Is real power of unitary matrix unitary?

Question

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? — little o, Mar 14 '19 at 04:37
I think your first problem will be defining $X^k$ for all $k \in \Bbb R$. — Robert Lewis, Mar 14 '19 at 04:42
For $k=-2$ use the fact that $(A^{-1})^* = (A^)^{-1}$ for any matrix $A,$ where $A^$ is the conjugate transpose of $A.$ — little o, Mar 14 '19 at 04:45
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. — little o, Mar 14 '19 at 04:49
I edited my question. Can anyone have a look at it? @RobertLewis — Lucia Guzheim, Mar 14 '19 at 05:15

Travis Willse · Accepted Answer · 2019-03-14T05:41:12.893

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.

score 0 · Answer 2 · 2019-03-15T11:35:22.317

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.

Is real power of unitary matrix unitary?

2 Answers2