TITLE: Inner product and orthogonality in non orthogonal basis
QUESTION [7 upvotes]: According to the definition of orthogonality (on finite vector spaces),
Given an inner product space, two vectors are orthogonal if their inner product is zero.
So as an example, assuming the inner product is the "the standard" Euclidean inner product, two vectors (1,0) and (0,1), in $\mathbb{R}^2$, are obviously orthogonal in standard basis. However, if we change the basis to another basis of two linearly independent vectors which are not orthogonal in standard $\mathbb{R}^2$ basis, then these new basis vectors would be also represented as, (1,0) and (0,1) in the new basis. So keeping the same Euclidean inner product, they would be orthogonal in the new basis, although, they are not orthogonal in standard $\mathbb{R}^2$ basis. This would mean that orthogonality depends on the basis chosen, which seems "strange" to me? 
Can you please explain what is wrong with in the above reasoning (assuming it is wrong).
Thanks.
Note, I have read the thread with the very similar title, change of basis and and inner product in non orthogonal basis but that doesn't answer my question

REPLY [3 votes]: You seem to be confusing the notion of an inner product and a basis.
A (real) inner product on a vector space $V$ is a bilinear function $\langle \cdot | \cdot \rangle\colon V\times V\to {\bf R}$ satisfying the standard axioms of positive definiteness and symmetry. It is this inner product that we use to determine whether a given tuple of vectors is orthogonal or orthonormal, not any basis.
Now, if $V$ is finite dimensional and $(v_1,\ldots, v_n)$ is a basis of $V$, there is a unique inner product such that $(v_1,\ldots, v_n)$ is an orthonormal basis. But the converse is far from truth: given any inner product on a space of dimension at least two, there are infinitely many orthonormal bases.
Now, if you have a space $V$ with a fixed inner product $\langle \cdot | \cdot \rangle_1$, you can take a basis $(v_1,\ldots, v_n)$ of $V$ and find the unique inner product $\langle \cdot | \cdot \rangle_2$ such that $(v_1,\ldots,v_n)$ are orthonormal with respect to it, but unless $(v_1,\ldots,v_n)$ are orthonormal with respect to $\langle \cdot | \cdot \rangle_1$, the two inner products are distinct, so it should come as no surprise that the notions of "orthonormal" differ (precisely because of what I have mentioned in the last paragraph about the inner product being unique given a basis which we want to be orthonormal).
Interestingly, it is actually true that any two inner products have a common orthogonal basis, but nonetheless, the notions of orthogonality do not coincide for distinct inner products, either. In other words, while there are special bases which are orthogonal jointly for both inner products, there are always bases which are orthogonal in one, but not in the other basis (for distinct inner products).