Inner Product, Orthogonality, and Coordinate Systems 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 $\mathbb{R^2}$?
 A: 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.
A: 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.
A: 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 $(\cdot,\cdot)_{B}$. $\;\;\Box$
