Why is a scalar product in a vector space necessary to determine if two vectors v, w are orthogonal?

Question

Two vectors $v,w \in V$ in a vector space $V$ with a scalar product $\langle \cdot,\cdot\rangle $ (we assume that $\langle v,w\rangle =\langle w,v\rangle $) are defined to be orthogonal if the scalar product $\langle v,w\rangle =0$. Lets suppose we have a normed vectorspace where the norm is defined with our scalar product in the following way

$$ \|x\|:= \sqrt{\langle x,x\rangle } \label{1}\tag{1}$$ It is easy to prove that $$\|v-w\|^2=\|v\|^2+\|w\|^ 2 \iff \langle v,w\rangle =0 \label{2}\tag{2}$$ if the norm is defined like in (\ref{1}).

To prove it in the one direction:

$"\Rightarrow"$ we start with $$\|v-w\|^2=\|v\|^2+\|w\|^ 2$$ lets use (\ref{1}) and substitute the norm $$ \langle v-w,v-w\rangle =\langle v,v\rangle +\langle w,w\rangle $$ $$ \langle v,v-w\rangle -\langle w,v-w\rangle =\langle v,v\rangle +\langle w,w\rangle $$ $$ \langle v,v\rangle -\langle v,w\rangle -(-\langle w,w\rangle +\langle w,v\rangle )=\langle v,v\rangle +\langle w,w\rangle $$ $$ \langle v,v\rangle -2\langle v,w\rangle +\langle w,w\rangle = \langle v,v\rangle +\langle w,w\rangle $$ when we substract $\langle v,v\rangle $ and $\langle w,w\rangle $ from both side and divide by $-2$ there is left $$\langle v,w\rangle =0$$

$"\Leftarrow"$ the other way of the equivalence we prove it by contradiction suppose: $$\|v-w\|^2 \neq \|v\|^2+\|w\|^ 2-2\langle v,w\rangle $$ Then lets substitute the norm with the scalar product we get: $$\langle v,v\rangle +\langle w,w\rangle -2\langle v,w\rangle \neq \langle v,v\rangle +\langle w,w\rangle -2\langle v,w\rangle $$ so this is wrong so $$\|v-w\|^2 = \|v\|^2+\|w\|^ 2-2\langle v,w\rangle $$ is true. Since we supposed that also $\langle v,w\rangle =0$ is true then also $$\|v-w\|^2=\|v\|^2+\|w\|^2$$ is true. So the the equivalence (\ref{2}) is proven.

My question is now why use scalar product to define orthogonality? A vector space not necessarily needs a scalar product for orthogonality. One could also define orthogonality by saying two vectors are considered to be orthogonal if the equation $\|v-w\|^2=\|v\|^2+\|w\|^ 2$ holds.

We could define the concept orthogonality even more general for a metric space $(X,d)$.

With $x,y \in X$ while $X$ should be at least an unital magma with $0$ as the neutral element. So we would say $x,y$ are orthogonal when $$d(x,y)^2=d(0,x)^2+d(0,y)^2$$ Or is there a general problem with it?

Answer 1

(Edit: I have changed the norm from $1$-norm to $\infty$-norm, as suggested by Rob Arthan in a comment to the question.)

If your norm is not induced from an inner product, or more generally, if orthogonality is not defined via some bilinear/sesquilinear form, one issue it may raise is that orthogonality may no longer play along with linearity. E.g. two orthogonal vectors may not remain orthogonal if you scale one of them. To illustrate, consider $$ v=\pmatrix{3\\ 1},\quad w=\pmatrix{0\\ -4},\quad v-w=\pmatrix{3\\ 5},\quad 2v-w=\pmatrix{6\\ 6}. $$ In your definition, $v$ is orthogonal to $w$ with respect to the supremum norm: $$ \|v-w\|_\infty^2=5^2=3^2+4^2=\|v\|_\infty^2+\|w\|_\infty^2, $$ but $2v$ is not orthogonal to $w$: $$ \|2v-w\|_\infty^2=6^2\ne 6^2+4^2=\|2v\|_\infty^2+\|w\|_\infty^2. $$

Answer 2

Basically the answer is that if you have a norm that can be used to define orthogonality, it can also be used to define a dot product.

The definition for orthogonality you derived was $\|\bar v-\bar w\|=\|\bar v\|^2+\|\bar w\|^2$. You could similarly and equivalently derive $\|\bar v+\bar w\|=\|\bar v\|^2+\|\bar w\|^2$ which I'll be using. If this works as a definition for orthogonality it should also hold if we scale the lengths of these vectors by scalars, i.e. $\|a\bar v+b\bar w\|=a^2\|\bar v\|^2+b^2\|\bar w\|^2$, because just scaling lenghts of the vectors doesn't change the angle between them. Let's take take two arbitrary vectors $\bar x$ and $\bar y$ which are in the plane spanned by the orthogonal vectors $\bar v$ and $\bar w$. Thus we have $\bar x=a\bar v+b\bar w$ and $\bar y=c\bar v+d\bar w$ for some scalars $a$, $b$, $c$, and $d$. From this we can derive

$$\|\bar x\|^2=a^2\|\bar v\|^2+b^2\|\bar w\|^2,$$ $$\|\bar y\|^2=c^2\|\bar v\|^2+d^2\|\bar w\|^2.$$

We can also derive

\begin{align} \|\bar x+\bar y\|^2 &=\|(a+c)\bar v+(b+d)\bar w\| \\ &=(a+b)^2\|\bar v\|^2+(b+d)^2\|\bar w\|^2 \\ &=(a^2+c^2+2ab)\|\bar v\|^2+(b^2+d^2+2cd)\|\bar w\|^2 \\ &=\|\bar x\|^2+\|\bar y\|^2+2(ab\|\bar v\|^2+cd\|\bar w\|^2) \end{align}

and

\begin{align} \|\bar x-\bar y\|^2 &=\|(a-c)\bar v+(b-d)\bar w\|=(a-b)^2\|\bar v\|^2+(b-d)^2\|\bar w\|^2 \\ &=(a^2+c^2-2ab)\|\bar v\|^2+(b^2+d^2-2cd)\|\bar w\|^2 \\ &=\|\bar x\|^2+\|\bar y\|^2-2(ab\|\bar v\|^2+cd\|\bar w\|^2). \end{align}

From these we get

$$\|\bar x+\bar y\|^2+\|\bar x-\bar y\|=2(\|\bar x\|^2+\|\bar y\|^2).$$

This is called the parallelogram law and here we derived that for this kind of norm it must hold for all vectors (or at least those in the plane spanned by these orthogonal vectors). But if the norm satisfies this law we can easily prove that we define a dot product using the norm in one of the following ways:

\begin{align} \langle\bar v,\bar w\rangle &=\tfrac{1}{2}(\|\bar x\|^2+\|\bar y\|^2+\|\bar x+\bar y\|^2) \\ &=\tfrac{1}{2}(\|\bar x\|^2+\|\bar y\|^2-\|\bar x-\bar y\|^2) \\ &=\tfrac{1}{4}(\|\bar x+\bar y\|^2+\|\bar x+\bar y\|^2). \end{align}

You can reasonably easily check that a dot product defined this way satisfies all the properties required of a dot product when the norm satisfies the parallelogram law.

Here I'm assuming the the field of scalars is the real numbers. In case of complex scalars you can define an inner product using a norm in a similar though slightly more complicated way.