When is the Frobenius norm of a matrix equal to the 2-norm of a matrix?

  • #1
What conditions most be true for these two norms to be equal? Or are they always equal?
 

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  • #2
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What conditions most be true for these two norms to be equal? Or are they always equal?



I'm far from being a specialist in this, but it seems to me that "Frobenius norm of a matrix" is just the name given to the 2-norm...

Don
 
  • #3
Well, in the applied linear algebra course I'm taking currently, the Frobenius norm of a matrix A is defined as the square root of the trace of A'A and the 2-norm is defined as the square root of the largest eigenvalue of A'A. I'm just not sure if they're always the same.
 
  • #4
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The Frobenius and 2-norm of a matrix coincide if and only if the matrix has rank 1 (i.e. if and only if the matrix can be represented as A=c r, where r is a row and c is a column).

You can see that from the fact that Frobenius norm is [itex]\left( \sum_k s_k^2\right)^{1/2}[/itex] and the 2-norm is [itex]\max s_k[/itex], where [itex]s_k[/itex] are singular values. So equality happens if and only if there is only one non-zero singular value, which is equivalent to the fact that the rank is 1.
 
  • #5
The Frobenius and 2-norm of a matrix coincide if and only if the matrix has rank 1 (i.e. if and only if the matrix can be represented as A=c r, where r is a row and c is a column).

You can see that from the fact that Frobenius norm is [itex]\left( \sum_k s_k^2\right)^{1/2}[/itex] and the 2-norm is [itex]\max s_k[/itex], where [itex]s_k[/itex] are singular values. So equality happens if and only if there is only one non-zero singular value, which is equivalent to the fact that the rank is 1.

Excellent, thank you. The matrix in a proof I'm working on involves a rank 1 matrix, so this equality of the two norms applies perfectly.
 
  • #6
AlephZero
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The Frobenius and 2-norm of a matrix coincide if and only if the matrix has rank 1

More generally, ##||A||_2 \le ||A||_F \le \sqrt{r}||A||_2## where r is the rank of A.
 
  • #7
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More generally, ##||A||_2 \le ||A||_F \le \sqrt{r}||A||_2## where r is the rank of A.

May you shed some light on this? Or quote any possible reference? Thanks
 
  • #8
jbunniii
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May you shed some light on this? Or quote any possible reference? Thanks
Assuming you accept Hawkeye18's formulas, namely
$$\|A\|_F = \left( \sum_k s_k^2\right)^{1/2}$$
and
$$\|A\|_2 = \max{s_k}$$
then we have
$$\|A\|_2 = \max{s_k} = \left( (\max{s_k})^2\right)^{1/2} \leq \left( \sum_{k} s_k^2 \right)^{1/2} = \|A\|_F$$

For the second inequality, note that the rank of ##A## is precisely the number of nonzero singular values. Let's sort the singular values so that the nonzero ones all come first. Then for a rank ##r## matrix, we have
$$\|A\|_F = \left( \sum_{k=1}^{r} s_k^2\right) ^{1/2} \leq \left( \sum_{k=1}^{r} (\max s_k)^2 \right)^{1/2} = (r (\max s_k)^2)^{1/2} = \sqrt{r} \|A\|_2$$
Equality holds if and only if the ##r## nonzero singular values are all equal.
 
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  • #9
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when matrix A is Singular which means det(A)=0.
 

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