What is the difference between the Frobenius norm and the 2-norm of a matrix?
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Given a matrix, is the Frobenius norm of that matrix always equal to the 2-norm of it, or are there certain matrices where these two norm methods would produce different results?
If they are identical, then I suppose the only difference between them is the method of calculation, eh?
matrices norm matrix-norms
edited Aug 21 at 10:23
Rodrigo de Azevedo
12.6k41751
asked Apr 15 '11 at 1:57
Ricket
4012613
add a comment |Â
up vote
34
down vote
favorite
Given a matrix, is the Frobenius norm of that matrix always equal to the 2-norm of it, or are there certain matrices where these two norm methods would produce different results?
If they are identical, then I suppose the only difference between them is the method of calculation, eh?
matrices norm matrix-norms
edited Aug 21 at 10:23
Rodrigo de Azevedo
12.6k41751
asked Apr 15 '11 at 1:57
Ricket
4012613
1
What do you mean by 2-norm?
– Jonas Meyer
Apr 15 '11 at 1:59
The p-norm where p=2, also known as the Euclidean norm.
– Ricket
Apr 15 '11 at 2:02
If you mean the Euclidean norm when $M_n$ is treated like $mathbbC^n^2$, then yes they are the same; this is the definition of the Frobenius norm, as seen on Wikipedia: en.wikipedia.org/wiki/Matrix_norm#Frobenius_norm
– Jonas Meyer
Apr 15 '11 at 2:06
Ricket: Could you please give the precise definitions you are using (or precise references to these definitions)? The phrase "p-norm" for matrices is ambiguous, which is why I asked earlier. Many think of "2-norm" as meaning the operator norm when $M_n$ acts on $mathbbC^n$ with Euclidean norm, hence Yuval's answer. I think I now know what you mean, but then your question is answered by the Wikipedia link, right?
– Jonas Meyer
Apr 15 '11 at 2:26
add a comment |Â
up vote
34
down vote
favorite
up vote
34
down vote
favorite
Given a matrix, is the Frobenius norm of that matrix always equal to the 2-norm of it, or are there certain matrices where these two norm methods would produce different results?
If they are identical, then I suppose the only difference between them is the method of calculation, eh?
matrices norm matrix-norms
edited Aug 21 at 10:23
Rodrigo de Azevedo
12.6k41751
asked Apr 15 '11 at 1:57
Ricket
4012613
Given a matrix, is the Frobenius norm of that matrix always equal to the 2-norm of it, or are there certain matrices where these two norm methods would produce different results?
If they are identical, then I suppose the only difference between them is the method of calculation, eh?
matrices norm matrix-norms
edited Aug 21 at 10:23
Rodrigo de Azevedo
12.6k41751
asked Apr 15 '11 at 1:57
Ricket
4012613
edited Aug 21 at 10:23
Rodrigo de Azevedo
12.6k41751
edited Aug 21 at 10:23
Rodrigo de Azevedo
12.6k41751
edited Aug 21 at 10:23
Rodrigo de Azevedo
12.6k41751
12.6k41751
asked Apr 15 '11 at 1:57
Ricket
4012613
asked Apr 15 '11 at 1:57
Ricket
4012613
asked Apr 15 '11 at 1:57
Ricket
4012613
4012613
1
What do you mean by 2-norm?
– Jonas Meyer
Apr 15 '11 at 1:59
The p-norm where p=2, also known as the Euclidean norm.
– Ricket
Apr 15 '11 at 2:02
If you mean the Euclidean norm when $M_n$ is treated like $mathbbC^n^2$, then yes they are the same; this is the definition of the Frobenius norm, as seen on Wikipedia: en.wikipedia.org/wiki/Matrix_norm#Frobenius_norm
– Jonas Meyer
Apr 15 '11 at 2:06
Ricket: Could you please give the precise definitions you are using (or precise references to these definitions)? The phrase "p-norm" for matrices is ambiguous, which is why I asked earlier. Many think of "2-norm" as meaning the operator norm when $M_n$ acts on $mathbbC^n$ with Euclidean norm, hence Yuval's answer. I think I now know what you mean, but then your question is answered by the Wikipedia link, right?
