4
$\begingroup$

I remember my professor saying there are certain advantages to using an orthogonal basis. One is that it's easy to determine the coordinates of a given vector. For example, we are familiar with the canonical {$[0,1], [1,0]$} basis of $\mathbb{R^2}$. But doesn't the orthogonality (and the inner product) depend on the coordinate system?

To illustrate, let $V_1 = [0,1]$ and $V_2 = [1,0]$ in canonical coordinate system. Then, let's use the basis vectors {$[0,1],[0.5,0.5]$} and use coordinates relative to this new basis. Then, relative to the new basis, $V_1$ has coordinates $[1,0]$ and $V_2$ has coordinates $[2,-1]$. So the inner product $V_1 \cdot V_2 = 1*2 + 0*-1 = 1$ is not longer zero, and thus $V_1$ and $V_2$ are no longer orthogonal in the new coordinate system.

Questions:
1. Is my understanding correct? Is it true that the inner product depend on the basis vectors and the coordinate system?
2. If yes, then are there any deficiencies to using {$[0,1],[0.5,0.5]$} as the basis for $\mathbb{R^2}$?

$\endgroup$

3 Answers 3

3
$\begingroup$

The inner product does not in fact depend on a particular coordinate system. The error you are making is in the statement:

"Then, relative to the new basis, V1 has coordinates [1,0] and V2 has coordinates [2,−1]. So the inner product V1⋅V2=1∗2+0∗−1=1 "

Summing up the product of coordinates of two vectors gives the inner product only if the coordinates are taken with respect to some orthonormal basis. Since V1,V2 do not form an orthonormal basis, the system of coordinates induced by them cannot be used in the usual formula for the inner product.

$\endgroup$
4
  • $\begingroup$ Thanks @uniquesolution. I'm still trying to make sense of this. To determine if a basis is orthonormal, the inner product must be defined first...and that requires a coordinate system. So it seems we need to some how define the canonical basis first, to show my new system is not orthogonal. But why can't we define the coordinates using my new system in the first place? $\endgroup$
    – FreshAir
    Sep 4, 2015 at 0:45
  • $\begingroup$ @FreshAir You can define coordinates however you like, but if you also define them to be orthogonal, you (by doing that) also implicitly define the inner product which is not the same as the original. The thing is in your case, the canonical basis also defines the inner product, and you cannot then simply define a new basis to be orthogonal without also changing the inner product. $\endgroup$
    – 5xum
    Sep 4, 2015 at 6:20
  • $\begingroup$ @5xum, thanks for this; what you say is consistent with Kevin Discoll's answer. If I understand correctly, the issue is that the new inner product based on my new basis (which is notated as $(x,y)_{B}$ in TrialAndError's answer) conflicts with the canonical inner product. But what if we disregard the canonical basis, coordinate system along with the canonical inner product we're familiar with? In my comment to TrialAndError's answer, I noted that this new inner product no longer satisfies the formula $(x,y)_{B} = |x||y| \cos(\theta)$. $\endgroup$
    – FreshAir
    Sep 5, 2015 at 2:06
  • $\begingroup$ But if we are not concerned about geometry, are there other deficiencies with this new "system"? $\endgroup$
    – FreshAir
    Sep 5, 2015 at 2:07
3
$\begingroup$

What you're missing is the 'cross term' in the dot product. The dot product is defined as a distributive operation, so if we call $$[0, 1] = \vec{e}_1 , \ [1/2, 1/2] = \vec{e}_2,$$ then your vectors can be written $$V_1 = \vec{e}_1, \ V_2 = - \vec{e}_1 + 2\vec{e}_2$$ and so the dot product, $$V_1 \cdot V_2 = \vec{e}_1 \cdot (-\vec{e}_1 + 2 \vec{e}_2)= -\lvert \vec{e}_1 \rvert^2 + 2 \vec{e}_1 \cdot \vec{e}_2. $$

Since you gave a representation for these new basis vectors in terms of the standard basis, we can compute the dot product of our new basis vectors using their representation in the standard basis, so that we are sure that our definitions are consistent. $$\lvert \vec{e}_1 \rvert^2 = 1, \ \vec{e}_1 \cdot \vec{e}_2 = 1/2$$ and thus the dot product of $V_1$ and $V_2$ is $ -1 + 2 (1/2) = 0$, as before. If we use an orthonormal basis, then we do not have any such cross-terms.

