How to prove the inequality $lambda_{min{B^TAB}} geq lambda_{min{A}}lambda_{min{B^TB}}$?
Given an arbitrary positive definite symmetric matrix $A$ and an arbitrary non-singular matrix $B$, let $lambda_{min{X}}$ represents the minimum eigenvalue of the matrix $X$. Can we obtain the eigenvalues inequality
$lambda_{min{B^TAB}} geq lambda_{min{A}}lambda_{min{B^TB}}$?
I have tried many examples and find it indeed works, but I cannot figure it out.
Thanks a lot!
matrices eigenvalues-eigenvectors
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Given an arbitrary positive definite symmetric matrix $A$ and an arbitrary non-singular matrix $B$, let $lambda_{min{X}}$ represents the minimum eigenvalue of the matrix $X$. Can we obtain the eigenvalues inequality
$lambda_{min{B^TAB}} geq lambda_{min{A}}lambda_{min{B^TB}}$?
I have tried many examples and find it indeed works, but I cannot figure it out.
Thanks a lot!
matrices eigenvalues-eigenvectors
New contributor
add a comment |
Given an arbitrary positive definite symmetric matrix $A$ and an arbitrary non-singular matrix $B$, let $lambda_{min{X}}$ represents the minimum eigenvalue of the matrix $X$. Can we obtain the eigenvalues inequality
$lambda_{min{B^TAB}} geq lambda_{min{A}}lambda_{min{B^TB}}$?
I have tried many examples and find it indeed works, but I cannot figure it out.
Thanks a lot!
matrices eigenvalues-eigenvectors
New contributor
Given an arbitrary positive definite symmetric matrix $A$ and an arbitrary non-singular matrix $B$, let $lambda_{min{X}}$ represents the minimum eigenvalue of the matrix $X$. Can we obtain the eigenvalues inequality
$lambda_{min{B^TAB}} geq lambda_{min{A}}lambda_{min{B^TB}}$?
I have tried many examples and find it indeed works, but I cannot figure it out.
Thanks a lot!
matrices eigenvalues-eigenvectors
matrices eigenvalues-eigenvectors
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