– Jonas Meyer
Apr 15 '11 at 2:26
add a comment |Â
1
What do you mean by 2-norm?
– Jonas Meyer
Apr 15 '11 at 1:59
The p-norm where p=2, also known as the Euclidean norm.
– Ricket
Apr 15 '11 at 2:02
If you mean the Euclidean norm when $M_n$ is treated like $mathbbC^n^2$, then yes they are the same; this is the definition of the Frobenius norm, as seen on Wikipedia: en.wikipedia.org/wiki/Matrix_norm#Frobenius_norm
– Jonas Meyer
Apr 15 '11 at 2:06
Ricket: Could you please give the precise definitions you are using (or precise references to these definitions)? The phrase "p-norm" for matrices is ambiguous, which is why I asked earlier. Many think of "2-norm" as meaning the operator norm when $M_n$ acts on $mathbbC^n$ with Euclidean norm, hence Yuval's answer. I think I now know what you mean, but then your question is answered by the Wikipedia link, right?
– Jonas Meyer
Apr 15 '11 at 2:26
1
1
What do you mean by 2-norm?
– Jonas Meyer
Apr 15 '11 at 1:59
What do you mean by 2-norm?
– Jonas Meyer
Apr 15 '11 at 1:59
The p-norm where p=2, also known as the Euclidean norm.
– Ricket
Apr 15 '11 at 2:02
The p-norm where p=2, also known as the Euclidean norm.
– Ricket
Apr 15 '11 at 2:02
If you mean the Euclidean norm when $M_n$ is treated like $mathbbC^n^2$, then yes they are the same; this is the definition of the Frobenius norm, as seen on Wikipedia: en.wikipedia.org/wiki/Matrix_norm#Frobenius_norm
– Jonas Meyer
Apr 15 '11 at 2:06
If you mean the Euclidean norm when $M_n$ is treated like $mathbbC^n^2$, then yes they are the same; this is the definition of the Frobenius norm, as seen on Wikipedia: en.wikipedia.org/wiki/Matrix_norm#Frobenius_norm
– Jonas Meyer
Apr 15 '11 at 2:06
Ricket: Could you please give the precise definitions you are using (or precise references to these definitions)? The phrase "p-norm" for matrices is ambiguous, which is why I asked earlier. Many think of "2-norm" as meaning the operator norm when $M_n$ acts on $mathbbC^n$ with Euclidean norm, hence Yuval's answer. I think I now know what you mean, but then your question is answered by the Wikipedia link, right?
– Jonas Meyer
Apr 15 '11 at 2:26
Ricket: Could you please give the precise definitions you are using (or precise references to these definitions)? The phrase "p-norm" for matrices is ambiguous, which is why I asked earlier. Many think of "2-norm" as meaning the operator norm when $M_n$ acts on $mathbbC^n$ with Euclidean norm, hence Yuval's answer. I think I now know what you mean, but then your question is answered by the Wikipedia link, right?
– Jonas Meyer
Apr 15 '11 at 2:26
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
23
down vote
There are three important types of matrix norms. For some matrix $A$
Induced norm, which measures what is the maximum of $frac$ for any $x neq 0$ (or, equivalently, the maximum of $|Ax|$ for $|x|=1$).
Element-wise norm, which is like unwrapping $A$ into a long vector, then calculating its vector norm.
Schatten norm, which measures the vector norm of the singular values of $A$.
So, to answer your question:
Frobenius norm = Element-wise 2-norm = Schatten 2-norm
Induced 2-norm = Schatten $infty$-norm. This is also called Spectral norm.
