I am trying to calculate the proximal mapping of the quadratic function:
$f(x)=\frac{1}{2}x^TAx+b^Tx+c$ , where $T$ is transpose.
The solution is: $(I+tA)^{-1}(x-tb)$.
taken from pg23 of: https://www.math.ucdavis.edu/~sqma/MAT258A_Files/6-prox-grad.pdf
My attempt
General form is:
$prox=argmin \{\frac{1}{2}||x-z||^{2}+f(x)\}$, giving:
$prox=argmin \{\frac{1}{2}x^2-xz+\frac{1}{2}z^2+\frac{1}{2}x^TAx+b^Tx\}$ , where $c$ has been removed as it is a constant.
Now take the gradient wrt to $x$, giving:
$x-z+\frac{1}{2}x^TA+\frac{1}{2}x^TA^T+b^T$
Now, set the above to zero and solve for x:
$x(I+\frac{1}{2}A+\frac{1}{2}A^T) = z-b^T$, where $I$ is the identity matrix
which gives:
$x=(I+A)^{-1}(z-b)$
What have I done wrong? I'm guessing I haven't been careful when dealing with the tranpose matrices but would appreciate it if someone could clarify this for me.