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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 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...

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Srinivas K
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  • The inner product also depends on a choice of basis, up to a unitary transformation. – Yuval Filmus Mar 06 '15 at 04:41
  • 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). – Janko Bracic Mar 06 '15 at 05:12
  • @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 ). – Srinivas K Mar 06 '15 at 05:16
  • 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... :) – Janko Bracic Mar 06 '15 at 05:29
  • @JankoBracic I'm not sure what prevents nilpotents from being represented as symmetric matrices? I know that $A^k=0$. So,if $A$ is symmetric why is this not possible ? – Srinivas K Mar 06 '15 at 07:14
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    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. – Gerry Myerson Mar 06 '15 at 08:19

3 Answers3

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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.

Gerry Myerson
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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 $(UAU^*)^* = UA^*U^* = UAU^*$.

Yuval Filmus
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A Hermitian matrix stays Hermitian only under a unitary transformation. If $A$ is Hermitian, then $A=A^*$. Under a unitary transformation, $B = UAU^*$. Now, $B^* = (UAU^*)^* = U^{**}A^*U^* = UAU^* = B$. Thus, proven. But if the transformation is non-unitary, then, $C=WAW^{-1}$. Then, $C^*= (WAW^{-1})^* = (W^{-1})^*AW^* \neq C$, the transformed matrix is not Hermitian. This is also proved by contradiction, for example, normal matrices have an eigenvalue decomposition, which is essentially a change of basis to have a diagonal representation(which is Hermitian). But normal matrices need not necessarily be Hermitian for this to be true. So, a Hermitian matrix(in this example a diagonal matrix) under a change of basis need not always be Hermitian(a non-Hermitian normal matrix). However, in the case of a Hermitian matrix, the transformation is unitary.

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