An inequality for symmetric positive semi-definite matrices
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Suppose that $C$ and $D$ are (real) symmetric positive definite matrices and that $A$ is a matrix such that
$$
H = (C + A^T D A)^-1A^TD^2 A(C + A^T D A)^-1
$$
is well defined. Is the maximum eigenvalue of $H$ bounded as a function of $D$? For scalars it clearly holds, if $Aneq 0$, that $H = A^2D^2 / (C + A^2D)^2 to A^-2$ as $D to infty$ and $H to 0$ as $D to 0$.
linear-algebra matrices eigenvalues-eigenvectors norm
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up vote
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Suppose that $C$ and $D$ are (real) symmetric positive definite matrices and that $A$ is a matrix such that
$$
H = (C + A^T D A)^-1A^TD^2 A(C + A^T D A)^-1
$$
is well defined. Is the maximum eigenvalue of $H$ bounded as a function of $D$? For scalars it clearly holds, if $Aneq 0$, that $H = A^2D^2 / (C + A^2D)^2 to A^-2$ as $D to infty$ and $H to 0$ as $D to 0$.
linear-algebra matrices eigenvalues-eigenvectors norm
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose that $C$ and $D$ are (real) symmetric positive definite matrices and that $A$ is a matrix such that
$$
H = (C + A^T D A)^-1A^TD^2 A(C + A^T D A)^-1
$$
is well defined. Is the maximum eigenvalue of $H$ bounded as a function of $D$? For scalars it clearly holds, if $Aneq 0$, that $H = A^2D^2 / (C + A^2D)^2 to A^-2$ as $D to infty$ and $H to 0$ as $D to 0$.
linear-algebra matrices eigenvalues-eigenvectors norm
Suppose that $C$ and $D$ are (real) symmetric positive definite matrices and that $A$ is a matrix such that
$$
H = (C + A^T D A)^-1A^TD^2 A(C + A^T D A)^-1
$$
is well defined. Is the maximum eigenvalue of $H$ bounded as a function of $D$? For scalars it clearly holds, if $Aneq 0$, that $H = A^2D^2 / (C + A^2D)^2 to A^-2$ as $D to infty$ and $H to 0$ as $D to 0$.
linear-algebra matrices eigenvalues-eigenvectors norm
asked Aug 3 at 15:10
ekvall
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211116
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