So if by "2-norm" you mean element-wise or Schatten norm, then they are identical to Frobenius norm. If you mean induced 2-norm, you get spectral 2-norm, which is $le$ Frobenius norm. (It should be less than or equal to)
As far as I can tell, if you don't clarify which type you're talking about, induced norm is usually implied. For example, in matlab, norm(A,2) gives you induced 2-norm, which they simply call the 2-norm. So in that sense, the answer to your question is that the (induced) matrix 2-norm is $le$ than Frobenius norm, and the two are only equal when all of the matrix's eigenvalues have equal magnitude.
edited Nov 8 '17 at 15:17
Brandon J DeHart
1035
answered Sep 17 '13 at 19:47
Kevin Holt
97186
add a comment |Â
up vote
11
down vote
See Wikipedia for all definitions. Take this matrix:
$$ beginpmatrix 2 & -1 \ -1 & 2 endpmatrix $$
Its Frobenius norm is $sqrt10$, but its eigenvalues are $3,1$ so its $2$-norm (or spectral radius) is $3$. The Frobenius norm is always at least as large as the spectral radius. The Frobenius norm is at most $sqrtr$ as much as the spectral radius, and this is probably tight (see the section on equivalence of norms in Wikipedia).
Note that the Schatten $2$-norm is equal to the Frobenius norm.
answered Apr 15 '11 at 2:04
Yuval Filmus
47.3k469142
Ohh, I was just using the vector 2-norm (Euclidean norm) operation on the matrix, not the correct matrix 2-norm. The vector 2-norm (piecewise square, sum all elements, square root) when extended to a matrix would be the Schatten 2-norm I guess.
– Ricket
Apr 15 '11 at 2:15
I think you got the $sqrtr$ bound backwards, no?
– Rahul
Apr 15 '11 at 2:23
You're right. Corrected.
– Yuval Filmus
Apr 18 '11 at 18:20
1
Also, it is true that both bounds are tight: the former is attained by a matrix with all but one entries being zero; the latter is attained by the identity matrix.
– Rahul
Apr 18 '11 at 18:36
add a comment |Â
up vote
-1
down vote
The L2 (or L^2) norm is the Euclidian norm of a vector.
The Frobenius norm is the Euclidian norm of a matrix.
answered May 10 at 8:11
Jay Griff
1
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
23
down vote
There are three important types of matrix norms. For some matrix $A$
Induced norm, which measures what is the maximum of $frac$ for any $x neq 0$ (or, equivalently, the maximum of $|Ax|$ for $|x|=1$).
Element-wise norm, which is like unwrapping $A$ into a long vector, then calculating its vector norm.
Schatten norm, which measures the vector norm of the singular values of $A$.
So, to answer your question:
Frobenius norm = Element-wise 2-norm = Schatten 2-norm
Induced 2-norm = Schatten $infty$-norm. This is also called Spectral norm.
So if by "2-norm" you mean element-wise or Schatten norm, then they are identical to Frobenius norm. If you mean induced 2-norm, you get spectral 2-norm, which is $le$ Frobenius norm. (It should be less than or equal to)
As far as I can tell, if you don't clarify which type you're talking about, induced norm is usually implied. For example, in matlab, norm(A,2) gives you induced 2-norm, which they simply call the 2-norm. So in that sense, the answer to your question is that the (induced) matrix 2-norm is $le$ than Frobenius norm, and the two are only equal when all of the matrix's eigenvalues have equal magnitude.
edited Nov 8 '17 at 15:17
Brandon J DeHart
1035
answered Sep 17 '13 at 19:47
Kevin Holt
97186
add a comment |Â
up vote
23
down vote
There are three important types of matrix norms. For some matrix $A$
Induced norm, which measures what is the maximum of $frac$ for any $x neq 0$ (or, equivalently, the maximum of $|Ax|$ for $|x|=1$).
Element-wise norm, which is like unwrapping $A$ into a long vector, then calculating its vector norm.
Schatten norm, which measures the vector norm of the singular values of $A$.
So, to answer your question:
Frobenius norm = Element-wise 2-norm = Schatten 2-norm
Induced 2-norm = Schatten $infty$-norm. This is also called Spectral norm.
