Submultiplicavity of spectral normProving definition of norms induced by vector normsIs the spectral norm submultiplicative?Spectral norm of block and square matricesSpectral norm minimizationIs the spectral norm of a matrix a strictly convex function?Solve matrix $2$-norm problem with diagonal matrix constraintWhat is the Hessian of the spectral norm?About a spectral norm estimationDerive Lipschitz norm equality.Spectral norm of matrices with complex eigenvaluesShow that the spectral norm of one matrix is smaller than the other.

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Submultiplicavity of spectral norm


Proving definition of norms induced by vector normsIs the spectral norm submultiplicative?Spectral norm of block and square matricesSpectral norm minimizationIs the spectral norm of a matrix a strictly convex function?Solve matrix $2$-norm problem with diagonal matrix constraintWhat is the Hessian of the spectral norm?About a spectral norm estimationDerive Lipschitz norm equality.Spectral norm of matrices with complex eigenvaluesShow that the spectral norm of one matrix is smaller than the other.













0












$begingroup$


Is spectral norm (i.e. the maximum singular value of a matrix) submultiplicative? I am absolutely confused. How to express the singular value of the product of matrices in terms of that of the original matrices? Please help.










share|cite|improve this question









$endgroup$











  • $begingroup$
    Is the spectral norm an operator norm?
    $endgroup$
    – kimchi lover
    Mar 22 at 15:20










  • $begingroup$
    I don't know about operator norms?
    $endgroup$
    – Martund
    Mar 22 at 15:22










  • $begingroup$
    Try the wikipedia page en.wikipedia.org/wiki/Operator_norm
    $endgroup$
    – kimchi lover
    Mar 22 at 15:23















0












$begingroup$


Is spectral norm (i.e. the maximum singular value of a matrix) submultiplicative? I am absolutely confused. How to express the singular value of the product of matrices in terms of that of the original matrices? Please help.










share|cite|improve this question









$endgroup$











  • $begingroup$
    Is the spectral norm an operator norm?
    $endgroup$
    – kimchi lover
    Mar 22 at 15:20










  • $begingroup$
    I don't know about operator norms?
    $endgroup$
    – Martund
    Mar 22 at 15:22










  • $begingroup$
    Try the wikipedia page en.wikipedia.org/wiki/Operator_norm
    $endgroup$
    – kimchi lover
    Mar 22 at 15:23













0












0








0





$begingroup$


Is spectral norm (i.e. the maximum singular value of a matrix) submultiplicative? I am absolutely confused. How to express the singular value of the product of matrices in terms of that of the original matrices? Please help.










share|cite|improve this question









$endgroup$




Is spectral norm (i.e. the maximum singular value of a matrix) submultiplicative? I am absolutely confused. How to express the singular value of the product of matrices in terms of that of the original matrices? Please help.







singularvalues matrix-norms spectral-norm






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 22 at 15:02









MartundMartund

1,945213




1,945213











  • $begingroup$
    Is the spectral norm an operator norm?
    $endgroup$
    – kimchi lover
    Mar 22 at 15:20










  • $begingroup$
    I don't know about operator norms?
    $endgroup$
    – Martund
    Mar 22 at 15:22










  • $begingroup$
    Try the wikipedia page en.wikipedia.org/wiki/Operator_norm
    $endgroup$
    – kimchi lover
    Mar 22 at 15:23
















  • $begingroup$
    Is the spectral norm an operator norm?
    $endgroup$
    – kimchi lover
    Mar 22 at 15:20










  • $begingroup$
    I don't know about operator norms?
    $endgroup$
    – Martund
    Mar 22 at 15:22










  • $begingroup$
    Try the wikipedia page en.wikipedia.org/wiki/Operator_norm
    $endgroup$
    – kimchi lover
    Mar 22 at 15:23















$begingroup$
Is the spectral norm an operator norm?
$endgroup$
– kimchi lover
Mar 22 at 15:20




$begingroup$
Is the spectral norm an operator norm?
$endgroup$
– kimchi lover
Mar 22 at 15:20












$begingroup$
I don't know about operator norms?
$endgroup$
– Martund
Mar 22 at 15:22




