Why is this equivalent?

Question

I came across this in a reading: $2\vec{x}'' \cdot \vec{x} =0$ $\implies$ $\vec{x}'' \cdot \vec{x}' = 0.$

How can you just drop the constant?

Answer 1

I will use the notation $\langle x,y \rangle$ instead of $x\cdot y$ as I think it becomes more clear then. I would assume there either is a typographic error in your book, or there is information you have not given us, because what you wrote does not seem correct, and I will explain why.

In general, if you have a real (but complex works just as well) inner product space $V$, then the inner product, by definition, satisfies linearity in the first argument. In your case this means that if $x,y\in V$ and $\lambda\in\mathbb{R}$, then

$$\langle \lambda x,y\rangle = \lambda \langle x,y\rangle.$$

In your case I am guessing that your inner product space is $\mathbb{R}^n$, and that you have $\vec{x} : \mathbb{R}\supset\kern-1.7pt\to \mathbb{R}^n$. Then also $\vec{x}',\vec{x}'':\mathbb{R}\supset\kern-1.7pt\to \mathbb{R}^n$, and so you would have that for any $t\in \operatorname{dom}(\vec{x})\cap\operatorname{dom}(\vec{x}''),$

$$\langle 2 \vec{x}''(t),\vec{x}(t)\rangle=2\langle \vec{x}''(t),\vec{x}(t)\rangle,$$

and so what you book most likely is intended to say is that

$$\langle 2 \vec{x}''(t),\vec{x}(t)\rangle=0 \implies \langle \vec{x}''(t),\vec{x}(t)\rangle=0.$$

However if we look at what it is currently saying then I can give a counterexample to why it does not hold in general. I will assume we are dealing with the standard inner product on $\mathbb{R}^n$, i.e. the scalar product given by the relation

$$\langle x,y \rangle = \sum_{j=1}^n x_jy_j.$$

Let us specifically consider the function

$$\vec{x} : \mathbb{R}\to\mathbb{R}^2, \quad t\mapsto (t^2-2t,t^2).$$

Then $\vec{x}'$ and $\vec{x}''$ are given by

$$\vec{x}'(t)=2(t-1,t), \quad \vec{x}''(t)=(2,2).$$

We have that

$$\langle 2\vec{x}''(1),\vec{x}(1)\rangle=\langle (4,4), (-1,1)\rangle=-4+4=0,$$

but also

$$\langle \vec{x}''(1),\vec{x}'(1)\rangle=\langle (4,4), (0,2)\rangle=4.$$

So even though $\langle 2\vec{x}''(1),\vec{x}(1)\rangle$ is zero, we do not have that $\langle \vec{x}''(1),\vec{x}'(1)\rangle$ is zero. And this is why I believe there is simply a typographic error in your book.