# Vector variable and single-variable gradient

How do I show that

$$f(\overrightarrow{x}) = \nabla f( \overrightarrow{0} ) \cdot \overrightarrow{x}$$ for every $x$, given that $f \space (t \cdot \overrightarrow{v})= t \space f( \overrightarrow{v})$?

I'm confused by the notion of the gradient of a single-variable function and the vector variable. I assume that since $f \space (t \cdot \overrightarrow{v})= t \space f( \overrightarrow{v})$, $\space$ $f({x})$ has the form $f({x})=ax$, but that's all I got so far. Any hints?

• Why do you assume that? What if $f(x)$ was rotation by say $\pi/3$ in $\mathbb{R}^2$? – David Peterson Sep 28 '14 at 6:31

Consider the function $g(t)=f(tx)$ then $g'(t)=\nabla f(tx)\cdot x$ (chain rule), but also $g'(t)= f(x)$.