Suppose we have a matrix $A$ with is its $LU$-decomposition such that $A=LU$ and suppose that $U$ is ill conditioned ($\left \| U \right \|\left \| U^{-1} \right \|$ is large) , does it mean that $A$ is ill conditioned ?
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Does the double bar denote spectral norm? – paulinho Jul 24 '20 at 16:45
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It is an induced norm , i think they're the same , spectral and induced – Pedro Alvarès Jul 24 '20 at 16:51
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Are you only interested for square matrices $A$? – paulinho Jul 24 '20 at 16:54
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yes square matrices – Pedro Alvarès Jul 24 '20 at 17:01
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Sorry last question - Is there some sort of constraint on $L$ and $U$? Sometimes people take $U$ to be unipotent (diagonal entries are all one). – paulinho Jul 24 '20 at 17:32
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no , there is no such thing – Pedro Alvarès Jul 24 '20 at 18:29
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What if simply we consider $$A=\left[\matrix{1&0\cr0&1}\right],\quad L=\left[\matrix{10^n&0\cr0&10^{-n}}\right],\quad U=\left[\matrix{10^{-n}&0\cr0&10^{n}}\right]?$$ Clearly $L$ is lower triangular, $U$ is upper triangular, $A=LU$ and $$\hbox{cond}_2(A)=1,\quad\hbox{cond}_2(L)=\hbox{cond}_2(U)=10^{2n}.$$
Omran Kouba
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