If $Q_kR_k$ converges to $QR$, where this represents their respective $QR$ decompositions, then $Q_k...
Suppose $Q_kR_k rightarrow QR$ as $k rightarrow infty$, where $Q_k, Q$ are orthogonal matrices and $R_k, R$ are upper triangular with positive diagonal entries, then would the uniqueness of the $QR$ decomposition imply that $Q_k rightarrow Q$ and $R_k rightarrow R?$
I need this detail for a proof, but I wasn't able to prove it.
It would suffice to show that if $Q_kR_k rightarrow I$, then $Q_k rightarrow I$ and $R_k rightarrow I$.
Edit: If the limit of $Q_k$ and $R_k$ exist, then they must be $I$, but how would one show that these limits exist, if they do?
matrices matrix-decomposition
asked Nov 25 at 6:37
Anu
665215
add a comment |
Suppose $Q_kR_k rightarrow QR$ as $k rightarrow infty$, where $Q_k, Q$ are orthogonal matrices and $R_k, R$ are upper triangular with positive diagonal entries, then would the uniqueness of the $QR$ decomposition imply that $Q_k rightarrow Q$ and $R_k rightarrow R?$
I need this detail for a proof, but I wasn't able to prove it.
It would suffice to show that if $Q_kR_k rightarrow I$, then $Q_k rightarrow I$ and $R_k rightarrow I$.
Edit: If the limit of $Q_k$ and $R_k$ exist, then they must be $I$, but how would one show that these limits exist, if they do?
matrices matrix-decomposition
asked Nov 25 at 6:37
Anu
665215
add a comment |
Suppose $Q_kR_k rightarrow QR$ as $k rightarrow infty$, where $Q_k, Q$ are orthogonal matrices and $R_k, R$ are upper triangular with positive diagonal entries, then would the uniqueness of the $QR$ decomposition imply that $Q_k rightarrow Q$ and $R_k rightarrow R?$
I need this detail for a proof, but I wasn't able to prove it.
It would suffice to show that if $Q_kR_k rightarrow I$, then $Q_k rightarrow I$ and $R_k rightarrow I$.
Edit: If the limit of $Q_k$ and $R_k$ exist, then they must be $I$, but how would one show that these limits exist, if they do?
matrices matrix-decomposition
asked Nov 25 at 6:37
Anu
665215
Suppose $Q_kR_k rightarrow QR$ as $k rightarrow infty$, where $Q_k, Q$ are orthogonal matrices and $R_k, R$ are upper triangular with positive diagonal entries, then would the uniqueness of the $QR$ decomposition imply that $Q_k rightarrow Q$ and $R_k rightarrow R?$
I need this detail for a proof, but I wasn't able to prove it.
It would suffice to show that if $Q_kR_k rightarrow I$, then $Q_k rightarrow I$ and $R_k rightarrow I$.
Edit: If the limit of $Q_k$ and $R_k$ exist, then they must be $I$, but how would one show that these limits exist, if they do?
matrices matrix-decomposition
matrices matrix-decomposition
asked Nov 25 at 6:37
Anu
665215
asked Nov 25 at 6:37
Anu
665215
edited Nov 25 at 6:49
asked Nov 25 at 6:37
Anu
665215
asked Nov 25 at 6:37
Anu
665215
asked Nov 25 at 6:37
Anu
665215
665215
add a comment |
add a comment |
1 Answer
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Since the $Q_k$ are orthogonal, they must be bounded, so we can apply Bolzano-Weierstrass. As you mentioned in your edit, any convergent subsequence would have to converge to $I$.
If $Q_k$ does not converge to $I$, then we can take the subsequence of all $Q_k$ that are at least $epsilon$ away from $I$. Bolzano-Weierstrass then tells us there would be a subsequence that converges to something other than $I$, which leads to a contradiction.
Convergence of $Q_k$ and $Q_kR_k$, along with invertibility of the $Q_k$, then implies convergence of the $R_k$.
answered Nov 25 at 9:54
Carmeister
2,5842920
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Since the $Q_k$ are orthogonal, they must be bounded, so we can apply Bolzano-Weierstrass. As you mentioned in your edit, any convergent subsequence would have to converge to $I$.
If $Q_k$ does not converge to $I$, then we can take the subsequence of all $Q_k$ that are at least $epsilon$ away from $I$. Bolzano-Weierstrass then tells us there would be a subsequence that converges to something other than $I$, which leads to a contradiction.
Convergence of $Q_k$ and $Q_kR_k$, along with invertibility of the $Q_k$, then implies convergence of the $R_k$.
answered Nov 25 at 9:54
Carmeister
2,5842920
add a comment |
Since the $Q_k$ are orthogonal, they must be bounded, so we can apply Bolzano-Weierstrass. As you mentioned in your edit, any convergent subsequence would have to converge to $I$.
If $Q_k$ does not converge to $I$, then we can take the subsequence of all $Q_k$ that are at least $epsilon$ away from $I$. Bolzano-Weierstrass then tells us there would be a subsequence that converges to something other than $I$, which leads to a contradiction.
Convergence of $Q_k$ and $Q_kR_k$, along with invertibility of the $Q_k$, then implies convergence of the $R_k$.
answered Nov 25 at 9:54
Carmeister
2,5842920
add a comment |
Since the $Q_k$ are orthogonal, they must be bounded, so we can apply Bolzano-Weierstrass. As you mentioned in your edit, any convergent subsequence would have to converge to $I$.
If $Q_k$ does not converge to $I$, then we can take the subsequence of all $Q_k$ that are at least $epsilon$ away from $I$. Bolzano-Weierstrass then tells us there would be a subsequence that converges to something other than $I$, which leads to a contradiction.
Convergence of $Q_k$ and $Q_kR_k$, along with invertibility of the $Q_k$, then implies convergence of the $R_k$.
answered Nov 25 at 9:54
Carmeister
2,5842920
Since the $Q_k$ are orthogonal, they must be bounded, so we can apply Bolzano-Weierstrass. As you mentioned in your edit, any convergent subsequence would have to converge to $I$.
If $Q_k$ does not converge to $I$, then we can take the subsequence of all $Q_k$ that are at least $epsilon$ away from $I$. Bolzano-Weierstrass then tells us there would be a subsequence that converges to something other than $I$, which leads to a contradiction.
Convergence of $Q_k$ and $Q_kR_k$, along with invertibility of the $Q_k$, then implies convergence of the $R_k$.
answered Nov 25 at 9:54
Carmeister
2,5842920
answered Nov 25 at 9:54
Carmeister
2,5842920
answered Nov 25 at 9:54
Carmeister
2,5842920
answered Nov 25 at 9:54
Carmeister
2,5842920
2,5842920
add a comment |
add a comment |
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