# Is real power of unitary matrix unitary?

Suppose $$X$$ is a unitary matrix. Would $$X^k$$ also be unitary, where $$k \in \mathbb{R}$$ (negative as well)? What if $$X$$ has infinite dimension?

I think for $$k \in \mathbb{Q}$$ the question can be restated as:

For $$k=k_1/k_2$$, $$k_1 \in \mathbb{Z}$$, $$k_2 \in \mathbb{Z}/0$$ and unitary $$X$$, does unitary $$Y$$ exist such that $$X^{k_1} = Y^{k_2}$$?

I am not sure how I am to generalize this to $$k$$ that is irrational.

• For $k=2$ observe that $$X^2 (X^2)^* = X^2 (X^*)^2 = X(XX^*)X^* = XIX^* = XX^* = I.$$ Can you now proceed? – Dbchatto67 Mar 14 at 4:37
• I think your first problem will be defining $X^k$ for all $k \in \Bbb R$. – Robert Lewis Mar 14 at 4:42
• For $k=-2$ use the fact that $(A^{-1})^* = (A^*)^{-1}$ for any matrix $A,$ where $A^*$ is the conjugate transpose of $A.$ – Dbchatto67 Mar 14 at 4:45
• Because $$X^{-2} (X^{-2})^* = (X^2)^{-1} ((X^2)^{-1})^* =(X^2)^{-1} ((X^2)^*)^{-1} = ((X^2)^* X^2)^{-1} = I^{-1} = I.$$ Note that here we have used that the result is true for $k=2.$ Now use induction to proceed. – Dbchatto67 Mar 14 at 4:49
• I edited my question. Can anyone have a look at it? @RobertLewis – Lucia Guzheim Mar 14 at 5:15

## 2 Answers

The unitary matrices of size $$n$$ form a group under matrix multiplication: For $$X, Y \in U(n)$$, we have $$(XX')^* (XX') = (X')^* X^* X X' = (X')^* I_n X' = (X')^* X' = I_n ,$$ and a similar claim shows that $$U(n)$$ is closed under inversion. Thus, $$X \in U(n)$$ implies $$X^k \in U(n)$$ for all $$k \in \Bbb Z$$.

In fact, it's straightforward to show that $$U(n)$$ is a compact Lie group. Thus, the exponential map $$\exp : \mathfrak u(n) \to U(n)$$ is surjective and hence $$U(n)$$ is divisible: For any element $$Z \in U(n)$$ and any positive integer $$s$$, there is an element $$Y \in U(s)$$ such that $$Z = Y^s$$. If we set $$Z = X^r$$, then we get $$\boxed{X^r = Y^s}$$ as desired. Except when $$s = 1$$ (i.e., when $$k$$ is an integer), however, the element $$Y$$ is not unique: For any integer $$t$$, we have $$(e^{2 \pi i t / s} Y)^s = e^{2 \pi i t} Y^s = Y^s$$.

The case of irrational exponent is more delicate, and one needs to be precise about what $$X^\alpha$$ means for nonintegral $$\alpha$$. (The non-uniqueness of $$Y$$ in the rational case already hints at this issue.) One option is to choose in some appropriate neighborhood $$V$$ in $$U(n)$$ of the identity matrix $$I_n$$ a matrix logarithm $$\log$$, that is, an inverse for $${\exp}\vert_V$$. Then, we can declare (for $$X \in V$$) that $$X^{\alpha} = \exp(\alpha \log X)$$, and in particular $$X^{\alpha} \in U(n)$$, but the quantity $$X^\alpha$$ depends on the choice of $$\log$$, which is not unique.

Let $$U_{-}(n)$$ be the set of unitary matrices that have not $$-1$$ as eigenvalue and $$u\in\mathbb{R}^*$$. When $$X\in U_{-}(n)$$ we put $$X^u=\exp(u\log(X))$$ where $$\log$$ denotes the principal logarithm.

$$\textbf{Proposition}$$. $$X^u$$ is unitary.

$$\textbf{Proof}$$. $$X=UDU^*$$ where $$D=diag(e^{¡\theta_j})$$ , $$\theta_j\in (-\pi,\pi)$$ and $$U$$ is unitary.

Then $$X^u=Udiag(\exp(i u\theta_j))U^*$$ and we are done. $$\square$$

i) Clearly, the result does not work when $$u$$ is a complex number.

ii) Note that, in general, $$\log(X^u)$$ is not $$u\log(X)$$ except when $$|u|<1$$.

iii) Note that, via the Cayley transform, $$U_{-}(n)$$ can be flattened.