Assuming that $f(x)=x^Tx-k^2=0$ holds for some $k$ and vector $x$, is it possible to derive that $$ u \nabla f = uIx $$ where $I$ is the identity matrix and $u$ is a lagrange multiplier?
If I simply derive $f$ with respect to $x$, I get $$ u\left( Ix + x^T\right) $$ where I use that $\frac{d}{dx}x^T=I$, but it gives me that annoying extra term $ux^T$. I don't know if I'm doing something wrong or missing a trick where you can somehow ignore the last term.
I'm a bit insecure in all of this, so any help is greatly appreciated!