We know that a hermitian matrix is a matrix which satisfies $A=A^*$, where $A^*$ is the conjugate transpose. A symmetric matrix (special case of hermitian – with real entries) is one for which $A=A^T$.
Observation: this property is dependent on choice of basis.
We know that we can even choose a basis where these matrices are diagonal (spectral theorem). So, my question is:
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Is this observation correct?
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Can we choose a basis where such matrices are not hermitian or symmetric?
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If so, is there a characterization of operators whose matrices can be hermitian or symmetric in some basis?
The following is a paragraph from wiki page which I'm unable to understand. Can someone shed light on this?
… Denote by $ \langle \cdot,\cdot \rangle $ the standard inner product on $R^n$. The real $n-by-n$ matrix $A$ is symmetric if and only if
$\langle Ax,y \rangle = \langle x, Ay\rangle \quad \forall x,y\in\Bbb{R}^n$. Since this definition is independent of the choice of basis, symmetry is a property that depends only on the linear operator A and a choice of inner product. This characterization of symmetry is useful, for example, in differential geometry, for each tangent space to a manifold may be endowed with an inner product, giving rise to what is called a Riemannian manifold. Another area where this formulation is used is in Hilbert spaces…
Best Answer
The matrices $\pmatrix{1&3\cr0&2\cr}$ and $\pmatrix{1&0\cr0&2\cr}$ are similar, so there is a change of basis that transforms one into the other, but one is symmetric and the other is not, so, yes, there are transformations that have a symmetric matrix with respect to one basis and not to another basis.