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

[GridironCPJ]

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

 

[DonAntonio]

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

 

[GridironCPJ]

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.

 

[Hawkeye18]

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 $\left( \sum_k s_k^2\right)^{1/2}$ and the 2-norm is $\max s_k$, where $s_k$ 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.

 

[GridironCPJ]

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 $\left( \sum_k s_k^2\right)^{1/2}$ and the 2-norm is $\max s_k$, where $s_k$ 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.

 

[AlephZero]

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.

 

[tomz]

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

 

[jbunniii]

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.

 

[armina]

when matrix A is Singular which means det(A)=0.