How do you express the Frobenius norm of a Matrix as the squared norm of its singular values? Let the Frobenius norm of an m by n ($m \times n$) matrix M be:
$$|| M ||_{F} = \sqrt{\sum_{i,j} M^2_{i,j}}$$
I was told that it can be proved that, if M can be expressed as follows (which we can because of SVD):
$$ M = \sum^{r}_{i=1} \sigma_i u_i v^T_i$$
Then one can show that the Frobenius norm equivalently be expressed as:
$$ || M ||_{F} = \sqrt{\sum_{i} \sigma_i^2} $$
I was a little stuck on how to do such a proof. This is what I had so far: 
I was thinking that maybe since the second expression is a linear combination of outer produced scaled by $\sigma_i$, then one could express each entry of M as follow: $M_{i,j} = \sum^{r}_{i=1} \sigma_i (u_i v^T_i)_{i,j}$. Thus we can substitute:
$$|| M ||^2_{F} = \sum_{i,j} M^2_{i,j} = \sum^n_{j=1} \sum^m_{i=1} (\sum^{r}_{i=1} \sigma_i (u_i v^T_i)_{i,j})^2 = \sum^n_{j=1} \sum^m_{i=1} (\sum^{r}_{i=1} \sigma_i (u_i v^T_i)_{i,j}) (\sum^{r}_{i=1} \sigma_i (u_i v^T_i)_{i,j}) $$
After that line I got kind of stuck. Though my intuition tells me that if I expand what I have somehow, something magical is going to happens with the combination of outer products of orthonormal vectors and get a bunch of zeros! Probably by re-arranging and forming inner products that evaluate to zero (due to orthogonality) ... Though, not sure how to expand that nasty little guy.
Anyway has any suggestion on how to move on or if maybe there is a better approach?

$\sum_{i}\sigma_i^2=Trace(\Lambda \Lambda^T)$ where $M=U\Lambda V^T$. Then, $$\|M\|_F^2=Trace(MM^T)=Trace(U\Lambda V^TV\Lambda^T U^T)=Trace(U\Lambda \Lambda^TU^T)=Trace(\Lambda\Lambda^T U^T U)=Trace(\Lambda\Lambda^T)$$ 
