Given is $\left | \left | A \right | \right |_{2} =\left ( \sum_{i,j=1}^{n} a_{ij}^{2} \right )^\frac{1}{2}$ for $A\in \mathbb{R}^{n\times n}$. Show that this defines a matrix norm.

I remember i've been reading something about Frobenius norm, but it was defined in a different way, if i am not wrong as $\left | \left | A \right | \right |_{2}=\sqrt{tr(A^{t}A)}$, where $A^{t}$ was the transposed matrix $A$.

I am sorry about the trivial question, but do i have to prove here that the given norm is Frobenius norm?

Thank you in advance!


Yes you're right. In fact if $A=(a_{ij})$ then $A^t=(b_{ij})$ where $b_{ij}=a_{ji}$ so if $C=(c_{ij})=A^tA$ then $$c_{ij}=\sum_{k=1}^n b_{ik}a_{kj}=\sum_{k=1}^na_{ki}a_{kj}$$ hence $$\operatorname{tr}(A^tA)=\sum_{i=1}^n c_{ii}=\sum_{i=1}^n\sum_{k=1}^na_{ki}^2$$ Now to prove the given expression defines a norm you can for example prove that the map $$\varphi:\mathcal{M}_n(\Bbb R)\times \mathcal{M}_n(\Bbb R)\rightarrow \Bbb R,\quad (A,B)\mapsto \operatorname{tr}(A^tB)$$ is an inner product and $||.||_2$ is the euclidean norm associated to this inner product.


This is indeed the Frobenius norm. So, if you've established that the Frobenius norm is indeed a matrix norm, it is sufficient to show that this is that, which is a norm.

In order to show that they are the same, note the following:

The $ith$ diagonal entry of $A^TA$ is given by $$ b_{ii}= \sum_{j=1}^n a_{ji}^2 $$ So that $$ \operatorname{trace}(A^TA) = \sum_{i=1}^n b_{ii} = \sum_{i=1}^n \sum_{j=1}^n a_{ji}^2 $$

  • $\begingroup$ Thank you very much, i was a little confused by the notation in my problem, but now is clear what i have to do. :) Have a nice day! $\endgroup$ – Lullaby Mar 2 '14 at 15:26

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