# Positive-definite 2x2 matrix invariant under conjugation with W -> W unitary?

I have the following problem for complex matrices: $G$ is a positive-definite, hermitian 2x2-matrix and $W$ an invertible 2x2 matrix. Moreover, $WGW^\dagger = G$ holds. Does this imply that $W$ is unitary (with respect to the standard scalar product)?

(Edit: The answer is NO. See the counter example below!) At first sight, it seemed obvious to me that the answer is YES. However, I still could not prove it. One should probably use the fact that $G = UDU^\dagger$ for unitary $U$ and a positive-definite diagonal matrix $D$. Furthermore, $\text{det}(WGW^\dagger) = \text{det}(W) \text{det}(G) \text{det}(W^\dagger) = \text{det}(G)$ and combined with $\text{det}(G) >0$ this yields $|\text{det}(G)| = 1$.

This is indication for the fact that $W$ could be unitary, but it's of course not enough.

Any help is welcome! Thanks a lot!

Cheers Qwerkopf

PS: What would the answer be in case of dxd-matrices?

This implies that $W$ is unitary with respect to the scalar product defined by $G$, i.e. $\langle v,w \rangle = v^T G \overline{w}.$

This is because the adjoint of $W$ is $G^{-1} \overline{W^T} G$: You can verify $$\langle Wv, w \rangle = v^T W^T G \overline{w} = v^T G\overline{G^{-1} \overline{W^T} G w} = \langle v, G^{-1} \overline{W^T} G w \rangle.$$

Typically this does not mean that $W$ will be unitary with respect to the Euclidean scalar product though.

• Thanks for this answer! I was thinking about the Euclidean (standard) scalar product (I added it now in the question). Jun 1, 2016 at 20:49

Write that $G = MM^\dagger$ (e.g. via the Cholesky decomposition). Then we have $$WGW^\dagger = G \iff\\ (WM)(WM)^\dagger = MM^\dagger \iff\\ (M^{-1}WM)(M^{-1}WM)^\dagger = I$$ That is, this is equivalent to saying that $M^{-1}WM$ is unitary, which is to say that $W = MUM^{-1}$ for some unitary $U$.

That is, in general: all we can know is that $M$ is similar to a unitary matrix (so it is diagonalizable with the correct eigenvalues, but won't necessarily satisfy $WW^\dagger = I$).

Counterexample: Take $$M = \pmatrix{1&1\\0&1}, \quad G = MM^\dagger = \pmatrix{2&1\\1&1}\\$$ Now, take $$W = M \pmatrix{1\\&-1}M^{-1} = \pmatrix{1&-2\\0&-1}$$

• Very elegant and a nice way to find a counter example. Problem solved. Thanks a lot!!! Jun 1, 2016 at 21:09
• No problem. The same idea applies for $d \times d$ matrices, of course. Jun 1, 2016 at 21:20