How to Derive this Matrix Equation - Converting Lagrangian in Proximal Operator Problem Form

Question

$$ \left< Z,X-L-S \right> \quad +\quad \frac { r }{ 2 } \left\| X-L-S \right\|_F^2 \quad =\quad \frac { r }{ 2 } { \left\| L-\left( X-S+\frac {Z}{r} \right) \right\| }_F^2 $$

I think $ \left< Z,X-L-S \right>$ should be $\operatorname{Tr}(Z^T(X-L-S))$. This equation is presented in a paper, but the author doesn't show the derivation. See Robust Principal Component Analysis on Graphs at page 11.

Answer 1

Note that the left side of

$$ \left< Z,X-L-S \right> \quad +\quad \frac { r }{ 2 } \left\| X-L-S \right\|_F^2 \quad =\quad \frac { r }{ 2 } { \left\| L-\left( X-S+\frac {Z}{r} \right) \right\| }_F^2 $$

can be written as

$$ \operatorname{tr}(Z^T(X-L-S)) + \frac { r }{ 2 } \operatorname{tr}( (X-L-S)^T(X-L-S) ).$$

And this is the same as

$$ \operatorname{tr}(Z^T(X-L-S)) + \operatorname{tr}(\frac { r }{ 2 } (X-L-S)^T(X-L-S) ) = \operatorname{tr}\left(Z^T(X-L-S)) + \frac { r }{ 2 } (X-L-S)^T(X-L-S) )\right) = \operatorname{tr}\left(\left((Z^T) + \frac { r }{ 2 } (X-L-S)^T\right)(X-L-S) )\right) = \frac { r }{ 2 }\operatorname{tr}\left(\left(\frac{2 Z^T}{r} + (X-L-S)^T\right)(X-L-S) )\right) = \frac { r }{ 2 }\operatorname{tr}\left(\left(\frac{2 Z}{r} + X-L-S\right)^T(X-L-S) )\right). $$

The right hand side on the other hand can be written as

$$\frac { r }{ 2 } { \left\| L-\left( X-S+\frac {Z}{r} \right) \right\| }_F^2 = \frac { r }{ 2 } \operatorname{tr} \left( \left(L-X+S-\frac {Z}{r} \right)^T \left(L-X+S-\frac{Z}{r} \right) \right) = \frac { r }{ 2 } \operatorname{tr} \left( \left(L-X+S-\frac {2Z}{r}+\frac {Z}{r} \right)^T \left(L-X+S-\frac{2Z}{r} +\frac{Z}{r} \right) \right) = \frac { r }{ 2 } \operatorname{tr} \left( \left(-L+X-S+\frac {2Z}{r}-\frac {Z}{r} \right)^T \left(-L+X-S+\frac{2Z}{r} -\frac{Z}{r} \right) \right) = \frac { r }{ 2 } \operatorname{tr} \left( \left((-L+X-S+\frac {2Z}{r})-\frac {Z}{r} \right)^T \left((-L+X-S)+(\frac{2Z}{r} -\frac{Z}{r}) \right) \right) = \frac { r }{ 2 } \operatorname{tr} \left((-L+X-S+\frac {2Z}{r})^T(-L+X-S)-\frac {Z^T}{r}(-L+X-S) + (-L+X-S+\frac {2Z}{r})^T\frac{Z}{r}-\frac {Z^T}{r}\frac{Z}{r} \right) = \frac { r }{ 2 } \operatorname{tr} \left((-L+X-S+\frac {2Z}{r})^T(-L+X-S) + (\frac {2Z}{r})^T\frac{Z}{r}-\frac {Z^T}{r}\frac{Z}{r} \right) $$

Answer 2

The assumed equality isn't true. The matrices {$L, S, X, Z$} must all be the same dimensions for the math (addition, subtraction, scalar product, etc) to make sense.

Yet, if you generate 4 random $3 \times 5$ matrices, and evaluate the above expression with random values of $r$, you will find that the RHS does not equal the LHS.

Try it again with random square ($5 \times 5$) matrices, and it fails to hold.

Try it again with square, positive matrices, and it fails.

There might be some unspecified constraint on the matrices that will make this thing work, but I can't figure it out.

Update

Let $Y=(X-L-S)$ then you can rewrite the expression in terms of the Frobenius (:) product as $$\eqalign{ Z:Y + \frac{r}{2}Y:Y &= \frac{r}{2}\Big(Y+\frac{Z}{r}\Big):\Big(Y+\frac{Z}{r}\Big) \cr &= \frac{r}{2}\Big(Y:Y + \frac{2}{r}Z:Y + \frac{1}{r^2}Z:Z\Big) \cr &= \frac{r}{2}Y:Y + Z:Y + \frac{1}{2r}Z:Z \cr\cr 0 &= \frac{1}{2r}Z:Z \cr\cr }$$ This is true when Z is zero, but what's point of it?