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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 :

1) Is this observation correct ?

2) Can we choose a basis where such matrices are not hermitian or symmetric ?

3) If so, is there a characterization of operators whose matrices can be hermitian or symmetric in some basis ?

The following is a para 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..."

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  • $\begingroup$ The inner product also depends on a choice of basis, up to a unitary transformation. $\endgroup$ – Yuval Filmus Mar 6 '15 at 4:41
  • $\begingroup$ I guess that there is some missunderstanding. One has to distinguish between matrices and linear transformations (operators). If one talks about ${\mathbb R}^n$ without mentioning of basis it is understood that we have standard basis and a matrix is then meant to present a linear transformation with respect to the standard basis. So, if we talk about properties of matrices, then they are meant to present a map with respect to the standard bases. If one changes bases the same linear transformation is presented with another matrix (which is similar to the previous one, of course). $\endgroup$ – Janko Bracic Mar 6 '15 at 5:12
  • $\begingroup$ @JankoBracic So, if we say a matrix is symmetric, clearly it is wrt a given basis. So, by changing the basis, can we make it not symmetric ? But wiki saya that symmetry is independent of the basis and is a property of the linear operator and the inner product we are using. since the inner product doesn't change with basis (by a unitary transformation ), the symmetry will be retained with changing the basis(by a unitary transformation ). $\endgroup$ – Srinivas K Mar 6 '15 at 5:16
  • $\begingroup$ I wanted to point out that matrix, say $T$, is not the same as the linear transformation, say $\tau$. If we talk about a matrix, and we think about the linear transformation which it presents, then we have some fixed bases. If we change bases, then $\tau$ will be presented by some other matrix and also matrix $T$ will present some other transformation. Example, calculate matrices for rotation of angle $\pi/2$ with respect to bases $\vec{i},\vec{j}$ and $\vec{e}=\vec{i}, \vec{f}=\vec{i}+\vec{j}$. Once you will get a symmetric matrix but once... :) $\endgroup$ – Janko Bracic Mar 6 '15 at 5:29
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    $\begingroup$ I think you meant to put that comment somewhere else, Srinivas. The point is that if $A$ is not diagonalizable, for example if $A=\pmatrix{0&1\cr0&0\cr}$, then it can't be similar to a symmetric matrix, since you can diagonalize a symmetric matrix. $\endgroup$ – Gerry Myerson Mar 6 '15 at 8:19
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Inner products also depend on a choice of basis. They are invariant under a unitary change of basis, and indeed Hermitian matrices stay Hermitian under unitary change of basis since $(UAU^*)^* = UA^*U^* = UAU^*$.

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  • $\begingroup$ Is change of basis possible only by unitary matrices ? $\endgroup$ – Srinivas K Mar 6 '15 at 4:45
  • $\begingroup$ Certainly not. But the inner product is invariant only under some changes of basis. $\endgroup$ – Yuval Filmus Mar 6 '15 at 4:46
  • $\begingroup$ So, by the defn in wiki, we are saying that since the inner product doesn't change ( depending on the basis ), the matrix will remain symmetric in any basis ? $\endgroup$ – Srinivas K Mar 6 '15 at 5:04
  • $\begingroup$ It won't remain symmetric in any basis – only in a unitary basis. In the same way, the inner product won't remain the same under any basis – only in a unitary basis. $\endgroup$ – Yuval Filmus Mar 6 '15 at 5:18
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    $\begingroup$ It's more usually known as an orthonormal basis. A basis is orthonormal if the change-of-basis matrix is unitary. $\endgroup$ – Yuval Filmus Mar 6 '15 at 5:23
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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.

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  • $\begingroup$ Is the change of basis matrix orthonormal ? bcoz' i learnt that under orthonormal change of basis we have the symmetry maintained. $\endgroup$ – Srinivas K Mar 6 '15 at 6:30
  • $\begingroup$ No. An orthonormal change of basis will preserve symmetry. $\endgroup$ – Gerry Myerson Mar 6 '15 at 6:31
  • $\begingroup$ I'm also curious to know if given an arbitrary linear transformation, can we find a basis in which it is represented by a symmetric matrix ? $\endgroup$ – Srinivas K Mar 6 '15 at 6:34
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    $\begingroup$ No, think about non-zero nilpotents. $\endgroup$ – Janko Bracic Mar 6 '15 at 6:46

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