# Direction vectors are linearly dependent, if $D_vf(x)=D_wf(x)=0$ with $\operatorname{grad} f(x)\neq 0$

Let $f \colon \mathbb{R}^2 \to \mathbb{R}$ be continuous differentiable and $x \in\mathbb{R}^2$ with $\operatorname{grad} f(x) \neq 0$. Let $v,w \in \mathbb{R}^2$ be direction vectors.

If $D_v f(x) = D_w f(x)=0$ then are $v$ and $w$ linearly dependent.

For my proof I use the following theorem:

Let $U \subset \mathbb{R}^m$ be open and $f \colon U \to \mathbb{R}$ continuous differentiable. Then it holds for $x \in U$ and $v \in \mathbb{R}^m$ with $\|v\| = 1$ that $D_v f(x) = \langle v, \operatorname{grad} f(x) \rangle$.

Proof:

With the theorem above and the assumption, we know that $$\langle v, \operatorname{grad} f(x) \rangle = \langle w, \operatorname{grad} f(x) \rangle.$$ It holds that \begin{align*} \langle v, \operatorname{grad} f(x) \rangle &\overset{\phantom{D_vf(x)=D_wf(x)}}{=} \sum_{i=1}^2 v_i \cdot \frac{\partial f}{\partial x_i} = v_1 \cdot \frac{\partial f}{\partial x_1} + v_2 \cdot \frac{\partial f}{\partial x_2} \\ &\overset{D_vf(x)=D_wf(x)}{=} \sum_{i=1}^2 w_i \cdot \frac{\partial f}{\partial x_i} = w_1 \cdot \frac{\partial f}{\partial x_1} + w_2 \cdot \frac{\partial f}{\partial x_2} = 0 \end{align*} Therefore \begin{align*} &\, v_1 \cdot \frac{\partial f}{\partial x_1} + v_2 \cdot \frac{\partial f}{\partial x_2} = w_1 \cdot \frac{\partial f}{\partial x_1} + w_2 \cdot \frac{\partial f}{\partial x_2} \\ \iff&\, (v_1-w_1) \frac{\partial f}{\partial x_1} + (v_2-w_2) \frac{\partial f}{\partial x_2} = 0 \end{align*}

Since $\operatorname{grad} f(x) = \left(\frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_w}\right) \neq 0$ this is a non trivial linear combination of $0$.

Hence $v$ and $w$ are linearly dependent.

Is this proof correct? I am actually not sure if I used the correct scalar product $\langle \cdot, \cdot \rangle$.

• $(v_1-w_1)\,f_1^{(1)}+(v_2-w_2)\,f^{(2)}$ is not a linear combination of the vectors $v$ and $w$, nor does it imply that they are linearly dependent. E.g., if $\nabla f=(1,1)$, then this quantity vanishes for $v=(1,0)$ and $w=(0,1)$. – amd Aug 25 '18 at 1:27
• Both $v$ and $w$ satisfy a particular homogeneous linear equation. What can you say about that equation’s solution space? – amd Aug 25 '18 at 2:34
• @amd I looked into my lecture notes, but I did not find anything about a homogeneous linear equation, which $v$ and $w$ satisfy. Do you have a source on this topic? Or do you just mean the quoted theorem, which gives: $\begin{pmatrix} v_1& v_2\\ w_1&w_2\end{pmatrix}\cdot\operatorname{grad}f(x)=\begin{pmatrix}0\\0\end{pmatrix}$ – Cornman Aug 25 '18 at 3:31
For given $x$ in the domain of $f$ the derivative $df(x)$ (also denoted by $f'(x)$, or similar) is a linear map acting as $$df(x):\quad {\mathbb R}^2\to{\mathbb R},\qquad X\mapsto\langle\nabla f(x),X\rangle\ ,$$ whereby $\langle\cdot,\cdot\rangle$ denotes the standard scalar product in ${\mathbb R}^2$, as in the question. Since $\nabla f(x)\ne0$ by assumption this map has rank $1$; hence its kernel $K$ has dimension $2-1=1$. We are told that both $v$ and $w$ are in $K$; therefore $v$ and $w$ have to be linearly dependent.
Note that this argument does not work for an $f:\>{\mathbb R}^n\to{\mathbb R}$ with $n>2$.
• How do we get $df(x).X=\langle\nabla f(x), X\rangle$ and how is $\langle\cdot, \cdot\rangle$ defined here? Standard scalar product on $\mathbb{R}^2$? – Cornman Aug 25 '18 at 15:06