So if by "2-norm" you mean element-wise or Schatten norm, then they are identical to Frobenius norm. If you mean induced 2-norm, you get spectral 2-norm, which is $le$ Frobenius norm. (It should be less than or equal to)
As far as I can tell, if you don't clarify which type you're talking about, induced norm is usually implied. For example, in matlab, norm(A,2) gives you induced 2-norm, which they simply call the 2-norm. So in that sense, the answer to your question is that the (induced) matrix 2-norm is $le$ than Frobenius norm, and the two are only equal when all of the matrix's eigenvalues have equal magnitude.
edited Nov 8 '17 at 15:17
Brandon J DeHart
1035
answered Sep 17 '13 at 19:47
Kevin Holt
97186
add a comment |Â
up vote
23
down vote
up vote
23
down vote
There are three important types of matrix norms. For some matrix $A$
Induced norm, which measures what is the maximum of $frac$ for any $x neq 0$ (or, equivalently, the maximum of $|Ax|$ for $|x|=1$).
Element-wise norm, which is like unwrapping $A$ into a long vector, then calculating its vector norm.
Schatten norm, which measures the vector norm of the singular values of $A$.
So, to answer your question:
Frobenius norm = Element-wise 2-norm = Schatten 2-norm
Induced 2-norm = Schatten $infty$-norm. This is also called Spectral norm.
So if by "2-norm" you mean element-wise or Schatten norm, then they are identical to Frobenius norm. If you mean induced 2-norm, you get spectral 2-norm, which is $le$ Frobenius norm. (It should be less than or equal to)
As far as I can tell, if you don't clarify which type you're talking about, induced norm is usually implied. For example, in matlab, norm(A,2) gives you induced 2-norm, which they simply call the 2-norm. So in that sense, the answer to your question is that the (induced) matrix 2-norm is $le$ than Frobenius norm, and the two are only equal when all of the matrix's eigenvalues have equal magnitude.
edited Nov 8 '17 at 15:17
Brandon J DeHart
1035
answered Sep 17 '13 at 19:47
Kevin Holt
97186
There are three important types of matrix norms. For some matrix $A$
Induced norm, which measures what is the maximum of $frac$ for any $x neq 0$ (or, equivalently, the maximum of $|Ax|$ for $|x|=1$).
Element-wise norm, which is like unwrapping $A$ into a long vector, then calculating its vector norm.
Schatten norm, which measures the vector norm of the singular values of $A$.
So, to answer your question:
Frobenius norm = Element-wise 2-norm = Schatten 2-norm
Induced 2-norm = Schatten $infty$-norm. This is also called Spectral norm.
So if by "2-norm" you mean element-wise or Schatten norm, then they are identical to Frobenius norm. If you mean induced 2-norm, you get spectral 2-norm, which is $le$ Frobenius norm. (It should be less than or equal to)
As far as I can tell, if you don't clarify which type you're talking about, induced norm is usually implied. For example, in matlab, norm(A,2) gives you induced 2-norm, which they simply call the 2-norm. So in that sense, the answer to your question is that the (induced) matrix 2-norm is $le$ than Frobenius norm, and the two are only equal when all of the matrix's eigenvalues have equal magnitude.
edited Nov 8 '17 at 15:17
Brandon J DeHart
1035
answered Sep 17 '13 at 19:47
Kevin Holt
97186
edited Nov 8 '17 at 15:17
Brandon J DeHart
1035
edited Nov 8 '17 at 15:17
Brandon J DeHart
1035
edited Nov 8 '17 at 15:17
Brandon J DeHart
1035
1035
answered Sep 17 '13 at 19:47
Kevin Holt
97186
answered Sep 17 '13 at 19:47
Kevin Holt
97186
answered Sep 17 '13 at 19:47
Kevin Holt
97186
97186
add a comment |Â
add a comment |Â
up vote
11
down vote
See Wikipedia for all definitions. Take this matrix:
$$ beginpmatrix 2 & -1 \ -1 & 2 endpmatrix $$
Its Frobenius norm is $sqrt10$, but its eigenvalues are $3,1$ so its $2$-norm (or spectral radius) is $3$. The Frobenius norm is always at least as large as the spectral radius. The Frobenius norm is at most $sqrtr$ as much as the spectral radius, and this is probably tight (see the section on equivalence of norms in Wikipedia).
Note that the Schatten $2$-norm is equal to the Frobenius norm.
answered Apr 15 '11 at 2:04
Yuval Filmus
47.3k469142
Ohh, I was just using the vector 2-norm (Euclidean norm) operation on the matrix, not the correct matrix 2-norm. The vector 2-norm (piecewise square, sum all elements, square root) when extended to a matrix would be the Schatten 2-norm I guess.
