# Can a symmetric matrix become non-symmetric by changing the basis?

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...

• The inner product also depends on a choice of basis, up to a unitary transformation. – Yuval Filmus Mar 6 '15 at 4:41
• 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 6 '15 at 5:12
• @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 6 '15 at 5:16
• 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 6 '15 at 5:29
• 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 6 '15 at 8:19

## 2 Answers

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

• Is change of basis possible only by unitary matrices ? – Srinivas K Mar 6 '15 at 4:45
• Certainly not. But the inner product is invariant only under some changes of basis. – Yuval Filmus Mar 6 '15 at 4:46
• 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 ? – Srinivas K Mar 6 '15 at 5:04
• 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. – Yuval Filmus Mar 6 '15 at 5:18
• It's more usually known as an orthonormal basis. A basis is orthonormal if the change-of-basis matrix is unitary. – Yuval Filmus Mar 6 '15 at 5:23

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.

• Is the change of basis matrix orthonormal ? bcoz' i learnt that under orthonormal change of basis we have the symmetry maintained. – Srinivas K Mar 6 '15 at 6:30
• No. An orthonormal change of basis will preserve symmetry. – Gerry Myerson Mar 6 '15 at 6:31
• 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 ? – Srinivas K Mar 6 '15 at 6:34
• No, think about non-zero nilpotents. – Janko Bracic Mar 6 '15 at 6:46