$begingroup$
I don't know about operator norms?
$endgroup$
– Martund
Mar 22 at 15:22












$begingroup$
Try the wikipedia page en.wikipedia.org/wiki/Operator_norm
$endgroup$
– kimchi lover
Mar 22 at 15:23




$begingroup$
Try the wikipedia page en.wikipedia.org/wiki/Operator_norm
$endgroup$
– kimchi lover
Mar 22 at 15:23










1 Answer
1






active

oldest

votes


















1












$begingroup$

Well, the spetral norm is then norm induced by the euclidean norm for vectors. I'm not sure if you are already supposed to know/use that fact, vs the definition using singular values. That means, if $A,B$ are square matrices and $x$ is a compatible vector, then



$$lVert (AB)xrVert_2 le lVert (AB)rVert_2 lVert xrVert_2 $$



is true for all $A,B,x$ and when fixing $AB$, there is an $x_0neq vec0$ where equality holds.



OTOH, we also have



$$lVert (AB)xrVert_2 = lVert A(Bx)rVert_2lVert le lVert ArVert_2lVert (Bx)rVert_2 le lVert ArVert_2lVert BrVert_2lVert xrVert_2$$



for all $A,B,x$. Setting $x=x_0$ with the above given $x_0$, we finally have



$$lVert (AB)rVert_2 lVert x_0rVert_2 = lVert (AB)x_0rVert_2 le lVert ArVert_2lVert BrVert_2lVert x_0rVert_2.$$



If we now divide by $lVert x_0rVert_2 neq 0$, we get the submultiplicativity.



Note that this does not use anything special about the euclidean vector norm $lVert xrVert_2$, the proof is valid for any vector norm and the induced matrix norm.



If you want a proof of the fact that the matrix norm induced by the euclidean vector norm is indeed the spectral norm, take a look here: Proving definition of norms induced by vector norms






share|cite|improve this answer









$endgroup$












  • $begingroup$
    very good solution, Thank You
    $endgroup$
    – Martund
    Mar 22 at 16:00











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Well, the spetral norm is then norm induced by the euclidean norm for vectors. I'm not sure if you are already supposed to know/use that fact, vs the definition using singular values. That means, if $A,B$ are square matrices and $x$ is a compatible vector, then



$$lVert (AB)xrVert_2 le lVert (AB)rVert_2 lVert xrVert_2 $$



is true for all $A,B,x$ and when fixing $AB$, there is an $x_0neq vec0$ where equality holds.



OTOH, we also have



$$lVert (AB)xrVert_2 = lVert A(Bx)rVert_2lVert le lVert ArVert_2lVert (Bx)rVert_2 le lVert ArVert_2lVert BrVert_2lVert xrVert_2$$



for all $A,B,x$. Setting $x=x_0$ with the above given $x_0$, we finally have



$$lVert (AB)rVert_2 lVert x_0rVert_2 = lVert (AB)x_0rVert_2 le lVert ArVert_2lVert BrVert_2lVert x_0rVert_2.$$



If we now divide by $lVert x_0rVert_2 neq 0$, we get the submultiplicativity.



Note that this does not use anything special about the euclidean vector norm $lVert xrVert_2$, the proof is valid for any vector norm and the induced matrix norm.



If you want a proof of the fact that the matrix norm induced by the euclidean vector norm is indeed the spectral norm, take a look here: Proving definition of norms induced by vector norms






share|cite|improve this answer









$endgroup$












  • $begingroup$
    very good solution, Thank You
    $endgroup$
    – Martund
    Mar 22 at 16:00















1












$begingroup$

Well, the spetral norm is then norm induced by the euclidean norm for vectors. I'm not sure if you are already supposed to know/use that fact, vs the definition using singular values. That means, if $A,B$ are square matrices and $x$ is a compatible vector, then



$$lVert (AB)xrVert_2 le lVert (AB)rVert_2 lVert xrVert_2 $$



is true for all $A,B,x$ and when fixing $AB$, there is an $x_0neq vec0$ where equality holds.



