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In the area of bilinear forms, my lecture notes say that there is a basis $\{e_i\}$ of $V$ with respect to which $\tau (e_i, e_j) = \delta_{ij}$ where $$\delta_{ij} = \begin{cases} 1 & i=j \\ 0 & i \neq j \end{cases}$$ and that a basis of a Euclidean space $V$ with this property is called an orthonormal basis of $V$. Does this mean that there is only one 'orthonormal basis' which is $(1,0,\dots,0), (0,1,\dots,0),\dots$ etc?

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No. For example, $\{(1/\sqrt2, 1/\sqrt 2)(1/\sqrt2, -1/\sqrt 2)\}$ is an orthonormal basis of $\Bbb R^2$. – David Mitra Apr 6 '12 at 15:02
Note the condition $\tau(e_i,e_j)=\delta_{ij}$ describes how the basis elements act on each other; it is not describing what the basis elements look like exactly. – David Mitra Apr 6 '12 at 15:08 – user2468 Apr 6 '12 at 15:18

The general feeling is, that an orthonormal basis consists of vectors that are orthogonal to one another and have length $1$. The standard basis is one example, but you can get any number of orthonormal bases by applying an isometric operation to this basis: For instance, the comment of David Mitra follows by applying the matrix $$ M := \frac{1}{\sqrt{2}} \cdot \begin{pmatrix} 1 & \hphantom{-} 1 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} \cos(\pi/4) & \sin(\pi/4) \\ \sin(\pi/4) & -\cos(\pi/4) \end{pmatrix} $$ to the standard basis. Observe that $M$ is just a reflection along the line with angle $\pi/8$. You can also obtain other orthonormal bases by applying rotations or, more general, by applying orthogonal transformations to the standard basis.

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