Problem with eigenvalue as a limit
$begingroup$
Can anyone point out how to prove this?
Let $A$ be a semipositive square matrix, and let $x_m^T [λ_mI-A]=0^T$, where $λ_m$ is the greatest real positive eigenvalue of $A$. Let there be a solution $y^T>0$ for $lim (λ→λ_m )x^T (λ)[λI-A]=y^T$, where $λ>λ_m$. Then $x_m^T≯0$.
matrices eigenvalues-eigenvectors matrix-calculus
asked Jan 14 at 15:49
a.giannela.giannel
163
$endgroup$
add a comment |
$begingroup$
Can anyone point out how to prove this?
Let $A$ be a semipositive square matrix, and let $x_m^T [λ_mI-A]=0^T$, where $λ_m$ is the greatest real positive eigenvalue of $A$. Let there be a solution $y^T>0$ for $lim (λ→λ_m )x^T (λ)[λI-A]=y^T$, where $λ>λ_m$. Then $x_m^T≯0$.
matrices eigenvalues-eigenvectors matrix-calculus
asked Jan 14 at 15:49
a.giannela.giannel
163
$endgroup$
add a comment |
$begingroup$
Can anyone point out how to prove this?
Let $A$ be a semipositive square matrix, and let $x_m^T [λ_mI-A]=0^T$, where $λ_m$ is the greatest real positive eigenvalue of $A$. Let there be a solution $y^T>0$ for $lim (λ→λ_m )x^T (λ)[λI-A]=y^T$, where $λ>λ_m$. Then $x_m^T≯0$.
matrices eigenvalues-eigenvectors matrix-calculus
asked Jan 14 at 15:49
a.giannela.giannel
163
$endgroup$
Can anyone point out how to prove this?
Let $A$ be a semipositive square matrix, and let $x_m^T [λ_mI-A]=0^T$, where $λ_m$ is the greatest real positive eigenvalue of $A$. Let there be a solution $y^T>0$ for $lim (λ→λ_m )x^T (λ)[λI-A]=y^T$, where $λ>λ_m$. Then $x_m^T≯0$.
matrices eigenvalues-eigenvectors matrix-calculus
matrices eigenvalues-eigenvectors matrix-calculus
asked Jan 14 at 15:49
a.giannela.giannel
163
asked Jan 14 at 15:49
a.giannela.giannel
163
asked Jan 14 at 15:49
a.giannela.giannel
163
asked Jan 14 at 15:49
a.giannela.giannel
163
asked Jan 14 at 15:49
a.giannela.giannel
163
163
add a comment |
add a comment |
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