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.
$begingroup$
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.
singularvalues matrix-norms spectral-norm
asked Mar 22 at 15:02
MartundMartund
1,945213
$endgroup$
add a comment |
$begingroup$
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.
singularvalues matrix-norms spectral-norm
asked Mar 22 at 15:02
MartundMartund
1,945213
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
singularvalues matrix-norms spectral-norm
asked Mar 22 at 15:02
MartundMartund
1,945213
$endgroup$
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
singularvalues matrix-norms spectral-norm
asked Mar 22 at 15:02
MartundMartund
1,945213
asked Mar 22 at 15:02
MartundMartund
1,945213
asked Mar 22 at 15:02
MartundMartund
1,945213
asked Mar 22 at 15:02
MartundMartund
1,945213
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
answered Mar 22 at 15:28
IngixIngix
5,137159
$endgroup$
$begingroup$
very good solution, Thank You
$endgroup$
– Martund
Mar 22 at 16:00
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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
answered Mar 22 at 15:28
IngixIngix
5,137159
$endgroup$
$begingroup$
very good solution, Thank You
$endgroup$
– Martund
Mar 22 at 16:00
add a comment |
$begingroup$
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
answered Mar 22 at 15:28
IngixIngix
5,137159
$endgroup$
$begingroup$
very good solution, Thank You
$endgroup$
– Martund
Mar 22 at 16:00
add a comment |
$begingroup$
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
answered Mar 22 at 15:28
IngixIngix
5,137159
$endgroup$
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
answered Mar 22 at 15:28
IngixIngix
5,137159
answered Mar 22 at 15:28
IngixIngix
5,137159
answered Mar 22 at 15:28
IngixIngix
5,137159
answered Mar 22 at 15:28
IngixIngix
5,137159
5,137159
$begingroup$
very good solution, Thank You
$endgroup$
– Martund
Mar 22 at 16:00
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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