– Ricket
Apr 15 '11 at 2:15
I think you got the $sqrtr$ bound backwards, no?
– Rahul
Apr 15 '11 at 2:23
You're right. Corrected.
– Yuval Filmus
Apr 18 '11 at 18:20
1
Also, it is true that both bounds are tight: the former is attained by a matrix with all but one entries being zero; the latter is attained by the identity matrix.
– Rahul
Apr 18 '11 at 18:36
add a comment |Â
up vote
11
down vote
See Wikipedia for all definitions. Take this matrix:
$$ beginpmatrix 2 & -1 \ -1 & 2 endpmatrix $$
Its Frobenius norm is $sqrt10$, but its eigenvalues are $3,1$ so its $2$-norm (or spectral radius) is $3$. The Frobenius norm is always at least as large as the spectral radius. The Frobenius norm is at most $sqrtr$ as much as the spectral radius, and this is probably tight (see the section on equivalence of norms in Wikipedia).
Note that the Schatten $2$-norm is equal to the Frobenius norm.
answered Apr 15 '11 at 2:04
Yuval Filmus
47.3k469142
Ohh, I was just using the vector 2-norm (Euclidean norm) operation on the matrix, not the correct matrix 2-norm. The vector 2-norm (piecewise square, sum all elements, square root) when extended to a matrix would be the Schatten 2-norm I guess.
– Ricket
Apr 15 '11 at 2:15
I think you got the $sqrtr$ bound backwards, no?
– Rahul
Apr 15 '11 at 2:23
You're right. Corrected.
– Yuval Filmus
Apr 18 '11 at 18:20
1
Also, it is true that both bounds are tight: the former is attained by a matrix with all but one entries being zero; the latter is attained by the identity matrix.
– Rahul
Apr 18 '11 at 18:36
add a comment |Â
up vote
11
down vote
up vote
11
down vote
See Wikipedia for all definitions. Take this matrix:
$$ beginpmatrix 2 & -1 \ -1 & 2 endpmatrix $$
Its Frobenius norm is $sqrt10$, but its eigenvalues are $3,1$ so its $2$-norm (or spectral radius) is $3$. The Frobenius norm is always at least as large as the spectral radius. The Frobenius norm is at most $sqrtr$ as much as the spectral radius, and this is probably tight (see the section on equivalence of norms in Wikipedia).
Note that the Schatten $2$-norm is equal to the Frobenius norm.
answered Apr 15 '11 at 2:04
Yuval Filmus
47.3k469142
See Wikipedia for all definitions. Take this matrix:
$$ beginpmatrix 2 & -1 \ -1 & 2 endpmatrix $$
Its Frobenius norm is $sqrt10$, but its eigenvalues are $3,1$ so its $2$-norm (or spectral radius) is $3$. The Frobenius norm is always at least as large as the spectral radius. The Frobenius norm is at most $sqrtr$ as much as the spectral radius, and this is probably tight (see the section on equivalence of norms in Wikipedia).
Note that the Schatten $2$-norm is equal to the Frobenius norm.
answered Apr 15 '11 at 2:04
Yuval Filmus
47.3k469142
edited Apr 18 '11 at 18:20
answered Apr 15 '11 at 2:04
Yuval Filmus
47.3k469142
answered Apr 15 '11 at 2:04
Yuval Filmus
47.3k469142
answered Apr 15 '11 at 2:04
Yuval Filmus
47.3k469142
47.3k469142
Ohh, I was just using the vector 2-norm (Euclidean norm) operation on the matrix, not the correct matrix 2-norm. The vector 2-norm (piecewise square, sum all elements, square root) when extended to a matrix would be the Schatten 2-norm I guess.
– Ricket
Apr 15 '11 at 2:15
I think you got the $sqrtr$ bound backwards, no?
– Rahul
Apr 15 '11 at 2:23
You're right. Corrected.
– Yuval Filmus
Apr 18 '11 at 18:20
1
Also, it is true that both bounds are tight: the former is attained by a matrix with all but one entries being zero; the latter is attained by the identity matrix.
