Sum-of-least-square loss:
$$ \ell\left(\mathbf{\tilde W}\right) = \sum_{n=1}^N \left\| \mathbf{\tilde W}^T\mathbf{\tilde x^{(n)}} -\mathbf{t}^{(n}) \right\|^2 = \left\|\mathbf{\tilde X\tilde W-T}\right\|^2_F $$
- the $n$-th row of $\mathbf{\tilde X}$ is $\left[\mathbf{\tilde x}^{(n)}\right]^T$
- the $n$-th row of $\mathbf{T}$ is $\left[\mathbf{t}^{(n)}\right]^T$
The subscript $F$ denotes the Frobenius Norm.
The expression above seems missed a step from the transpose.
This is what I would write:
$\displaystyle\sum_{n=1}^N\|\mathbf{\tilde W}^T\mathbf{\tilde x^{(n)}}-\mathbf{t}^{(n})\|^2=\displaystyle\sum_{n=1}^N\| [[\mathbf{\tilde W}^T\mathbf{\tilde x^{(n)}}-\mathbf{t}^{(n})]^T]^T\|^2=\displaystyle\sum_{n=1}^N\| [[\mathbf{\tilde x}^{(n)}]^T\mathbf{\tilde W}-[\mathbf{t}^{(n)}]^T]^T\|^2=\|[\mathbf{\tilde X\tilde W-T}]^T\|^2_F$
We need 2 transposes to keep the expression unchanged, don't we?
The title may not describe my question well. Any modification would be appreciated!
Note that we have $$\|A\|_F^2=\|A^T\|_F^2$$
in general since both expression just sum up the square of each entry.