In general, one must define how to compute the dot product in some basis and then how to compute the dot product in other bases will be determined by the condition that the dot product be independent of the choice of basis. If we don't satisfy this property then what we've defined isn't a true inner product.

$\endgroup$
1
$\begingroup$

An inner product is a lot less unique than one might first suspect. As you have essentially noted:

Theorem: Let $V$ be a finite-dimensional linear space over the field $\mathcal{F}$ of real or complex numbers with basis $B=\{ b_1,b_2,\cdots,b_n\}$. Then there exists an inner product $(\cdot,\cdot)_{B}$ on $V$ with respect to which $B$ is an orthonormal basis of $V$.

Proof: Assume $\mathcal{F}=\mathbb{R}$ (the proof is essentially the same for $\mathcal{F}=\mathbb{C}$.) Define a linear map $L : \mathbb{R}^{n}\rightarrow V$ by $$ L(\alpha_1,\alpha_2,\cdots,\alpha_n)=\alpha_1b_1+\alpha_2b_2+\cdots\alpha_nb_n. $$ Then $L$ is a linear bijection from $\mathbb{R}^{n}$ onto $V$. It is easy to verify that the following defines an inner product $(\cdot,\cdot)_{B}$ on $V$: $$ (x,y)_{B}=(L^{-1}x,L^{-1}y)_{\mathbb{R}^{n}}. $$ If $e_1=(1,0,0,\cdots,1)$, $e_2=(0,1,0,\cdots,0)$, $\cdots$, $e_n=(0,0,\cdots,1)$ are the standard basis elements of $\mathbb{R}^{n}$, then $L^{-1}b_k = e_k$ and, therefore, $(b_l,b_m)_{B}=(e_l,e_m)_{\mathbb{R}^{n}}=\delta_{l,m}$, which makes $B$ an orthonormal basis of $V$ in the inner product $(\cdot,\cdot)_{B}$. $\;\;\Box$

$\endgroup$
5
  • $\begingroup$ Thanks @TrialAndError for the theorem and proof. You say "verify that the following defines an inner product", but I'm unclear on the requirements - is this a function satisfying all the properties listed here en.wikipedia.org/wiki/Dot_product#Properties? $\endgroup$
    – FreshAir
    Sep 5, 2015 at 1:56
  • $\begingroup$ Secondly, let us take $L$ to be $\mathbb{R}^{2}$, use the two vectors stated in my Question as basis $B$ (one along the "x-axis" and one along the 45 deg. line) with the inner product $(x,y)_{B}$ you defined. If my understanding is sound, we have now created a new coordinate system with a new inner product. I understand for this inner product, the usual $\cos(\theta)$ formula no longer holds, i.e. $(x,y)_{B} \neq |x||y| \cos(\theta)$, with $\theta$ being the angle between $x$ and $y$, at least according to our definition of the angle. $\endgroup$
    – FreshAir
    Sep 5, 2015 at 1:57
  • $\begingroup$ But if we don't care about these geometric constructs, this new coordinate system, based on the new basis $B$ with the new inner product, is just as good as the canonical one, right? Are there any deficiencies I overlooked? $\endgroup$
    – FreshAir
    Sep 5, 2015 at 1:57
  • $\begingroup$ I'm thinking about a modeling problem, where $b_1 \in \mathbb{R}^{2}$ is temperature and $b_2 \in \mathbb{R}^{2}$ represents pressure, and $U(b_1, b_2)=\gamma_1 b_1 + \gamma_2 b_2$ is a linear combination of temperature and pressure. In this case, any geometric concepts (angles between $b_1$ and $b_2$ and their projections are not too meaningful. $\endgroup$
    – FreshAir
    Sep 5, 2015 at 1:57
  • $\begingroup$ @FreshAir : A real inner product $(\cdot,\cdot)$ is linear in both coordinates, $(x,x) \ge 0$ for all $x$, with equality iff $x=0$. As for $(x,y)_{B}=|x||y|\cos(\theta)$, that's how the angle $\theta$ is defined for any inner product. There's nothing Mathematical to distinguish an inner product from another on a vector space. The Math for one inner product is the same and just as good for any other inner product. We have our favorite inner product on $\mathbb{R}^{n}$, but there's nothing Mathematical to distinguish. $\endgroup$ Sep 5, 2015 at 2:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.