# Is this a theorem?

1. Mar 12, 2013

### dEdt

Let A be any real invertible matrix. There exists a non-zero diagonal matrix D such that $A^T D A=D$.

I'm pretty sure this is true (maybe with some conditions on A), but I need some help proving it.

2. Mar 13, 2013

### fzero

This does not seem to be true as stated. The matrix

$$A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$$

is a counter-example.

What is true is that, given a nondegenerate (no zero eigenvalue) matrix $Q$ (not necessarily diagonal), there is a matrix $A$ such that $A^T Q A = Q$. The set of matrices $A$ should actually form a group, which is the orthogonal group of $Q$ viewed as a bilinear form. This gives us some insight as to why the original proposition is false. For $Q = I$, the matrices $A$ actually satisfy $A^{-1} =A^T$, so this is a fairly restrictive condition. So it is not surprising that a generic invertible matrix $A$ should fail to preserve a bilinear form.

3. Mar 13, 2013

### dEdt

After thinking about it a bit more, it's clear that few real invertible matrices could satisfy that condition simply because
$$A^T DA=A\rightarrow \mathrm{det}(A)\mathrm{det}(D) \mathrm{det}(A)=\mathrm{det}(D) \rightarrow \mathrm{det}A=\pm 1.$$

So, let A be any real invertible matrix with $\mathrm{det}A=\pm 1$. There exists a non-zero diagonal matrix D such that $A^T DA=A$. Is this a theorem?

4. Mar 13, 2013

### fzero

I didn't mention this because you didn't specify that $\mathrm{det}(D)\neq 0$. You'll note that this is implied by the nondegeneracy property specified in the theorem I mentioned.

$$A = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$$

is a counterexample with $\mathrm{det}A=- 1$.

5. Mar 14, 2013

### micromass

6. Mar 15, 2013

### dEdt

Of course, how stupid of me!

It was interesting, thank you for the link. Unfortunately it didn't help me find what I was looking for.

For simplicity, consider a two dimensional vector space. One class of coordinate transformations are the Euclidean rotations
$$R(\theta)=\left( \begin{array}{cc} \cos \theta & \sin \theta\\ -\sin \theta & \cos \theta \end{array} \right).$$
These transformations preserve $x^2+y^2$, and hence $R^T D R=D$ for
$$D=\left( \begin{array}{cc} 1 & 0\\ 0 & 1 \end{array} \right).$$
Another two types of transformations are the Galilean and Lorentz transformations. There are also transformations that scale the coordinates by some factor. All of these transformations satisfy the condition above.

I'm pretty confident that these four classes of transformations are the only permissible linear coordinate transformations, and I want to prove it by showing that any linear coordinate transformation $A$ must satisfy $A^T D A=D$ for some diagonal matrix $D$.

7. Mar 15, 2013

### fzero

Linear coordinate transformations modulo translations form a group known as the general linear group, which is the group of invertible matrices (over the reals). In some cases, a linear transformation does preserve a bilinear form, for instance, for the Euclidean metric $\delta = \mathrm{diag}(1, \ldots 1)$ we get the rotation group. For the Minkowski metric, $\eta = \mathrm{diag}(-1,1, \ldots 1)$, we find the Lorentz transformations. Both of these groups are orthogonal groups, with the Lorentz group an example of an indefinite orthogonal group.

In general, there won't be a preserved bilinear form. For instance, the Galilean transformations don't preserve one. Rescalings of coordinates also do not preserve a quadratic form. Under a uniform rescaling of all coordinates, the Euclidean and Minkowski metrics are rescaled by the square of the parameter.

Therefore, the tractable question is really, given a bilinear form or matrix $Q$, to describe the properties of the matrices $A$ that leave it invariant. This is a classical problem and it reduces to the properties of $Q$: the symmetry properties, the signature, etc. Over the reals, we find the orthogonal and symplectic groups, see here.

8. Mar 15, 2013

### dEdt

The Galilean transformations may not preserve a bilinear form, but they preserve $x^2 +y^2+z^2$ and $t^2$ separately, so there's still a D satisfying those conditions.

9. Mar 16, 2013

### fzero

Yes, I was forgetting about the origin. So it's a pair of bilinear forms that are preserved independently. You could derive the form of the Galilean transformations in an analogous fashion to the previous cases.