Aren't All Bases Othonormal? It's probably a very stupid question, but someone isn't all basis orthonormal ? For example, let $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$ the canonical basis of $\mathbb R^2$. Let $v_1=\begin{pmatrix}2\\0\end{pmatrix}$ and $v_2=\begin{pmatrix}4\\2\end{pmatrix}$. Of course $\{v_1,v_2\}$ is a basis of $\mathbb R^2$. But in $\mathbb R^2=Span\{v_1,v_2\}$, we have that $v_1=\begin{pmatrix}1\\0\end{pmatrix}$ and $v_2=\begin{pmatrix}0\\1\end{pmatrix}$. So with respect to the basis $\{v_1,v_2\}$, the family $\{v_1,v_2\}$ is orthonormal, no ? I'm a bit confuse now... If yes, Gramm-schmidt is useless so I guess something is wrong in what I'm saying. Could someone enlighten me ?
 A: You can't just switch bases and expect the dot product to compute the same thing, as your example shows. The coordinate representation of the inner product has to change away from the dot product to something else.
The dot product's Gram matrix reads $\begin{bmatrix}1&0\\0&1\end{bmatrix}$ but after the change of basis which converts from these coordinates to coordinates of the second basis you gave, the new Gram matrix is 
$\begin{bmatrix}4&8\\8&20\end{bmatrix}$, thus the inner product, computed both with the old and new coordinates matches:
$\begin{bmatrix}2&0\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}4\\2\end{bmatrix}=\begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix}4&8\\8&20\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}=8$.
The mistake you were making was thinking that the Gram matrix stayed the same between the two expressions, but it does change.
And what about the original basis?
Well $[1,0]$ was sent to $[1/2, 0]$ and $[0,1]$ was sent to $[-1, 1/2]$, and 
$\begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}1/2&0\end{bmatrix}\begin{bmatrix}4&8\\8&20\end{bmatrix}\begin{bmatrix}-1\\1/2\end{bmatrix}=0$
$\begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}1/2&0\end{bmatrix}\begin{bmatrix}4&8\\8&20\end{bmatrix}\begin{bmatrix}1/2\\0\end{bmatrix}=1$
$\begin{bmatrix}0&1\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}-1&1/2\end{bmatrix}\begin{bmatrix}4&8\\8&20\end{bmatrix}\begin{bmatrix}-1\\1/2\end{bmatrix}=1$
so it is still orthonormal.
Remember: Inner products have an existence outside of coordinates, and the dot product depends on a basis. We pick a basis and switch to coordinates to do computation sometimes, but choice of basis is immaterial, and won't affect the end result that the inner product computes.  So, changing the basis out from under the inner product changes the way the quantity is computed.
The same goes for linear transformations and the matrices we choose to represent them.
To give an example of a similar mistake you could try with transformations: 

"$\begin{bmatrix}1\\1\end{bmatrix}$ is in the kernel of the linear transformation $\begin{bmatrix}1&-1\\1&-1\end{bmatrix}$, but when I use $v=\begin{bmatrix}1\\1\end{bmatrix}$ in a basis so that $v=\begin{bmatrix}1\\0\end{bmatrix}$ it is not in the kernel anymore... why?"

The answer is, of course, that the matrix has to change when the basis changes too.
A: A "vector space", in the general "Linear Algebra" sense, does not have an inner product attached and that is what is needed for "orthogonality".  
In any finite dimensional space, given a basis $\{v_1, v_2,\cdot\cdot\cdot, v_n\}$, we can define "standard" inner product: If $u= a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n$ and $v= b_1v_1+ b_2v_2+ \cdot\cdot\cdot+ b_nv_n$ then the "standard" inner product is $a_1b_1+ a_2b_2+ \cdot\cdot\cdot+ a_nb_n$.
The "usual basis" for $R^2$ is {<1, 0>, <0, 1>} but any two independent vectors will form a basis.  For example, {<1, 0>, <1, 1>} is a perfectly good basis and the vectors <1, 0>, <1, 1> are not "orthogonal" with the "standard" definition of inner product which then gives the usual definiton of "angle"- they have angle 45 degrees.
However, given any basis for a vector space it is always possible to redefine the inner product so that the basis vectors are orthogonal.  Is that what you mean?
A: The basis is orthonormal respect to a inner product $\cdot$ if 


*

*$|v_i|=1,\forall i$

*$v_i\cdot v_j=0,\forall i\neq j$
The vectors of the basis you showed do not have norm equal to 1, and if we use the common inner product you have that $v_1\cdot v_2 = 8\neq 0$, so it is not orthonormal.