OTOH, we also have



$$lVert (AB)xrVert_2 = lVert A(Bx)rVert_2lVert le lVert ArVert_2lVert (Bx)rVert_2 le lVert ArVert_2lVert BrVert_2lVert xrVert_2$$



for all $A,B,x$. Setting $x=x_0$ with the above given $x_0$, we finally have



$$lVert (AB)rVert_2 lVert x_0rVert_2 = lVert (AB)x_0rVert_2 le lVert ArVert_2lVert BrVert_2lVert x_0rVert_2.$$



If we now divide by $lVert x_0rVert_2 neq 0$, we get the submultiplicativity.



Note that this does not use anything special about the euclidean vector norm $lVert xrVert_2$, the proof is valid for any vector norm and the induced matrix norm.



If you want a proof of the fact that the matrix norm induced by the euclidean vector norm is indeed the spectral norm, take a look here: Proving definition of norms induced by vector norms






share|cite|improve this answer









$endgroup$












  • $begingroup$
    very good solution, Thank You
    $endgroup$
    – Martund
    Mar 22 at 16:00













1












1








1





$begingroup$

Well, the spetral norm is then norm induced by the euclidean norm for vectors. I'm not sure if you are already supposed to know/use that fact, vs the definition using singular values. That means, if $A,B$ are square matrices and $x$ is a compatible vector, then



$$lVert (AB)xrVert_2 le lVert (AB)rVert_2 lVert xrVert_2 $$



is true for all $A,B,x$ and when fixing $AB$, there is an $x_0neq vec0$ where equality holds.



OTOH, we also have



$$lVert (AB)xrVert_2 = lVert A(Bx)rVert_2lVert le lVert ArVert_2lVert (Bx)rVert_2 le lVert ArVert_2lVert BrVert_2lVert xrVert_2$$



for all $A,B,x$. Setting $x=x_0$ with the above given $x_0$, we finally have



$$lVert (AB)rVert_2 lVert x_0rVert_2 = lVert (AB)x_0rVert_2 le lVert ArVert_2lVert BrVert_2lVert x_0rVert_2.$$



If we now divide by $lVert x_0rVert_2 neq 0$, we get the submultiplicativity.



Note that this does not use anything special about the euclidean vector norm $lVert xrVert_2$, the proof is valid for any vector norm and the induced matrix norm.



If you want a proof of the fact that the matrix norm induced by the euclidean vector norm is indeed the spectral norm, take a look here: Proving definition of norms induced by vector norms






share|cite|improve this answer









$endgroup$



Well, the spetral norm is then norm induced by the euclidean norm for vectors. I'm not sure if you are already supposed to know/use that fact, vs the definition using singular values. That means, if $A,B$ are square matrices and $x$ is a compatible vector, then



$$lVert (AB)xrVert_2 le lVert (AB)rVert_2 lVert xrVert_2 $$



is true for all $A,B,x$ and when fixing $AB$, there is an $x_0neq vec0$ where equality holds.



OTOH, we also have



$$lVert (AB)xrVert_2 = lVert A(Bx)rVert_2lVert le lVert ArVert_2lVert (Bx)rVert_2 le lVert ArVert_2lVert BrVert_2lVert xrVert_2$$



for all $A,B,x$. Setting $x=x_0$ with the above given $x_0$, we finally have



$$lVert (AB)rVert_2 lVert x_0rVert_2 = lVert (AB)x_0rVert_2 le lVert ArVert_2lVert BrVert_2lVert x_0rVert_2.$$



If we now divide by $lVert x_0rVert_2 neq 0$, we get the submultiplicativity.



Note that this does not use anything special about the euclidean vector norm $lVert xrVert_2$, the proof is valid for any vector norm and the induced matrix norm.



If you want a proof of the fact that the matrix norm induced by the euclidean vector norm is indeed the spectral norm, take a look here: Proving definition of norms induced by vector norms







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 22 at 15:28









IngixIngix

5,137159




5,137159











  • $begingroup$
    very good solution, Thank You
    $endgroup$
    – Martund
    Mar 22 at 16:00
















  • $begingroup$
    very good solution, Thank You
    $endgroup$
    – Martund
    Mar 22 at 16:00















$begingroup$
very good solution, Thank You
$endgroup$
– Martund
Mar 22 at 16:00




$begingroup$
very good solution, Thank You
$endgroup$
– Martund
Mar 22 at 16:00

















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