Are there any properties for matrix A (A= USVt) in Singular Value Decomposition?

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I have seen a lot of explanations about SVD with examples figure how it works. like for example figure in this link. by looking the figure it seems that m must be bigger than n (m>n) for matrix A with mxn. what about in real cases do we have to have a matrix which the rows must be wider than the columns?




linear-algebra svd linear-regression




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    I have seen a lot of explanations about SVD with examples figure how it works. like for example figure in this link. by looking the figure it seems that m must be bigger than n (m>n) for matrix A with mxn. what about in real cases do we have to have a matrix which the rows must be wider than the columns?




    linear-algebra svd linear-regression




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      up vote
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      up vote
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      down vote

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      I have seen a lot of explanations about SVD with examples figure how it works. like for example figure in this link. by looking the figure it seems that m must be bigger than n (m>n) for matrix A with mxn. what about in real cases do we have to have a matrix which the rows must be wider than the columns?




      linear-algebra svd linear-regression




      share|cite|improve this question











      I have seen a lot of explanations about SVD with examples figure how it works. like for example figure in this link. by looking the figure it seems that m must be bigger than n (m>n) for matrix A with mxn. what about in real cases do we have to have a matrix which the rows must be wider than the columns?





      linear-algebra svd linear-regression





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      asked Jul 29 at 10:06









      bernard

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          SVD works for every matrix $Ain mathbbR^n times m$ and arbitrary $n, m$.



          As a comment to the case $mleq n$: Consider this case and let be given a SVD for $A^T$, i.e.



          $$ A^T = U S V^T. $$



          Then $$A = V S^T U^T $$ is a SVD for $A$.






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            SVD works for every matrix $Ain mathbbR^n times m$ and arbitrary $n, m$.



            As a comment to the case $mleq n$: Consider this case and let be given a SVD for $A^T$, i.e.



            $$ A^T = U S V^T. $$



            Then $$A = V S^T U^T $$ is a SVD for $A$.






            share|cite|improve this answer

























              up vote
              0
              down vote













              SVD works for every matrix $Ain mathbbR^n times m$ and arbitrary $n, m$.



              As a comment to the case $mleq n$: Consider this case and let be given a SVD for $A^T$, i.e.



              $$ A^T = U S V^T. $$



              Then $$A = V S^T U^T $$ is a SVD for $A$.






              share|cite|improve this answer























                up vote
                0
                down vote










                up vote
                0
                down vote









                SVD works for every matrix $Ain mathbbR^n times m$ and arbitrary $n, m$.



                As a comment to the case $mleq n$: Consider this case and let be given a SVD for $A^T$, i.e.



                $$ A^T = U S V^T. $$



                Then $$A = V S^T U^T $$ is a SVD for $A$.






                share|cite|improve this answer













                SVD works for every matrix $Ain mathbbR^n times m$ and arbitrary $n, m$.



                As a comment to the case $mleq n$: Consider this case and let be given a SVD for $A^T$, i.e.



                $$ A^T = U S V^T. $$



                Then $$A = V S^T U^T $$ is a SVD for $A$.







                share|cite|improve this answer













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                share|cite|improve this answer











                answered Aug 1 at 12:09









                til

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