# 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. Mar 6, 2015 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). Mar 6, 2015 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 ). Mar 6, 2015 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... :) Mar 6, 2015 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. Mar 6, 2015 at 8:19

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. Mar 6, 2015 at 6:30
• No. An orthonormal change of basis will preserve symmetry. Mar 6, 2015 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 ? Mar 6, 2015 at 6:34
• No, think about non-zero nilpotents. Mar 6, 2015 at 6:46

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 ? Mar 6, 2015 at 4:45
• Certainly not. But the inner product is invariant only under some changes of basis. Mar 6, 2015 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 ? Mar 6, 2015 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. Mar 6, 2015 at 5:18
• It's more usually known as an orthonormal basis. A basis is orthonormal if the change-of-basis matrix is unitary. Mar 6, 2015 at 5:23