# Thread: What is the gradient of u(Ox) when O is orthogonal.

1. ## What is the gradient of u(Ox) when O is orthogonal.

Suppose we know $\displaystyle \nabla{u(x)}$. Let O be an orthogonal matrix.

What is $\displaystyle \nabla{u(Ox)}$?

Thank you!

2. See here.

3. Yea, I don't understand exactly what that means though.

Is it $\displaystyle (O^{-1}) \nabla u \cdot (Ox)$?

4. I don't think there's a dot product there (except implicitly, in the definition of the gradient.) Looking at the equation applied to your case, we'd have

$\displaystyle \nabla(u(O\mathbf{x}))=O^{T}(\nabla u)(O\mathbf{x}).$

As I see it, the way to interpret the LHS of this equation is to take the vector $\displaystyle \mathbf{x}$, rotate by left-multiplying by $\displaystyle O$, then you stuff the result of the rotation into the function $\displaystyle u$, and finally, you take the gradient.

I interpret the RHS of this equation as follows: first, you take the gradient of $\displaystyle u$, and then you evaluate that function at the point $\displaystyle Ox$, and finally you left-multiply by $\displaystyle O^{T}.$

If I were you, I'd work out a simple example in, say, 2 dimensions. See how it works out.