Suppose $g:\mathbb{R}^{n}\times\mathbb{R}^{n}\rightarrow\mathbb{R}$ is a real inner-product and $\vec{v}_{1}, \ldots, \vec{v}_{n}$ are vectors such that $g(\vec{v}_{i}, \vec{v}_{j}) = \delta_{ij}$ for all $i, j$. In the standard basis, we can represent $g$ as a real, symmetric, positive-definite matrix $A$ so that $g(\vec{v}, \vec{w}) = \vec{v}\,^{T}\!A\vec{w}$.
By the spectral theorem, there are eigenvectors $\vec{p}_{1}, \ldots, \vec{p}_{n}$ with eigenvalues $\lambda_{1}, \ldots, \lambda_{n}$ such that $\vec{p}_{i}\cdot\vec{p}_{j} = \delta_{ij}$, and there is an orthogonal matrix $P$ such that $P^{T}AP = L$ where $L = \text{diag}(\lambda_{1}, \ldots, \lambda_{n})$. Now presumably, $P$ can be interpreted as the change of basis matrix from the standard basis $(\vec{e}_{1}, \ldots, \vec{e}_{n})$ to $(\vec{v}_{1}, \ldots, \vec{v}_{n})$. In this new basis, the matrix representation of $g$ is diagonalized, and so $\vec{v}_{1}, \ldots, \vec{v}_{n}$ are orthogonal.
Now my confusion here is that I can in principle construct an inner-product $g$ for which the vectors $\vec{v}_{1}, \ldots, \vec{v}_{n}$ are not orthogonal, yet the transformation $(\vec{e}_{1}, \ldots, \vec{e}_{n})\rightarrow (\vec{v}_{1}, \ldots, \vec{v}_{n})$ is done by an orthogonal matrix. To put it briefly, where is the mistake in my thinking regarding this?
Best Answer
Why should the vectors $v_i$ coincide with the eigenvectors $p_i$ of $A$?
More explicitly, consider an inner product such that $e_1$ and $e_1+e_2$ are an orthonormal basis. You can check that this corresponds to $x^tAy$ where $A=\left(\begin{array}{cc} 1 & -1\\ -1 & 2\end{array}\right)$. But neither $e_1$ nor $e_1+e_2$ are eigenvectors of this matrix.