I'm trying to show that the frobenius norm is a norm. however it appears as if triangle inequality isnt met.
$$||A+B||_F = \sqrt{\sum_{i,j=1}^n |a_{ij}+b_{ij}|^2} \leq \sqrt{\sum_{i,j=1}^n |a_{ij}|^2+2|a_{ij}||b_{ij}|+|b_{ij}|^2}$$
This follows from triangle inequality on absolute value.
$|a+b| \leq |a|+|b|$ and so it follows that $|a+b|^2 \leq |a|^2+2|a||b|+|b|^2$
So we have:
$$\sqrt{\sum_{i,j=1}^n |a_{ij}|^2+2|a_{ij}||b_{ij}|+|b_{ij}|^2} =\sqrt{\sum_{i,j=1}^n |a_{ij}|^2 + 2\sum_{i,j=1}^n |a_{ij}||b_{ij}|+\sum_{i,j=1}^n |b_{ij}|^2}$$
it isn't apparent to me why $$\sqrt{\sum_{i,j=1}^n |a_{ij}|^2 + 2\sum_{i,j=1}^n |a_{ij}||b_{ij}|+\sum_{i,j=1}^n |b_{ij}|^2} \leq \sqrt{\sum_{i,j=1}^n |a_{ij}|^2}+\sqrt{\sum_{i,j=1}^n |b_{ij}|^2} = ||A||_F+||B||_F$$
That would require that $$\sum_{i,j=1}^n |a_{ij}||b_{ij}| \leq \sqrt{\sum_{i,j=1}^n |a_{ij}|^2|b_{ij}|^2}$$ and i dont know why this is true.
Best Answer
Hint: Given a real or complex inner product $\langle \cdot, \cdot \rangle$, The map $v\mapsto \sqrt{\langle v, v\rangle}$ is a norm.
For completeness, the triangular inequality is proved below.
Take $u,v\in \mathbb C$. Proving that $\Vert u+v\Vert\leq \Vert u\Vert +\Vert v\Vert$ is equivalent to proving that $\Vert u+v\Vert^2\leq \left(\Vert u\Vert +\Vert v\Vert\right)^2.$ The following holds $$\begin{align} \Vert u+v\Vert^2&=\langle u+v, u+v\rangle\\ &=\langle u,u \rangle+\langle u,v\rangle+\langle v,u\rangle+\langle v,v\rangle\\ &=\Vert u\Vert ^2+\langle u,v\rangle+\overline{\langle u,v\rangle}+\Vert v\Vert ^2\\ &=\Vert u\Vert ^2+2\Re\left(\langle u,v\rangle\right)+\Vert v\Vert ^2\\ &\leq \Vert u\Vert ^2+2|\langle u,v\rangle|+\Vert v\Vert^2\\ &=\Vert u\Vert^2+2\Vert u\Vert\Vert v\Vert+\Vert v\Vert^2\\ &=\left(\Vert u\Vert +\Vert v\Vert\right)^2. \end{align}$$
In this problem, the underlying inner product is $\langle A, B\rangle=\text{tr}\left(B^*A\right)$.