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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?

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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 $\mathbb{C}^{n^2}$, 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 $\mathbb{C}^n$ 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

2 Answers 2

See Wikipedia for all definitions. Take this matrix: $$ \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix} $$ Its Frobenius norm is $\sqrt{10}$, 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 $\sqrt{r}$ 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.

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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 $\sqrt{r}$ bound backwards, no? –  Rahul Apr 15 '11 at 2:23
    
You're right. Corrected. –  Yuval Filmus Apr 18 '11 at 18:20
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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

There are three important types of matrix norms. For some matrix $A$

  • Induced norm, which measures what is the maximum of $\|Ax\|/\|x\|$ for any $x$.

  • Element-wise norm, which is like unwrapping $A$ into a long vector, then measuring 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 $\ge$ Frobenius norm.

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 $\ge$ than Frobenius norm, and the two are only equal when all of the matrix's eigenvalues have equal magnitude.

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