Can you raise a Matrix to a non integer number? [duplicate]

Sort of.

First, we have to ask what $f(x) = a^x$ means for $f: \mathbb{R} \to \mathbb{R}$, where $a \in \mathbb{R}$ is a constant. Assuming that $a > 0$, we can write: $$ f(x) = a^x = \left( e^{\ln(a)}\right) ^x = e^{(\ln a)x}, $$ so it boils down to having a sensible definition of the natural exponential and logarithm functions.

The exponential function can be defined by its Maclaurin series: $$ e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \tfrac{1}{2}x^2 + \tfrac{1}{6}x^3 + \cdots $$

One has to, of course, verify that this infinite series converges for every $x \in \mathbb{R}$ for this definition to make sense.

The same series can be used to define the matrix exponential function, interpreting $x$ now as an $n \times n$ matrix. (It also converges for all matrices.)

But, here the analogy begins to fall apart.

First of all, the matrix logarithm is a much more finicky creature. The standard Mercator series for the logarithm doesn't converge for all matrices. Even if we extend our ground field to the complex numbers, then a matrix has a logarithm if and only if it's invertible. Even when a logarithm exists, it is not unique!

I've never thought about this question, but what about the following. Recall that $a^b=e^{b \ln(a)}$. Define the log of a matrix A as a matrix B such that $e^B=A$. This exists, allowing complex numbers, iff A is invertible.

Also recall that we can define $e^A = \displaystyle \sum_{k=0}^\infty \frac{1}{k!} A^k$. So what about defining $A^{2.3} = e^{2.3 \ln{A}} = \displaystyle \sum_{k=0}^\infty \frac{1}{k!} (2.3 \ln(A))^k$ which converges iff $A$ is invertible.

To me (with the benefit of much hindsight), the most natural way to define $A^\beta$ where $\beta$ is a non-integer real number is to define $$ A^\beta = e^{\beta\log A}. $$ Here $\log A$ is the logarithm of the matrix $A$ (which exists whenever $A$ is invertible), and $e^C$ is the exponential of the matrix $C$ (which always exists).

When the power is a rational number, like $1/2$, you might be referring to a whole family of matrices. For instance $I^{1/2}$ could be interpreted as all of the matrices $A$ such that $A^2=I$. there are lots of these: any reflection will do.

In general $B^{m/n}$ could mean the collection of all matrices $A$ such that $A^n=B^m$.

This usage is not uncommon, so if you adopt @Josh, @Greg, and @Sammy's (Hi Sammy!) definitive meaning for $B^q$, then it's good to be aware of the other usage.