If $(\mathbb R,V,+,\langle\cdot,\cdot\rangle)$ is a finite-dimensional inproductspace and $L:V\rightarrow V$ is a linear transformation with matrix-representation $A$ that is symmetric with respect to some basis of $V$. Does the spectral theorem hold in this situation? And why (not)?
I'm asking this because I'm trying to generalize the spectral theorem as it is stated in my textbook, which, if I understand it correctly, assumes that $A$ is a symmetric matrix-representation of $L$ with respect to specifically the standard basis of $V=\mathbb R^n$ (or any orthonormal basis, it's unclear). It says that in this section that it is dealing with "symmetric transformations on a Euclidean space. If we're working with coordinates with respect to an orthonormal basis then this comes down to transformations of $R^n$ that are describable by a symmetric matrix." But it doesn't explain what 'symmetric transformations' are, why the matrix-representations are symmetric with respect to an orthonormal basis, etc.
You may generalize it further to Hermitic spaces if you are first of all capable of answering my questions clearly. Thank you.
No. You need a matrix representation with respect to an orthogonal basis with vectors of equal length, perhaps not $1$.
Take the matrix $$A=\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$$ representing a linear map on the plane with respect to the basis $v_1=\begin{bmatrix} 1\\0\end{bmatrix}$, $v_2=\begin{bmatrix} 1\\1\end{bmatrix}$.
Then the standard matrix of $A$ (i.e., the matrix with respect to the standard basis) is $$B=\begin{bmatrix} 3 & 0 \\ 2 & -1\end{bmatrix}.$$ As you can check, the eigenvalues are $-1$ and $3$, with corresponding non-orthogonal eigenvectors $\begin{bmatrix} 0\\1\end{bmatrix}$ and $\begin{bmatrix} 2\\1\end{bmatrix}$.