For any k deficient matrices there is full column rank matrix to converge
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I want to prove the existence of QR decomposition for rank deficient matrices and I am stuck on the step which says that: for any rank-deficient matrix there is a sequence of full-column rank matrices $A_k$ such that $A_k to A$ where $A$ is a full column rank matrix. I guess that the set of full column rank matrices is compact and therefore there exist a converging subsequence of matrices $A_k to A$ but is it correct?
linear-algebra matrices
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I want to prove the existence of QR decomposition for rank deficient matrices and I am stuck on the step which says that: for any rank-deficient matrix there is a sequence of full-column rank matrices $A_k$ such that $A_k to A$ where $A$ is a full column rank matrix. I guess that the set of full column rank matrices is compact and therefore there exist a converging subsequence of matrices $A_k to A$ but is it correct?
linear-algebra matrices
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to prove the existence of QR decomposition for rank deficient matrices and I am stuck on the step which says that: for any rank-deficient matrix there is a sequence of full-column rank matrices $A_k$ such that $A_k to A$ where $A$ is a full column rank matrix. I guess that the set of full column rank matrices is compact and therefore there exist a converging subsequence of matrices $A_k to A$ but is it correct?
linear-algebra matrices
I want to prove the existence of QR decomposition for rank deficient matrices and I am stuck on the step which says that: for any rank-deficient matrix there is a sequence of full-column rank matrices $A_k$ such that $A_k to A$ where $A$ is a full column rank matrix. I guess that the set of full column rank matrices is compact and therefore there exist a converging subsequence of matrices $A_k to A$ but is it correct?
linear-algebra matrices
linear-algebra matrices
asked Nov 17 at 6:59
user2660964
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