– Rahul
Apr 18 '11 at 18:36
add a comment |Â
Ohh, I was just using the vector 2-norm (Euclidean norm) operation on the matrix, not the correct matrix 2-norm. The vector 2-norm (piecewise square, sum all elements, square root) when extended to a matrix would be the Schatten 2-norm I guess.
– Ricket
Apr 15 '11 at 2:15
I think you got the $sqrtr$ bound backwards, no?
– Rahul
Apr 15 '11 at 2:23
You're right. Corrected.
– Yuval Filmus
Apr 18 '11 at 18:20
1
Also, it is true that both bounds are tight: the former is attained by a matrix with all but one entries being zero; the latter is attained by the identity matrix.
– Rahul
Apr 18 '11 at 18:36
Ohh, I was just using the vector 2-norm (Euclidean norm) operation on the matrix, not the correct matrix 2-norm. The vector 2-norm (piecewise square, sum all elements, square root) when extended to a matrix would be the Schatten 2-norm I guess.
– Ricket
Apr 15 '11 at 2:15
Ohh, I was just using the vector 2-norm (Euclidean norm) operation on the matrix, not the correct matrix 2-norm. The vector 2-norm (piecewise square, sum all elements, square root) when extended to a matrix would be the Schatten 2-norm I guess.
– Ricket
Apr 15 '11 at 2:15
I think you got the $sqrtr$ bound backwards, no?
– Rahul
Apr 15 '11 at 2:23
I think you got the $sqrtr$ bound backwards, no?
– Rahul
Apr 15 '11 at 2:23
You're right. Corrected.
– Yuval Filmus
Apr 18 '11 at 18:20
You're right. Corrected.
– Yuval Filmus
Apr 18 '11 at 18:20
1
1
Also, it is true that both bounds are tight: the former is attained by a matrix with all but one entries being zero; the latter is attained by the identity matrix.
– Rahul
Apr 18 '11 at 18:36
Also, it is true that both bounds are tight: the former is attained by a matrix with all but one entries being zero; the latter is attained by the identity matrix.
– Rahul
Apr 18 '11 at 18:36
add a comment |Â
up vote
-1
down vote
The L2 (or L^2) norm is the Euclidian norm of a vector.
The Frobenius norm is the Euclidian norm of a matrix.
answered May 10 at 8:11
Jay Griff
1
add a comment |Â
up vote
-1
down vote
The L2 (or L^2) norm is the Euclidian norm of a vector.
The Frobenius norm is the Euclidian norm of a matrix.
answered May 10 at 8:11
Jay Griff
1
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
The L2 (or L^2) norm is the Euclidian norm of a vector.
The Frobenius norm is the Euclidian norm of a matrix.
answered May 10 at 8:11
Jay Griff
1
The L2 (or L^2) norm is the Euclidian norm of a vector.
The Frobenius norm is the Euclidian norm of a matrix.
answered May 10 at 8:11
Jay Griff
1
answered May 10 at 8:11
Jay Griff
1
answered May 10 at 8:11
Jay Griff
1
answered May 10 at 8:11
Jay Griff
1
1
add a comment |Â
add a comment |Â
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1
What do you mean by 2-norm?
– Jonas Meyer
Apr 15 '11 at 1:59
The p-norm where p=2, also known as the Euclidean norm.
– Ricket
Apr 15 '11 at 2:02
If you mean the Euclidean norm when $M_n$ is treated like $mathbbC^n^2$, then yes they are the same; this is the definition of the Frobenius norm, as seen on Wikipedia: en.wikipedia.org/wiki/Matrix_norm#Frobenius_norm
– Jonas Meyer
Apr 15 '11 at 2:06
Ricket: Could you please give the precise definitions you are using (or precise references to these definitions)? The phrase "p-norm" for matrices is ambiguous, which is why I asked earlier. Many think of "2-norm" as meaning the operator norm when $M_n$ acts on $mathbbC^n$ with Euclidean norm, hence Yuval's answer. I think I now know what you mean, but then your question is answered by the Wikipedia link, right?
– Jonas Meyer
Apr 15 '11 at 2:26