment |
I have the following matrix as a biproduct of inverting a matrix sum by the Woodbury matrix identity:
$$ mathbf{A} = -(gmathbf{G})^{-1} + mathbf{W}^T mathbf{K}^{-1} mathbf{W} $$
where $g$ is an arbitrary but positive scalar, $mathbf{G}$ is a $mtimes m$ diagonal matrix with positive entities, $mathbf{W}$ is an $n times m$ array with each column having a single $1$ or $-1$ entity and otherwise zeros and $mathbf{K}$ is a $n times n$ symmetric, real and positive definite matrix, representing the stiffness matrix of an elastic solid. Further more $m << n$, typically $10^4<n$ and $m < 20$.
$mathbf{K}$ has a structure such that for very small values of $g$ $mathbf{A}$ will be positive definite. When increasing $g$ towards $g_0$ ($0<gleq g_0$) the matrix $mathbf{A}$ will at some point become singular and the system loses its stability. I am to determine $g_0$ which indentifies instability.
I am trying to set up a robust way to determine $mathbf{A}$'s transition from non-singular to singular. Right now I am calculating the determinant of $mathbf{A}$ with Matlab, and it seems that as $mathbf{A}$ becomes 'less and less' positive definite, the determinant decreases and becomes $0$ when singular. For $g_0<g$ it seems that the determinant becomes negative, which I guess is a result of $mathbf{A}$ then being negative definite?
Is it correct that as $g rightarrow infty$ $mathbf{A}$ goes from being positive definite to singular to negative definite along with the determinant going from positive through $0$ to negative with $det(mathbf{A})=0$ identifying the point when $mathbf{A}$ becomes singular? Is there a precise mathematical argument for the determinant to behave in that way?
Thanks for any comments.
linear-algebra matrices determinant singularity
asked 5 hours ago
DavidH
1
New contributor
DavidH is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have the following matrix as a biproduct of inverting a matrix sum by the Woodbury matrix identity:
$$ mathbf{A} = -(gmathbf{G})^{-1} + mathbf{W}^T mathbf{K}^{-1} mathbf{W} $$
where $g$ is an arbitrary but positive scalar, $mathbf{G}$ is a $mtimes m$ diagonal matrix with positive entities, $mathbf{W}$ is an $n times m$ array with each column having a single $1$ or $-1$ entity and otherwise zeros and $mathbf{K}$ is a $n times n$ symmetric, real and positive definite matrix, representing the stiffness matrix of an elastic solid. Further more $m << n$, typically $10^4<n$ and $m < 20$.
$mathbf{K}$ has a structure such that for very small values of $g$ $mathbf{A}$ will be positive definite. When increasing $g$ towards $g_0$ ($0<gleq g_0$) the matrix $mathbf{A}$ will at some point become singular and the system loses its stability. I am to determine $g_0$ which indentifies instability.
I am trying to set up a robust way to determine $mathbf{A}$'s transition from non-singular to singular. Right now I am calculating the determinant of $mathbf{A}$ with Matlab, and it seems that as $mathbf{A}$ becomes 'less and less' positive definite, the determinant decreases and becomes $0$ when singular. For $g_0<g$ it seems that the determinant becomes negative, which I guess is a result of $mathbf{A}$ then being negative definite?
Is it correct that as $g rightarrow infty$ $mathbf{A}$ goes from being positive definite to singular to negative definite along with the determinant going from positive through $0$ to negative with $det(mathbf{A})=0$ identifying the point when $mathbf{A}$ becomes singular? Is there a precise mathematical argument for the determinant to behave in that way?
Thanks for any comments.
linear-algebra matrices determinant singularity
linear-algebra matrices determinant singularity
asked 5 hours ago
DavidH
1
New contributor
DavidH is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
DavidH
1
New contributor
DavidH is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
DavidH
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New contributor
DavidH is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
DavidH
1
asked 5 hours ago
DavidH
1
1
New contributor
DavidH is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
DavidH is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
DavidH is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Look at the corner case: $W=G=I_2$ and $K = a,I_2$. The determinant is always positive or null.
– Damien
3 hours ago
add a comment |
Look at the corner case: $W=G=I_2$ and $K = a,I_2$. The determinant is always positive or null.
– Damien
3 hours ago
Look at the corner case: $W=G=I_2$ and $K = a,I_2$. The determinant is always positive or null.
– Damien
3 hours ago
Look at the corner case: $W=G=I_2$ and $K = a,I_2$. The determinant is always positive or null.
– Damien
3 hours ago
add a comment |
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Look at the corner case: $W=G=I_2$ and $K = a,I_2$. The determinant is always positive or null.
– Damien
3 hours ago